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Reconstruction of the measurable sets in the twodimensional plane by two projectionsTo cite this article Takashi Takiguchi 2007 J Phys Conf Ser 73 012022

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Reconstruction of the measurable sets in the two

dimensional plane by two projections

Takashi TakiguchiDepartment of Mathematics National Defense Academy of Japan1-10-20 Hashirimizu Yokosuka Kanagawa 239-8686 JAPAN

E-mail takashindaacjp

Abstract We discuss the reconstruction of the measurable plane sets from their twoprojections Reconstruction of a measurable plane set F sub R2 with λ2(F ) lt infin from itsorthogonal projections is well studied where λi is the Lebesgue measure on Ri i = 1 2 Inthis paper we first discuss generalization of the known results in the frame of general twoprojections The main purpose is to study for any measurable plane set F whether there aresuitable two angles such that F is uniquely reconstructed from the pair of its projections in thattwo directions We give an example to show that this problem is negatively solved

1 IntroductionIn this paper we discuss the reconstruction of the measurable plane sets from their twoprojections Let us consider a problem to reconstruct the characteristic function of a measurableplane set from its line integrals Since this is a problem in the Radon transform if we have the allline integrals then the reconstruction is an easy corollary of the results in the Radon transformOur object however is too simple to take its all line integrals It is better to reconstructthe object with as less data as possible In this paper we try the reconstruction with lineintegrals only in the two directions Though this problem is well studied for the case wherethese two projections are orthogonal few sets can be reconstructed in such a frame Thereforein this paper we take projections in two directions which are not necessarily orthogonal in thereconstruction of the measurable plane sets Let us formulate our problem

Definition 11 Let F sub R2 be a measurable plane set such that λ2(F ) lt infin where λi is theLebesgue measure on Ri i = 1 2 Let f(x y) be the characteristic function of F For α βsatisfying minusπ2 le α + β le π2 minusπ2 lt α minus β lt π2 we rotate the x-axis by the angle αand y-axis by the angle β which we call the x(α)-axis and the y(β)-axis respectively Define thecoordinates

(x y)(αβ) = (x cosαminus y sinβ x sinα + y cosβ)(00) = (x cosαminus y sinβ x sinα + y cosβ) (1)

where the coordinates in the right hand side are in the usual frame of the orthogonal coordinatesWe call this generalized frame (middot middot)(αβ) the frame of the (α β) coordinates For such α β wealso define

f(αβ)1 (yprime) =

intinfinminusinfin f(minusyprime sinβ + t cosα yprime cosβ + t sinα)dt

f(αβ)2 (xprime) =

intinfinminusinfin f(xprime cosαminus t sinβ xprime sinα + t cosβ)dt

(2)

Inverse Problems in Applied Sciencesmdashtowards breakthrough IOP PublishingJournal of Physics Conference Series 73 (2007) 012022 doi1010881742-6596731012022

ccopy 2007 IOP Publishing Ltd 1

Figure 1 The (α β) projections

We call these projections (f (αβ)1 f

(αβ)2 ) the (α β) projections

Remark that if α = β 6= 0 then the projections (f (αα)1 f

(αα)2 ) are orthogonal but we only

call them the (α α) projections In this paper by ldquothe orthogonal projectionsrdquo we mean onlythe (0 0) projections

The reconstruction problem of the measurable plane sets from their (α β) projections is asfollows

Problem 11 Given two non-negative integrable functions P (x) and Q(y) having the sameL1 norm find a measurable plane set F with characteristic function f(x y) such that P (x) =f

(αβ)2 (x) and Q(y) = f

(αβ)1 (y) almost everywhere

As an example of the application of Problem 11 consider a homogeneous object in the threedimensional space which contains a hole in its interior We would like to detect the hole withoutdestructing the object itself Study this problem on the section by a two dimensional plane andapply the X-ray tomography for the reconstruction Then the results in Problem 11 are appliedto solve this problem

Problem 11 is well studied for the orthogonal case where (α β) = (0 0) It was GGLorentz [6] who first studied Problem 11 for (α β) = (0 0) in 1949 in view of which let uscall this problem the Lorentz X-ray problem There are a number of studies on this problem for(α β) = (0 0) ([1 2 4 5 6 8] and so on)

In the next section we review these known results on Problem 11 for (α β) = (0 0) If westudy Problem 11 only in the frame of the orthogonal coordinates then we can reconstructonly a few sets Most sets are non-unique in this frame Therefore there arises the necessity tostudy Problem 11 in the frame of the general coordinates which are not necessary orthogonal


2

In the third section we discuss this generalization of the frame in the coordinates In the frameof the generalized coordinates many non-unique sets in the frame of orthogonal coordinates aremade uniquely reconstructed by their two projections in a suitable frame in view of which itis natural to pose the following problem

Problem 12 For any measurable plane set F are there any pair of angles α and β such thatF is uniquely reconstructed from the pair of projections f

(αβ)1 and f

(αβ)2

It is our main theorem to give an answer to this problem We are very sorry that theanswer to Problem 12 is negative We prove the existence of the non-unique set by any pairof its projections in the fourth section (cf Theorem 41 below) which is our main theorem inthis paper In the last section we conclude the conclusion and introduce the important openproblems left to be solved for further development

Throughout this paper unless mentioned otherwise all discussions are made up to the setsof the measure zero

2 Known resultsIn this section we review the known results on Problem 11 for (α β) = (0 0) which are relatedto our main theorem in this paper For (α β) = (0 0) we omit to write ldquo(0 0)rdquo for examplewe denote f1 = f

(00)1 and so on

Definition 21 For non-negative integrable functions f1 and f2 on R define the projection off1 by

f12(x) = λ1(y | f1(y) ge x) (3)

for x ge 0 and the projection of f2 by

f21(y) = λ1(x | f2(x) ge y) (4)

for y ge 0 Similarly the functions f121 and f212 are defined respectively as

f121(y) = λ1(x | f12(x) ge y) (5)

f212(x) = λ1(y | f21(y) ge x) (6)

The answer to Problem 11 for the orthogonal case was classified by GG Lorentz into thethree cases The following theorem is a reformulation by A Kuba and A Volcic [4] of thetheorem by GG Lorentz [6]

Theorem 21 (cf [4] and [6]) Let f1(y) and f2(x) be non-negative integrable functionssatisfying int infin

minusinfinf1(y)dy =

int infin

minusinfinf2(x)dx (7)

(i) The unique caseThere exists a unique set which has (f1 f2) as the orthogonal projections if and only if

int c

0f12(x)dx =

int c

0f212(x)dx for any c gt 0 (8)

(ii) The non-unique caseThere exist non-unique sets having (f1 f2) as the orthogonal projections if and only if

int c

0f12(x)dx ge

int c


and there is a constant c gt 0 for which the strict inequality holds


3

(iii) The inconsistent caseThere exists no set having (f1 f2) as the orthogonal projections if and only if

int c

0f12(x)dx lt

int c

0f212(x)dx for some c gt 0 (10)

In the next section we generalize this theorem in the frame of the general coordinates(Theorem 31 below) which serves as a key lemma to prove our main theorem

We introduce the following theorems (Theorems 22-28) since by their generalizations(Theorems 32-38) we explain the importance of Problem 12 In 1988 A Kuba and A Volcic[4] gave a reconstruction formula for the uniquely determined sets

Theorem 22 (cf [4]) If a measurable set F is uniquely determined by the pair of its orthogonalprojections (f1 f2) then

F = (x y) | f2(x) ge f12(f1(y)) (11)

In the same paper [4] they also gave a characterization of the non-uniquely determined setsFor a measurable set P sub R2 define the parallel translation of P by

P(st) = (x y) | (xminus s y minus t) isin P (12)

A plane set F is called to have (P P 12P 1 P 2) as a switching component if there exist four setsP P 1 P 2 P 12 of the positive measure and two real numbers s t 6= 0 such that

P 1 = P(s0) P 2 = P(0t) P 12 = P(st) (P cup P 12) sube F (P 1 cup P 2) cap F = empty (13)

Let a set F have (f1 f2) as the orthogonal projections and (P P 12 P 1 P 2) as a switchingcomponent By to switch the switching components (P P 12 P 1 P 2) in F we mean the procedureof making another set (F cup P 1 cup P 2) (P cup P 12) which has the same orthogonal projections asF

Theorem 23 (cf [4]) A measurable plane set having a finite measure is non-uniquelydetermined by its orthogonal projections if and only if it has a switching component

They also studied the structure of non-uniquely determined sets (cf [5]) In 1998 LHuang and T Takiguchi [1] proved the following theorem which proves the L1 stability inthe reconstruction of the measurable plane sets from the pair of their orthogonal projections inthe class of the uniquely determined bounded sets

Theorem 24 (cf [1]) Let f1 f2 g1 and g2 be non-negative essentially bounded integrablefunctions Assume that the pairs of the orthogonal projections (f1 f2) and (g1 g2) uniquelydetermine the measurable plane sets F and G (with characteristic functions f(x y) and g(x y))respectively Then we have

||f minus g||L1(R2) le C middotmax||f1 minus g1||Linfin(R) ||f2 minus g2||Linfin(R) (14)

where C = C(||f1||Linfin(R) ||g1||Linfin(R) ||f2||Linfin(R) ||g2||Linfin(R)) is a constant

As an application of this theorem they also gave an algorithm to construct approximatesolutions for the uniquely determined sets from their orthogonal projections possibly containingnoise and error By these results we can approximately reconstruct the solution from theorthogonal projections possibly containing noise and error if the set to be reconstructed is apriori known to be uniquely determined In practical applications however it hardly happensthat the set to be reconstructed is a priori known to be uniquely determined Although it ispossible to judge whether the pair determines a set uniquely if the projections contain no errorit is impossible to obtain the projections without noise and error In view of this argumentthe best we can hope is to construct an approximate set of the original one from the pair of itsorthogonal projections with noise and error Therefore other problems arise


4

Problem 21

(i) Give a sufficient condition for a non-uniquely determined set F from its orthogonalprojections to be approximately reconstructed from the pair of its orthogonal projectionspossibly containing noise and error

(ii) Give an algorithm to approximate such non-unique sets

Since we may assume the existence of the solutions in practical applications what we haveto study is the approximation of the non-unique solutions A solution to Problem 21 was givenin [8]

Theorem 25 (cf [8]) Assume that a pair of orthogonal projections (f1 f2) determines setsnon-uniquely Take a set F whose orthogonal projections are (f1 f2) Then any set having(f1 f2) as the orthogonal projections is obtained by switching the switching components in Fcountable times

Theorem 26 (cf [8]) Assume that a non-unique pair (f1 f2) by the orthogonal projectionsf1 f2 isin L1 cap Linfin(R) satisfies

f12 minus f212L1(R) lt ε (15)

Then there exists a uniquely determined set G such that

λ2(F ordfG) lt 2ε minf1Linfin(R) f2Linfin(R) (16)

where F ordfG is the symmetric difference of F and G

Theorem 27 (cf [8]) Assume that two sets F1 and F2 have the same orthogonal projections(f1 f2) which satisfy (15) Then there holds

λ2(F1 ordf F2) le 4ε minf1Linfin(R) f2Linfin(R) (17)

By virtue of Theorems 26 and 27 if the orthogonal projections possibly containing noiseand error satisfy (15) then we obtain an approximate solution by approximating the uniquelydetermined set G in Theorem 26 As was mentioned in [8] the construction of G and itsorthogonal projections (g1 g2) is not difficult Let

g212(x) = minf12(x) f212(x) (18)

andϕfi

(x) = λ1(ξ|fi(ξ) gt fi(x)) + λ1(ξ|fi(ξ) = fi(x) ξ gt x) (19)

for i = 1 2 By Propositions 1-3 in [7] (confer also [4]) the function ϕf is measure preserving(for the definition of the term ldquomeasure preservingrdquo confer also [7]) and satisfies f212(ϕf2(x)) equivf2(x) We define

g2(x) = g212(ϕf2(x)) (20)

By the definition g212 is the non-increasing rearrangements of g2 (confer [4 7] for the definitionof the rearrangements) In the same way we define

g121(y) = minf21(y) f121(y) (21)

andg1(y) = g121(ϕf1(y)) (22)

By the definition we obtain g212 equiv g12 ae hence the pair of the orthogonal projections (g1 g2)determines the set G uniquely Thus we obtain the unique set G and the pair of its orthogonalprojections (g1 g2)


5

For the approximation of the set G we can apply the algorithm by L Huang and T Takiguchisince the set G is uniquely determined by the pair of its orthogonal projections Note that no apriori information on the set itself is required in this approximation Therefore we obtain thefollowing theorem

Theorem 28 (an answer to Problem 21) (cf [8])In order to approximately reconstruct the non-unique sets whose orthogonal projections satisfy(15) we have only to approximate the set G defined above Since G is uniquely determined by thepair of its orthogonal projections we can apply the algorithm by Huang-Takiguchi to approximateit Note that no a priori information on the set itself is required in this approximation

3 Generalizations in the frame of the generalized coordinatesIn the previous section we reviewed the known results in the frame of the orthogonal coordinatesone of which claims that if the orthogonal projections satisfy (15) then we can reconstruct thesolution approximately To our regret only limited sets satisfy (15) For instance consider theparallelogram

F = (x y) | y lt x lt y + 1 0 lt y lt 1 (23)

This set would not satisfy (15) for small ε gt 0 This set is a typical example of the non-uniquelydetermined ones in the frame of the orthogonal coordinates however we can reconstruct ituniquely from its two projections if we rotate the y-axis by minusπ4 As well as the example (23)the set

F = 0 lt x lt 1 0 lt y lt 1 cup 1 lt x lt 2 1 lt y lt 2 (24)

is not unique in the orthogonal coordinates but is made unique in the frame of the (π4 π4)coordinates For the uniqueness of these two examples in the suitable coordinates conferTheorem 31 below Besides of these examples there are a number of the measurable plane setswhich are not uniquely reconstructed in the frame of the orthogonal coordinates but are madeunique by a suitable affine transform of the coordinates By Theorem 23 we can claim that theessential character of the non-unique set is that they have the switching components Thereforein the examples (23) and (24) we claim that some sets having the switching components in theframe of orthogonal coordinates would not have ones in the frame of the suitable coordinates(cf Theorem 33 below) In this section we study the reconstruction of the measurable planesets from their two projections in the frame of such generalized coordinates We note that AKuba [3] discussed the reconstruction of the measurable plane sets from their non-orthogonal twoprojections however our approach is different from his idea He fixed the x-axis and studied therotation in the y-axis hence the frames of the type namely (π4 π4) coordinates mentionedin the example (24) was not studied by his idea In order to give generalizations of the theoremsmentioned in the previous section we prepare several definitions

Definition 31 For f(αβ)1 f

(αβ)2 defined in Definition 11 we define the projections of them by

f(αβ)12 (x) = λ1(y | f

(αβ)1 (y) ge x) f

(αβ)21 (y) = λ1(x | f

(αβ)2 (x) ge y)

f(αβ)121 (y) = λ1(x | f

(αβ)12 (x) ge y) f

(αβ)212 (x) = λ1(y | f

(αβ)21 (y) ge x) (25)

Definition 32 Let us define the measure in the (α β) coordinatesint infin

minusinfinf (αβ)(τ)dτ(αβ) =

int infin

minusinfinf (αβ)(t) cos(β minus α)dt (26)


6

Definition 33 For a measurable set P sub R2 we define the (α β) parallel translation of P by

P(αβ)(st) = (x y)(αβ)|(xminus s y minus t)(αβ) isin P (27)

Definition 34 A measurable plane set F is defined to have (P P 12 P 1 P 2) as an (α β)switching component if P P 1 P 2 P 12 are of the positive measure and if there exist two realnumbers s t 6= 0 such that

P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P

(αβ)(st) (P cup P 12) sube F (P 1 cup P 2) cap F = empty (28)

hold By to switch the (α β) switching components (P P 12P 1 P 2) in F we mean the procedureof making another set

F = (F cup P 1 cup P 2) (P cup P 12) (29)

which has the same (α β) projections as F

We claim that all the theorems in the previous section are generalized in the frame of the(α β) coordinates

Theorem 31 Let f(αβ)1 (y) and f

(αβ)2 (x) be non-negative integrable functions such that

int infin

minusinfinf

(αβ)1 (y)dy(αβ) =

int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case

There exists a unique set which has (f (αβ)1 f

(αβ)2 ) as the (α β) projections if and only if

int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f

(αβ)212 (x)dx(αβ) for any c gt 0 (31)

(ii) The non-unique case

There exist non-unique sets having (f (αβ)1 f


int c

0f

(αβ)12 (x)dx(αβ) ge

int c

0f


and there is a constant c gt 0 for which the strict inequality holds(iii) The inconsistent case

There exists no set having (f (αβ)1 f


int c

0f

(αβ)12 (x)dx(αβ) lt

int c

0f

(αβ)212 (x)dx(αβ) for some c gt 0 (33)

In order to prove this theorem we modify the proof in [6] For this modification is very longand contains some important properties in the reconstruction we divide it into several lemmas

Lemma 31 If (f (αβ)1 f

(αβ)2 ) are the (α β) projections of some measurable plane set F then

(32) holds

Proof Let us define

F(αβ)x = (x y)(αβ) | 0 lt y lt f

(αβ)2 (x) F

(αβ)y = (x y)(αβ) | 0 lt x lt f

(αβ)1 (y)

F(αβ)yxy = (x y)(αβ) | 0 lt y lt f

(αβ)212 (x) F

(αβ)xyx = (x y)(αβ) | 0 lt x lt f

(αβ)121 (y) (34)


7

The set F(αβ)y defined in (34) has (f (αβ)

1 f(αβ)12 ) as its (α β) projections

We first prove that int

If

(αβ)2 (x)dx(αβ) le

int k

0f

(αβ)12 (x)dx(αβ) (35)

for any interval I sub R λ1(I) le k Let

F = F cap (x y)(αβ) | x isin I F = F (αβ)y cap (x y)(αβ) | 0 lt x lt k (36)

We note that what we want to prove is equivalent to λ2(F ) le λ2(F ) Denote by (f (αβ)1 f

(αβ)2 )

the (α β) projections of F and by (f (αβ)1 f

(αβ)2 ) the (α β) projections of F By their definitions

f(αβ)1 (y) le f

(αβ)1 (y) and f

(αβ)1 (y) le k hence

f(αβ)1 (y) le min(f (αβ)

1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f

(αβ)1 (y)dy(αβ) le

int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)

which proves (35) for any interval I sub R λ1(I) le k Since the interval I sub R λ1(I) le k isarbitrary we also have for the the non-increasing rearrangement f

(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)

for any k gt 0 which is (32)

Lemma 32 For any pair of the functions (f (αβ)1 f

(αβ)2 ) satisfying (32) there exists a

measurable set F having (f (αβ)1 f

(αβ)2 ) as its (α β) projections

Proof For x gt 0 y gt 0 we define

D = (x y)(αβ) | y lt f(αβ)12 (x) y lt f

(αβ)212 (x)

D+ = (x y)(αβ) | f(αβ)12 (x) lt y lt f

(αβ)212 (x)

Dminus = (x y)(αβ) | f(αβ)212 (x) lt y lt f

(αβ)12 (x)

(40)

We first prove that D+ and Dminus are represented as

D+ =sum

L+i Dminus =

sumLminusi (41)

where i = 1 2 middot middot middot L+i rsquos and Lminusi rsquos are the open net lozenges of the kind

(x y)(αβ) | 2minusmlx lt x lt 2minusm(lx + 1) 2minusmly lt y lt 2minusm(ly + 1) (42)

for lx ly m = 0 1 2 middot middot middot L+i equiv Lminusi Lminusi = L+

i(αβ)

[minusst] for some s gt 0 t gt 0 and λ2(Lplusmn1 ) geλ2(Lplusmn2 ) ge middot middot middot By the definition λ2(D+) and λ2(Dminus) must be equal We assume thatλ2(D+) = λ2(Dminus) gt 0 otherwise we have nothing to do for the representation of D+ andDminus


8

Remark that (32) is equivalent to

λ2(D+ cap (x y)(αβ) | 0 lt x lt k) le λ2(Dminus cap (x y)(αβ) | 0 lt x lt k) (43)

for any k gt 0 We also note that (32) is represented in another formint c

0f

(αβ)21 (y)dy(αβ) ge

int c

0f

(αβ)121 (y)dy(αβ) for any c gt 0 (44)

which is equivalent to

λ2(Dminus cap (x y)(αβ) | 0 lt y lt k) le λ2(D+ cap (x y)(αβ) | 0 lt y lt k) (45)

for any k gt 0In order to obtain the representation of D+ and Dminus we first represent Dminus as

sumLi where Lirsquos

are open net lozenges By virtue of the assumption λ2(D+) = λ2(Dminus) gt 0 and (32) we can takeL1 lying in the strip (x y)(αβ) | kminus lt y lt lminus such that D+ cap (x y)(αβ) | kminus lt y lt lminus = emptyBy (45) there is a k+ 0 lt k+ lt kminus such that λ2(D+cap(x y)(αβ) | k+ lt y lt kminus = λ2(L1) Forany pair of points (x1 y1)(αβ) isin L1 (x2 y2)(αβ) isin D+ cap (x y)(αβ) | k+ lt y lt kminus there arepositive numbers s t such that x1 = x2minus s y1 = y2 + t Therefore we obtain the representation(41) for L1 and D+ cap (x y)(αβ) | k+ lt y lt kminus by decomposing them into suitable open netlozenges We apply the same method to L2 L3 middot middot middot to obtain the representation (41) for D+ andDminus

The set F(αβ)yxy defined in (34) has (f (αβ)

21 f(αβ)212 ) as its (α β) projections We inductively

construct the set F having (f (αβ)121 f

(αβ)212 ) as its (α β) projections Let

Lplusmni = (x y)(αβ) | kiplusmnx lt x lt liplusmnx kiplusmn

y lt y lt liplusmny (46)

for i = 1 2 middot middot middotAs the first step we make a set (F (αβ)

yxy )1 by either of the following two procedures

(F (αβ)yxy )1 = (F (αβ)

yxy cup Lminus1 ) (x y)(αβ) | k1minusx lt x lt l1minusx k1+

y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)

yxy cup (x y)(αβ) | k1+x lt x lt l1+

x k1minusy lt y lt l1minusy ) L+

1 (47)

We note that for the first step either of the procedures in (47) is possible and (F (αβ)yxy )1 has the

same projection f(αβ)212 (x) as F

(αβ)yxy

Let us assume that the 1 2 middot middot middot nminus 1-st steps are taken and we obtain the sets (F (αβ)yxy )1 middot middot middot

(F (αβ)yxy )nminus1 all of which have the same projection f

(αβ)212 (x) Denote by ((f (αβ)

1 )j f(αβ)212 ) the

(α β) projections of (F (αβ)yxy )j j = 1 2 middot middot middot nminus 1 Let rn be the length of the sides of L+

n equiv Lminusn There hold by the definition that

(f (αβ)1 )nminus1(ln+

y ) ge f(αβ)21 (ln+

y ) + rn ge f(αβ)21 (knminus

y ) + rn

(f (αβ)1 )nminus1(knminus

y ) le f(αβ)21 (ln+

y )minus rn(48)

Therefore the section (F (αβ)yxy )nminus1 cap (x y)(αβ) | y = ln+

y contains at least [f (αβ)21 (ln+

y )rn] + 1net-intervals of the length rn where [middot] is the Gauss bracket In the same way the section(F (αβ)

yxy )nminus1 cap (x y)(αβ) | y = knminusy is contained in [f (αβ)

21 (ln+y )rn] net-intervals of the length

rn Therefore there exists 0 lt knx such that

(x y)(αβ) | knx lt x lt lnx kn+

y lt y lt ln+y sub (F (αβ)

yxy )nminus1

(x y)(αβ) | knx lt x lt lnx knminus

y lt y lt lnminusy cap (F (αβ)yxy )nminus1 = empty (49)


9

where lnx = knx + rn Therefore as the n-th step we construct

((F (αβ)yxy )n = (F (αβ)

yxy )nminus1 cup (x y)(αβ) | knx lt x lt lnx knminus

y lt y lt lnminusy )(x y)(αβ) | kn

x lt x lt lnx kn+y lt y lt ln+

y (50)

Note that the set (F (αβ)yxy )n also has the same projection f

(αβ)212 (x) as F

(αβ)yxy

By this inductive procedure we obtain the set F Denote by (f (αβ)1 f

(αβ)212 ) the (α β)

projections of F Since

f(αβ)1 (y0) = f

(αβ)21 (y0)minus λ1(D+ cap (x y)(αβ) | y = y0) + λ1(Dminus cap (x y)(αβ) | y = y0)

= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f

(αβ)121 Therefore F has (f (αβ)


Let us define the set F x by

F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)

where ϕ is defined in (19) F x has (f (αβ)121 f

(αβ)2 ) as its (α β) projections Then we obtain the

set F having (f (αβ)1 f

(αβ)2 ) as its (α β) projections by the following

F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)

which proves the lemma

Lemma 33 If the pair of functions (f (αβ)1 f

(αβ)2 ) satisfies (32) but not (31) then there exist

non-unique sets having (f (αβ)1 f

(αβ)2 ) as their (α β) projections

Proof By virtue of the previous lemma we obtain the proof easily In the first step in theproof of Lemma 32 two ways of the construction of the set (F (αβ)

yxy )1 in (47) are possible Bythe following steps these two (F (αβ)

yxy )1rsquos yield different sets having the same (α β) projectionsTherefore we have at fewest two sets having (f (αβ)

1 f(αβ)2 ) as their (α β) projections

Lemma 34 Assume that a measurable set F has (f (αβ)1 f

(αβ)2 ) as its pair of (α β) projections

If for any y1 y2 isin R

f(αβ)1 (y1) le f

(αβ)1 (y2) hArr F (αβ)[y1] sub F (αβ)[y2] (54)

where for fixed yprime isin R

F (αβ)[yprime] = (x yprime)(αβ) | (x yprime)(αβ) isin F (55)

then no set other than F has (f (αβ)1 f


Proof Assume that there exists another set F prime than F which has (f (αβ)1 f

(αβ)2 ) as its pair of

(α β) projections We can choose ε gt 0 and a set J sub R λ1(J) gt 0 such that

λ1(F (αβ)[y0]minus F prime(αβ)[y0]) gt ε for y0 isin J (56)


10

Take a point of continuity y0 isin J of f(αβ)1 and fix it since we may assume the existence of such

points Let the sets I(y0) C be

I(y0) = x isin R | (x y0)(αβ) isin F C = (x y)(αβ) | x isin I(y0) (57)

Thenλ2(F cap C) =

int

I(y0)f

(αβ)2 (x)dx(αβ) = λ2(F prime cap C) (58)

therefore int infin

minusinfinλ1(C cap F (αβ)[y])dy(αβ) =

int infin

minusinfinλ1(C cap F prime(αβ)[y])dy(αβ) (59)

If f(αβ)1 (y) le f

(αβ)1 (y0) then by (54) we have F (αβ)[y] sub C therefore

λ1(C cap F (αβ)[y]) = f(αβ)1 (y) = λ1(F prime(αβ)[y]) ge λ1(C cap F prime(αβ)[y]) (60)

If f(αβ)1 (y) ge f

(αβ)1 (y0) then by (54) we have F (αβ)[y] sup C(αβ)[y0] therefore

λ1(C cap F (αβ)[y]) = λ1(C(αβ)[y0]) ge λ1(C cap F prime(αβ)[y]) (61)

Hence λ1(C cap F (αβ)[y]) ge λ1(C cap F prime(αβ)[y]) which with (59) yields

λ1(C cap F (αβ)[y]) = λ1(C cap F prime(αβ)[y]) (62)

Let us prove that (56) contradicts (62) For any ε gt 0 there exists δ gt 0 such that∣∣∣λ1(F (αβ)[y])minus λ1(F (αβ)[y0])

∣∣∣ ltε

2 for forally isin J |y minus y0| lt δ (63)

Hence we have

λ1(F (αβ)[y] cap C) gt λ1(F (αβ)[y0])minus ε


since if f(αβ)1 (y) ge f

(αβ)1 (y0) then by (54) F (αβ)[y] capC = F (αβ)[y0] otherwise F (αβ)[y] capC =

F (αβ)[y] Let the set D be

D = (x y)(αβ) | (x y)(αβ) isin F or (x y0)(αβ) isin F (65)

then we have ∣∣∣λ1(D(αβ)[y]minus F (αβ)[y0])∣∣∣ le

∣∣∣λ1(F (αβ)[y]minus F (αβ)[y0])∣∣∣ lt

ε

2 (66)

therefore there holds

λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε


By (56) and the definition of D we have

λ1(D(αβ)[y]minus F prime(αβ)[y]) = λ1(D(αβ)[y]minusD cap F prime(αβ)[y]) ge λ1(F (αβ)[y]minus F prime(αβ)[y]) gt ε (68)

Therefore By (64) (67) and (68) we obtain

λ1(C cap F prime(αβ)[y]) le λ1(D cap F prime(αβ)[y]) lt λ1(D(αβ)[y])minus ε

lt λ1(F (αβ)[y0])minus ε2 lt λ1(C cap F (αβ)[y])

(69)

for any y isin J |y minus y0| lt δ which contradicts (62) Hence the lemma is proved


11

Lemma 35 If a pair of (α β) projections (f (αβ)1 f

(αβ)2 ) satisfies (31) then the set F having

(f (αβ)1 f

(αβ)2 ) as its (α β) projections satisfies the condition (54)

Proof Since we have by the assumption D+ = Dminus = empty in (40) the set F constructed in theproof of Lemma 32 is represented as F = F

(αβ)yxy = F

(αβ)xyx which has (f (αβ)

121 f(αβ)212 ) as its (α β)

projections By its definition F satisfies the condition (54) therefore the set F x defined in (52)has the same property so does the set F defined in (53)

Lemmas 31-35 prove Theorem 31 This theorem takes an important role to prove our maintheorem in the next section (cf the proof of Lemma 41 below) In view of Theorem 31 the set(23) is uniquely reconstructed in the frame of the (0minusπ4) coordinates and so is the set (24)in the frame of the (π4 π4) coordinates

The following Theorems 32-38 are generalization of the theorems introduced in the secondsection Though they are proved by modifying the proofs of the original theorems (Theorem22-28) other proofs are possible which we shall discuss in our another paper In this paperwe would not mention the proofs of the following theorems (32-38) since there are neither ourmain purpose here nor applied to prove our main theorem We however introduce them inorder to explain the importance of Problem 12 (cf also Problem 31 below)

Theorem 32 If a measurable set F is uniquely determined by the pair of its (α β) projections(f (αβ)

1 f(αβ)2 ) then

F = (x y)(αβ)|f (αβ)2 (x) ge f

(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)

Theorem 33 A measurable plane set having a finite measure is non-uniquely determined byits (α β) projections if and only if it has an (α β) switching component

Definition 35 For a measurable function f(x) on R define

||f ||L1(αβ)

(R) =intR |f(x)| cos(αminus β)dx

||f ||Linfin(αβ)

(R) = ||f ||Linfin(R) cos(αminus β) (71)

Theorem 34 Let f(αβ)1 f

(αβ)2 g

(αβ)1 and g

(αβ)2 be non-negative essentially bounded integrable

functions Assume that the pairs of the projections (f (αβ)1 f

(αβ)2 ) and (g(αβ)

1 g(αβ)2 ) uniquely

determine the measurable plane sets F and G (with characteristic functions f(x y) and g(x y))respectively in the (α β) coordinate Then we have

||f minus g||L1(R2) le C middotmax||f (αβ)1 minus g

(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)

2 minus g(αβ)2 ||Linfin

(αβ)(R) (72)

where

C = C(||f (αβ)1 ||Linfin

(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)

(R) ||f (αβ)2 ||Linfin

(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)

Theorem 35 Assume that a pair of the (α β) projections (f (αβ)1 f

(αβ)2 ) determines sets

non-uniquely Take a set F whose (α β) projections are (f (αβ)1 f

(αβ)2 ) Then any set having

(f (αβ)1 f

(αβ)2 ) as the (α β) projections is obtained by switching the (α β) switching components

in F countable times


12


(αβ)2 isin L1capLinfin(R) Assume that the non-unique pair (f (αβ)

1 f(αβ)2 )

of the (α β) projections satisfies

f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)

Then there exists an (α β) uniquely determined set G such that

λ2(F ordfG) lt 2ε minf (αβ)1 Linfin

(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)

for any set F having (f (αβ)1 f

(αβ)2 ) as the (α β) projections

Theorem 37 Assume that two sets F1 and F2 have the same (α β) projections (f (αβ)1 f

(αβ)2 )

which satisfy (74) Then there holds

λ2(F1 ordf F2) le 4εminf1Linfin(αβ)

(R) f2Linfin(αβ)

(R) (76)

Theorem 38 We can approximately reconstruct the measurable plane set F whose (α β)projections satisfy (74) by approximating the set G in Theorem 36 Since G is uniquelydetermined by the pair of its (α β) projections we can apply an algorithm by Huang-Takiguchifor its approximate reconstruction In this algorithm the original set is approximated by adirect sum of parallelograms Note that no a priori information on the set itself is required inthis approximation

In view of these generalizations if we find the angles α and β satisfying the condition (31)then reconstruction is possible by Theorem 32 At least if we find the angles α and β satisfyingthe condition (74) in view of Theorems 34-38 approximate reconstruction is possible for thebounded sets

It is a very interesting problem to find such angles α and β for an unknown measurable setby some data of the projections however before studying this problem it is necessary to studythe following problem

Problem 31 For any measurable plane set F do there exist directions α and β such that (31)or (74) for small ε gt 0 holds

If this problem is positively solved then (approximate) reconstruction of any measurable setby two projection is possible Unfortunately the answer to Problem 31 is negative In the nextsection we give an example of a measurable set which would satisfy neither (31) nor (74) forsmall ε gt 0 for any directions α and β which is our main purpose in this paper

4 Main theoremIn this section we prove that Problem 31 is negatively solved The following is our maintheorem in this paper

Theorem 41 For any δ gt 0 there exists such a bounded measurable set F having(f (αβ)

1 f(αβ)2 ) as the (α β) projections that

f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)

for any α β satisfying minusπ2 le α + β le π2 minusπ2 lt αminus β lt π2

For the preparation of the proof we introduce a lemma


13

Definition 41 Let

M(α β) = λ1(y | f(αβ)1 (y) gt 0) N(α β) = f (αβ)

2 Linfin(R) (78)

Lemma 41

(i) If a pair of the (α β) projections (f (αβ)1 f

(αβ)2 ) f

(αβ)1 f

(αβ)2 isin L1 cap Linfin(R) determines a

set uniquely thenM(α β) = N(α β) (79)

(ii) If F is a bounded measurable plane set having (f (αβ)1 f

(αβ)2 ) as the (α β) projections then

M(α β) ge N(α β) (80)

Proof The first statement is an easy corollary of Theorem 31 (i) By their definitions thefunctions f

(αβ)12 and f

(αβ)212 has their values at the origin and there hold that λ1(y | f

(αβ)1 (y) gt

0) = f(αβ)12 (0) and that f (αβ)

2 Linfin(R) = f(αβ)212 (0) By virtue of (31) f

(αβ)12 (0) = f

(αβ)212 (0)

which proves the first statement in the lemmaThe proof of the second statement depends on Theorem 31 (i) and (ii) The same idea to

prove (i) is applied for the proof

Let us prove our main theorem

Proof of Theorem 41 We prove the theorem by constructing a counterexample Let ε gt 0 bevery small and define the set F as follows

F =(1 2)times [

(1 1 + ε) cup (2minus ε

2 2 + ε2

) cup (3minus ε 3)]

cup(1 1 + ε)times (1 2 + ε

2

) cup (2minus ε 2)times (2 + ε

2 3)

(81)

We claim that the set defined in (81) is the counterexample We first find the pair (α β)such that M(α β) is minimal for minusπ2 le α + β le π2 minusπ2 lt αminus β lt π2

For the set F defined in (81) the minimal point of M is (α β) = (0 0) and

M(0 0) = 2 (82)

It is easy to verify that for any line l in the plane

λ1(F cap l) lt 1 + 4ε (83)

which implies thatN(α β) lt 1 + 4ε (84)

for any α β satisfying minusπ2 le α + β le π2 minusπ2 lt α minus β lt π2 By virtue of (82) (84) ifwe take 0 lt ε lt 110 for example then we obtain

M(α β)minusN(α β) gt 12 (85)

for any α β satisfying minusπ2 le α + β le π2 minusπ2 lt α minus β lt π2 Lemma 41 implies that Fis non-uniquely reconstructed by any pair of the (α β) projections Therefore there exists sucha constant δ0 gt 0 that

f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)

for any α β satisfying minusπ2 le α + β le π2 minusπ2 lt α minus β lt π2 The theorem is proved byexpanding the set (81) in accordance with the given value δ gt 0

The following theorem is a straightforward corollary of Theorem 41Theorem 42 There exist measurable plane sets which would not be uniquely reconstructedfrom any pair of their (α β) projections where minusπ2 le α + β le π2 minusπ2 lt αminus β lt π2


14

5 Conclusions and open problemsIn this section we mention the concluding remarks and the open problems left to be solved forfurther development We first conclude our conclusions

Conclusion 51

(i) We have proved that the known results on the reconstruction of the measurable plane setsfrom the orthogonal projections are extended in the frame of the generalized coordinates

(ii) We have also proved that Problem 31 is negatively solved that is it is not true that themethod of our (approximate) reconstruction would be effective for the all measurable sets

In view of our conclusions we pose the following problems to be solved for furtherdevelopment

Problem 51

(i) Classify the all bounded measurable plane sets into the two classes One is the class ofthe measurable plane sets for which there exist such angles α and β that they are uniquelyreconstructed from their (α β) projections The other one is the class of the measurableplane sets for which no such angles α and β exist that they are uniquely reconstructed fromtheir (α β) projections

(ii) Assume that for a measurable set F there exist such angles α and β that F is uniquelyreconstructed from (f (αβ)

1 f(αβ)2 ) Find such angles α and β from some data of the

projections(iii) How much more data of the suitable projections would make the all measurable plane sets

reconstructible

The problem (i) in Problem 51 is a natural one On this problem the author has the followingconjecture

Conjecture 51 For any bounded convex set there exist such angles α and β that it is uniquelyreconstructed by its (α β) projections

In this conjecture the convexity on the set is assumed which seems to be too strong sincethere are many non-convex sets which are uniquely reconstructed by their two projectionsAnyway it is important to solve the problem (i) in Problem 51 in view of both theoreticinterest and practical applications

The problem (ii) in Problem 51 is much more important for applications For examplethe solution of this problem enables us non-destructive reconstruction of the hole in a threedimensional homogeneous object with much less data than the usual X-ray transform

As for the problem (iii) in Problem 51 it is known that arbitrary finite projections woulddetermine not the all measurable sets which was proved by GG Lorentz ([6]) In our problemwe pose that how many suitable projections would determine the all measurable plane setsuniquely where the directions of the projections may change in accordance with the set onlythe number of the projections is fixed We have proved in this paper that the two projectionswould not determine the all measurable plane sets In (iii) in Problem 51 we pose that for anymeasurable plane set whether the suitable three projections or more make it uniquely determinedor not


15

References[1] Huang L and Takiguchi T 1998 J Inv Ill-Posed Problems vol 6 pp 217-242[2] Kaneko A and Huang L 1999 Discrete Tomography Eds Herman G T and Kuba A (Birkhauser) pp 115-135[3] Kuba A 1991 Inverse Problems vol 7 pp 101-107[4] Kuba A and Volcic A 1988 Inverse Problems vol 4 pp 513-527[5] Kuba A and Volcic A 1993 Acta Sci Math (Szeged) vol 58 pp 363-388[6] Lorentz G G Amer J Math vol 71 pp 417-426[7] Ryff J V 1970 J Math Anal Appl vol 31 pp 449-458[8] Takiguchi T 2004 Contemporary Mathematics vol 348 pp 199-208


16

Reconstruction of the measurable sets in the two

dimensional plane by two projections

Takashi TakiguchiDepartment of Mathematics National Defense Academy of Japan1-10-20 Hashirimizu Yokosuka Kanagawa 239-8686 JAPAN

E-mail takashindaacjp

Abstract We discuss the reconstruction of the measurable plane sets from their twoprojections Reconstruction of a measurable plane set F sub R2 with λ2(F ) lt infin from itsorthogonal projections is well studied where λi is the Lebesgue measure on Ri i = 1 2 Inthis paper we first discuss generalization of the known results in the frame of general twoprojections The main purpose is to study for any measurable plane set F whether there aresuitable two angles such that F is uniquely reconstructed from the pair of its projections in thattwo directions We give an example to show that this problem is negatively solved

1 IntroductionIn this paper we discuss the reconstruction of the measurable plane sets from their twoprojections Let us consider a problem to reconstruct the characteristic function of a measurableplane set from its line integrals Since this is a problem in the Radon transform if we have the allline integrals then the reconstruction is an easy corollary of the results in the Radon transformOur object however is too simple to take its all line integrals It is better to reconstructthe object with as less data as possible In this paper we try the reconstruction with lineintegrals only in the two directions Though this problem is well studied for the case wherethese two projections are orthogonal few sets can be reconstructed in such a frame Thereforein this paper we take projections in two directions which are not necessarily orthogonal in thereconstruction of the measurable plane sets Let us formulate our problem

Definition 11 Let F sub R2 be a measurable plane set such that λ2(F ) lt infin where λi is theLebesgue measure on Ri i = 1 2 Let f(x y) be the characteristic function of F For α βsatisfying minusπ2 le α + β le π2 minusπ2 lt α minus β lt π2 we rotate the x-axis by the angle αand y-axis by the angle β which we call the x(α)-axis and the y(β)-axis respectively Define thecoordinates

(x y)(αβ) = (x cosαminus y sinβ x sinα + y cosβ)(00) = (x cosαminus y sinβ x sinα + y cosβ) (1)

where the coordinates in the right hand side are in the usual frame of the orthogonal coordinatesWe call this generalized frame (middot middot)(αβ) the frame of the (α β) coordinates For such α β wealso define

f(αβ)1 (yprime) =

intinfinminusinfin f(minusyprime sinβ + t cosα yprime cosβ + t sinα)dt

f(αβ)2 (xprime) =

intinfinminusinfin f(xprime cosαminus t sinβ xprime sinα + t cosβ)dt

(2)


ccopy 2007 IOP Publishing Ltd 1















2



(αβ)1 and f

(αβ)2




(00)1 and so on


f12(x) = λ1(y | f1(y) ge x) (3)


f21(y) = λ1(x | f2(x) ge y) (4)


f121(y) = λ1(x | f12(x) ge y) (5)

f212(x) = λ1(y | f21(y) ge x) (6)



minusinfinf1(y)dy =

int infin



int c

0f12(x)dx =

int c



int c

0f12(x)dx ge

int c




3


int c

0f12(x)dx lt

int c





F = (x y) | f2(x) ge f12(f1(y)) (11)













4

Problem 21













g212(x) = minf12(x) f212(x) (18)

andϕfi



g2(x) = g212(ϕf2(x)) (20)


g121(y) = minf21(y) f121(y) (21)

andg1(y) = g121(ϕf1(y)) (22)



5










f(αβ)12 (x) = λ1(y | f

(αβ)1 (y) ge x) f

(αβ)21 (y) = λ1(x | f

(αβ)2 (x) ge y)

f(αβ)121 (y) = λ1(x | f

(αβ)12 (x) ge y) f

(αβ)212 (x) = λ1(y | f

(αβ)21 (y) ge x) (25)



int infin



6




P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P








int infin

minusinfinf


int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case



int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f





int c

0f


int c

0f





int c

0f


int c

0f





(32) holds

Proof Let us define


(αβ)2 (x) F


(αβ)1 (y)


(αβ)212 (x) F


(αβ)121 (y) (34)


7




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16















2



(αβ)1 and f

(αβ)2




(00)1 and so on


f12(x) = λ1(y | f1(y) ge x) (3)


f21(y) = λ1(x | f2(x) ge y) (4)


f121(y) = λ1(x | f12(x) ge y) (5)

f212(x) = λ1(y | f21(y) ge x) (6)



minusinfinf1(y)dy =

int infin



int c

0f12(x)dx =

int c



int c

0f12(x)dx ge

int c




3


int c

0f12(x)dx lt

int c





F = (x y) | f2(x) ge f12(f1(y)) (11)













4

Problem 21













g212(x) = minf12(x) f212(x) (18)

andϕfi



g2(x) = g212(ϕf2(x)) (20)


g121(y) = minf21(y) f121(y) (21)

andg1(y) = g121(ϕf1(y)) (22)



5










f(αβ)12 (x) = λ1(y | f

(αβ)1 (y) ge x) f

(αβ)21 (y) = λ1(x | f

(αβ)2 (x) ge y)

f(αβ)121 (y) = λ1(x | f

(αβ)12 (x) ge y) f

(αβ)212 (x) = λ1(y | f

(αβ)21 (y) ge x) (25)



int infin



6




P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P








int infin

minusinfinf


int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case



int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f





int c

0f


int c

0f





int c

0f


int c

0f





(32) holds

Proof Let us define


(αβ)2 (x) F


(αβ)1 (y)


(αβ)212 (x) F


(αβ)121 (y) (34)


7




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16



(αβ)1 and f

(αβ)2




(00)1 and so on


f12(x) = λ1(y | f1(y) ge x) (3)


f21(y) = λ1(x | f2(x) ge y) (4)


f121(y) = λ1(x | f12(x) ge y) (5)

f212(x) = λ1(y | f21(y) ge x) (6)



minusinfinf1(y)dy =

int infin



int c

0f12(x)dx =

int c



int c

0f12(x)dx ge

int c




3


int c

0f12(x)dx lt

int c





F = (x y) | f2(x) ge f12(f1(y)) (11)













4

Problem 21













g212(x) = minf12(x) f212(x) (18)

andϕfi



g2(x) = g212(ϕf2(x)) (20)


g121(y) = minf21(y) f121(y) (21)

andg1(y) = g121(ϕf1(y)) (22)



5










f(αβ)12 (x) = λ1(y | f

(αβ)1 (y) ge x) f

(αβ)21 (y) = λ1(x | f

(αβ)2 (x) ge y)

f(αβ)121 (y) = λ1(x | f

(αβ)12 (x) ge y) f

(αβ)212 (x) = λ1(y | f

(αβ)21 (y) ge x) (25)



int infin



6




P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P








int infin

minusinfinf


int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case



int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f





int c

0f


int c

0f





int c

0f


int c

0f





(32) holds

Proof Let us define


(αβ)2 (x) F


(αβ)1 (y)


(αβ)212 (x) F


(αβ)121 (y) (34)


7




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16


int c

0f12(x)dx lt

int c





F = (x y) | f2(x) ge f12(f1(y)) (11)













4

Problem 21













g212(x) = minf12(x) f212(x) (18)

andϕfi



g2(x) = g212(ϕf2(x)) (20)


g121(y) = minf21(y) f121(y) (21)

andg1(y) = g121(ϕf1(y)) (22)



5










f(αβ)12 (x) = λ1(y | f

(αβ)1 (y) ge x) f

(αβ)21 (y) = λ1(x | f

(αβ)2 (x) ge y)

f(αβ)121 (y) = λ1(x | f

(αβ)12 (x) ge y) f

(αβ)212 (x) = λ1(y | f

(αβ)21 (y) ge x) (25)



int infin



6




P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P








int infin

minusinfinf


int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case



int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f





int c

0f


int c

0f





int c

0f


int c

0f





(32) holds

Proof Let us define


(αβ)2 (x) F


(αβ)1 (y)


(αβ)212 (x) F


(αβ)121 (y) (34)


7




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16

Problem 21













g212(x) = minf12(x) f212(x) (18)

andϕfi



g2(x) = g212(ϕf2(x)) (20)


g121(y) = minf21(y) f121(y) (21)

andg1(y) = g121(ϕf1(y)) (22)



5










f(αβ)12 (x) = λ1(y | f

(αβ)1 (y) ge x) f

(αβ)21 (y) = λ1(x | f

(αβ)2 (x) ge y)

f(αβ)121 (y) = λ1(x | f

(αβ)12 (x) ge y) f

(αβ)212 (x) = λ1(y | f

(αβ)21 (y) ge x) (25)



int infin



6




P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P








int infin

minusinfinf


int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case



int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f





int c

0f


int c

0f





int c

0f


int c

0f





(32) holds

Proof Let us define


(αβ)2 (x) F


(αβ)1 (y)


(αβ)212 (x) F


(αβ)121 (y) (34)


7




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16










f(αβ)12 (x) = λ1(y | f

(αβ)1 (y) ge x) f

(αβ)21 (y) = λ1(x | f

(αβ)2 (x) ge y)

f(αβ)121 (y) = λ1(x | f

(αβ)12 (x) ge y) f

(αβ)212 (x) = λ1(y | f

(αβ)21 (y) ge x) (25)



int infin



6




P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P








int infin

minusinfinf


int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case



int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f





int c

0f


int c

0f





int c

0f


int c

0f





(32) holds

Proof Let us define


(αβ)2 (x) F


(αβ)1 (y)


(αβ)212 (x) F


(αβ)121 (y) (34)


7




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16




P 1 = P(αβ)(s0) P 2 = P

(αβ)(0t) P 12 = P








int infin

minusinfinf


int infin

minusinfinf

(αβ)2 (x)dx(αβ) (30)

(i) The unique case



int c

0f

(αβ)12 (x)dx(αβ) =

int c

0f





int c

0f


int c

0f





int c

0f


int c

0f





(32) holds

Proof Let us define


(αβ)2 (x) F


(αβ)1 (y)


(αβ)212 (x) F


(αβ)121 (y) (34)


7




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16




If


int k

0f

(αβ)12 (x)dx(αβ) (35)




(αβ)2 )



f(αβ)1 (y) le f

(αβ)1 (y) and f



1 (y) k) = f(αβ)1 (y) (37)

Therefore we obtain

λ2(F ) =int infin

0f


int infin

0f

(αβ)1 (y)dy(αβ) = λ2(F ) (38)


(αβ)212 of f

(αβ)2

int k

0f


int k

0f

(αβ)12 (x)dx(αβ) (39)








(αβ)212 (x)


(αβ)212 (x)


(αβ)12 (x)

(40)


D+ =sum

L+i Dminus =

sumLminusi (41)




i(αβ)



8




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16




0f


int c

0f















(F (αβ)yxy )1 = (F (αβ)


y lt y lt l1+y

(F (αβ)yxy )1 = (F (αβ)



1 (47)



(αβ)yxy




1 )j f(αβ)212 ) the






y ) + rn



y )minus rn(48)









yxy )nminus1




9






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16






y (50)


(αβ)212 (x) as F

(αβ)yxy


(αβ)212 ) the (α β)


f(αβ)1 (y0) = f


= f(αβ)121 (y0)

(51)we obtain f

(αβ)1 equiv f




F x = (x y)(αβ) | (ϕf(αβ)2

(x) y) isin F (52)





F = (x y)(αβ) | (x ϕf(αβ)1

(y)) isin F x (53)













f(αβ)1 (y1) le f











10




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16




Thenλ2(F cap C) =

int

I(y0)f


therefore int infin


int infin











∣∣∣ ltε


Hence we have









ε

2 (66)


λ1(D(αβ)[y]) lt λ1(F (αβ)[y0]) +ε







(69)



11



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16



(f (αβ)1 f



(αβ)yxy = F


121 f(αβ)212 ) as its (α β)





1 f(αβ)2 ) then


(αβ)12 (f (αβ)

1 (y)) = (x y)(αβ)|f (αβ)1 (y) ge f

(αβ)21 (f (αβ)

2 (x)) (70)



||f ||L1(αβ)


||f ||Linfin(αβ)



(αβ)2 g

(αβ)1 and g







(αβ)1 ||Linfin

(αβ)(R) ||f (αβ)


(αβ)(R) (72)

where


(αβ)(R) ||g(αβ)

1 ||Linfin(αβ)


(αβ)(R) ||g(αβ)

2 ||Linfin(αβ)

(R)) (73)





(f (αβ)1 f




12



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16



1 f(αβ)2 )


f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) lt ε (74)



(αβ)(R) f (αβ)

2 Linfin(αβ)

(R) (75)




(αβ)2 )



(R) f2Linfin(αβ)

(R) (76)









f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ (77)




13

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16

Definition 41 Let


2 Linfin(R) (78)

Lemma 41


(αβ)2 ) f

(αβ)1 f





M(α β) ge N(α β) (80)


(αβ)12 and f


(αβ)1 (y) gt



(αβ)12 (0) = f

(αβ)212 (0)





F =(1 2)times [


2 2 + ε2


cup(1 1 + ε)times (1 2 + ε

2


2 3)

(81)



M(0 0) = 2 (82)


λ1(F cap l) lt 1 + 4ε (83)





f (αβ)12 minus f

(αβ)212 L1

(αβ)(R) ge δ0 (86)




14


Conclusion 51




Problem 51





reconstructible







15



16


Conclusion 51




Problem 51





reconstructible







15



16



16

reconstruction of the measurable sets in the two - iopscience

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