520 CHAPTER 19. HAUSDORFF SPACES AND MEASURES

Proof: Let spt( f ) ⊆ [−R,R]n. It clearly suffices to show this for n = 2. Then by thedefinition of the Riemann integral,

∫ R−R∫ R−R f (x,y)dxdy =∫ R

−R∑

i

∫ xi+1

xi

f (s,y)dsdy =n−1

∑j=0

n−1

∑i=0

∫ yi+1

yi

∫ xi+1

xi

f (s, t)dsdt

=n−1

∑j=0

n−1

∑i=0

∫ yi+1

yi

f (si, t)(xi+1− xi)dt

=n−1

∑j=0

n−1

∑i=0

f (si, t j)(xi+1− xi)(y j+1− y j

)=

n−1

∑i=0

n−1

∑j=0

f (si, t j)(xi+1− xi)(y j+1− y j

)where−R = x0 < x1 < · · ·< xn = R is a uniform partition of [−R,R] with the yi also givinga uniform partition of [−R,R]. Similar reasoning implies∫ R

−R

∫ R

−Rf (x,y)dydx =

n−1

∑i=0

n−1

∑j=0

f (ŝi, t̂ j)(xi+1− xi)(y j+1− y j

).

Now (si, t j) ,(ŝi, t̂ j) are both in [xi,xi+1]×[y j,y j+1

]. Thus, by uniform continuity, if n is

large enough, ∣∣ f (si, t j)− f (ŝi, t̂ j)∣∣< ε

4R2

Then it follows that∣∣∣∫ R−R∫ R−R f (x,y)dxdy−

∫ R−R∫ R−R f (x,y)dydx

∣∣∣≤n−1

∑i=0

n−1

∑j=0

∣∣ f (ŝi, t̂ j)− f (si, t j)∣∣(xi+1− xi)

(y j+1− y j

)≤

n−1

∑i=0

n−1

∑j=0

ε

4R2 (xi+1− xi)(y j+1− y j

)= ε

Since ε is arbitrary, this shows that the two iterated integrals are the same. In case n> 2, youcan do exactly the same argument using the mean value theorem for integrals and obtainthe same result by a similar argument, or you could use this result on pairs of integrals. ■

19.8 Slicing MeasuresI saw this material first in the book [17]. It can be presented as an application of the theoryof differentiation of Radon measures and the Riesz representation theorem for positivelinear functionals. It is an amazing theorem and can be used to understand conditionalprobability However, here I will obtain it from Theorem 10.14.12.

Theorem 19.8.1 Let µ be a finite Radon measure on Rn+m defined on a σ algebra,F . Then there exists a unique finite Radon measure α, defined on a σ algebra S , of setsof Rn which satisfies

α (E) = µ (E×Rm) (19.12)

for all E Borel. There also exists a Borel set of α measure zero N, such that for each x /∈N,there exists a Radon probability measure νx such that if f is a nonnegative µ measurablefunction or a µ measurable function in L1 (µ),

y→ f (x,y) is νx measurable α a.e.

520 CHAPTER 19. HAUSDORFF SPACES AND MEASURESProof: Let spt(f) C [—R,R]". It clearly suffices to show this for n = 2. Then by thedefinition of the Riemann integral, [ x f x f (x,y) dxdy =R Xi4] xy 1 xy I Yi+1 Xi+1/ y? / f (s,y) dsdy rf f (s,t) dsdtRG Xi j= dd 0yyi+]= FEL rout —xyarJ=0 i=0n—\1n-1 n—In-1app Si,t;) (xi41 — i) (Yj — yj) SE Si,t;) (xi41 — x) (Yj41 — yj)j=0i i=0 jwhere —R = x9 < x1 <+++ <x, =R isa uniform partition of [—R, R] with the y; also givinga uniform partition of |—R,R]. Similar reasoning impliesrR PR n—1n—1[| farex= »y dL F( §1,8)) (xin1 — Hi) (Yn — ys) -TIN i=0 j=0Now (s;,t;),(5;,;) are both in [x;,xi+1] x [yj,yj41]. Thus, by uniform continuity, if n islarge enough,If (si.t;) — f (Si.F))| < raThen it follows that | [Bo [Ref (x.y) dxdy — [®p [ef (x.y) dydx| <n—\n—-1| f (8,8) — f (Sint; ) (xi41 — xi) (Yj —yj)i=0 j=0n—\|n-1 €<yy pz Cit! ~%i) (Viti ys) =i=0 j=0Since € is arbitrary, this shows that the two iterated integrals are the same. In case n > 2, youcan do exactly the same argument using the mean value theorem for integrals and obtainthe same result by a similar argument, or you could use this result on pairs of integrals.19.8 Slicing MeasuresI saw this material first in the book [17]. It can be presented as an application of the theoryof differentiation of Radon measures and the Riesz representation theorem for positivelinear functionals. It is an amazing theorem and can be used to understand conditionalprobability However, here I will obtain it from Theorem 10.14.12.Theorem 19.8.1 Lez ut be a finite Radon measure on R"*" defined on a © algebra,F. Then there exists a unique finite Radon measure a, defined on a © algebra -, of setsof IR” which satisfiesa(E)=uU(E xR") (19.12)for all E Borel. There also exists a Borel set of 0& measure zero N, such that for each x € N,there exists a Radon probability measure Vz such that if f is a nonnegative & measurablefunction or a LL measurable function in L! (1),y > f(x,y) is Vz measurable Ot a.e.