Page 89 - Vector Analysis

§3.4 The Fubini theorem 85

Since ε ą 0 is given arbitrarily, we conclude that

ż ż (ż )

f (z) dz ď f (x, y)dx dy .

AˆB BA

ż (ż )ż

Similarly, f (x, y)dy dx ď f (z) dz; thus (3.3) is concluded. ˝

AB AˆB

Corollary 3.26. Let S Ď Rn be a closed and bounded set such that B S has volume zero,

φ1, φ2 : S Ñ R be continuous maps such that φ1(x) ď φ2(x) for all x P S, A = ␣(x, y) P

Rn ˆ R ˇ x P S, φ1(x) ď y ď φ2(x)(, and f : A Ñ R be continuous. Then f is Riemann
ˇ

integrable over A, and

ż ż ( ż φ2(x) )

f (x, y) d(x, y) = f (x, y) dy dx . (3.5)

A S φ1(x)

Proof. To establish that f is Riemann integrable over A, by Theorem 3.23 it suffices to show

that B A has volume zero. Let m = min φ1(x) and M = max φ2(x). Since
xPS xPS

B A Ď ␣(x, φ1(x)) ˇ x P S( Y ␣(x, φ2(x)) ˇ x P S( Y ( S ˆ [m, M ]) ,
ˇ ˇ B

to see BA has volume zero it suffices to show that BS ˆ [m, M ], ␣(x, φ1(x)) ˇ x P S( and
ˇ

␣(x, φ2(x)) ˇ x P S( have volume zero because of Lemma 3.21 and 3.22. Note that Theorem
ˇ

3.23 implies that φ1 is Riemann integrable over S; thus for a given ε ą 0 there exists

partition P of S such that

U (φ1, P) ´ L(φ1, P) ă ε .

Let B = Ť [] Then C ” ␣(x, φ1 (x)) ˇ x P S( Ď B
∆ ˆ infxP∆ φ1S(x), supxP∆ φ1S(x) . ˇ

∆PP ,∆XS ‰H

and

żż ÿ ()
0 ď 1C(z) dz ď 1B(z) dz ď supφ1S(x) ´ inf φ1S(x) ˆ νn(∆)
xP∆
C B ∆PP,∆XS‰H xP∆

ď U (φ1, P) ´ L(φ1, P) ă ε .

Therefore, C = ␣(x, φ1(x)) ˇ x P S( has volume zero and similarly, ␣(x, φ2(x)) ˇ x P S( has
ˇ ˇ

volume zero.

Now we show that B S ˆ [m, M ] has volume zero. Since B S has volume zero in Rn, for a

given ε ą 0 there exists a partition P of B S such that

U (1S, P) ă M ε +1 .
´m