5.8. AN APPROACH TO THE INTEGRAL 127

Proposition 5.8.7 The above definition is well defined.

Proof: Letting b ≡ sup{x : f (x) ̸= 0} , it follows f (b) = 0 and f (x) = 0 for x > b.Similarly if a ≡ inf{x : f (x) ̸= 0} , it follows f (a) = 0 and f (x) = 0 for x < a. Thus, bythe mean value theorem, F ′= f requires F (x) =F (b) for x> b and F (x) =F (a) for x< a.It follows that the above definition is not dependent on the interval [a,b] containing spt( f ).■

Consider the iterated integral∫ b1

a1· · ·∫ bp

apαxα1

1 · · ·xα pp dxp · · ·dx1. It means just what it

meant in calculus. You do the integral with respect to xp first, keeping the other variablesconstant, obtaining a polynomial function of the other variables. Then you do this one withrespect to xp−1 and so forth. Thus, doing the computation, it reduces to

α

p

∏k=1

(∫ bk

ak

xαkk dxk

)= α

p

∏k=1

(bαk+1

αk +1− aαk+1

αk +1

)and the same thing would be obtained for any other order of the iterated integrals. Sinceeach of these integrals is linear, it follows that if (i1, · · · , ip) is any permutation of (1, · · · , p) ,then for any polynomial q,∫ b1

a1

· · ·∫ bp

ap

q(x1, ...,xp)dxp · · ·dx1 =∫ bi1

aip

· · ·∫ bip

aip

q(x1, ...,xp)dxip · · ·dxi1

Now let f : ∏pk=1 [ak,bk]→ R be continuous. Then each iterated integral results in a con-

tinuous function of the remaining variables and so the iterated integral makes sense. Forexample, by Proposition 5.8.5,

∣∣∣∫ dc f (x,y)dy−

∫ dc f (x̂,y)dy

∣∣∣=∣∣∣∣∫ d

c( f (x,y)− f (x̂,y))dy

∣∣∣∣≤ maxy∈[c,d]

| f (x,y)− f (x̂,y)|< ε

if |x− x̂| is sufficiently small, thanks to uniform continuity of f on the compact set [a,b]×[c,d]. Thus it makes perfect sense to consider the iterated integral

∫ ba∫ d

c f (x,y)dydx. Thenusing Proposition 5.8.5 on the iterated integrals along with Theorem 5.6.1, there exists asequence of polynomials which converges to f uniformly {pn} . Then applying Proposition5.8.5 repeatedly,∣∣∣∣∣∫ bi1

aip

· · ·∫ bip

aip

f (x)dxp · · ·dx1−∫ bi1

aip

· · ·∫ bip

aip

pn (x)dxp · · ·dx1

∣∣∣∣∣≤ ∥ f − pn∥p

∏k=1|bk−ak|

(5.4)With this, it is easy to prove a rudimentary Fubini theorem valid for continuous functions.

Theorem 5.8.8 f : ∏pk=1 [ak,bk]→ R be continuous. Then for (i1, · · · , ip) any per-

mutation of (1, · · · , p) ,∫ bi1

aip

· · ·∫ bip

aip

f (x)dxip · · ·dxi1 =∫ b1

a1

· · ·∫ bp

ap

f (x)dxp · · ·dx1

If f ≥ 0, then the iterated integrals are nonnegative if each ak ≤ bk. Also, we can define forf ∈Cc (Rp) ∫

Rpf ≡

∫ b1

a1

· · ·∫ bp

ap

f (x)dxp · · ·dx1