3Integration

II Probability and Measure

3.4 Integration and differentiation

In “normal” calculus, we had three results involving both integration and dif-

ferentiation. One was the fundamental theorem of calculus, which we already

stated. The others are the change of variables formula, and differentiating under

the integral sign.

We start by proving the change of variables formula.

Proposition

(Change of variables formula)

.

Let

φ

: [

a, b

]

→ R

be continuously

differentiable and increasing. Then for any bounded Borel function g, we have

Z

φ(b)

φ(a)

g(y) dy =

Z

b

a

g(φ(x))φ

0

(x) dx. (∗)

We will use the monotone class theorem.

Proof. We let

V = {Borel functions g such that (∗) holds}.

We will want to use the monotone class theorem to show that this includes all

bounded functions.

We already know that

(i) V

contains

1

A

for all

A

in the

π

-system of intervals of the form [

u, v

]

⊆

[

a, b

].

This is just the fundamental theorem of calculus.

(ii) By linearity of the integral, V is indeed a vector space.

(iii)

Finally, let (

g

n

) be a sequence in

V

, and

g

n

≥

0,

g

n

% g

. Then we know

that

Z

φ(b)

φ(a)

g

n

(y) dy =

Z

b

a

g

n

(φ(x))φ

0

(x) dx.

By the monotone convergence theorem, these converge to

Z

φ(b)

φ(a)

g(y) dy =

Z

b

a

g(φ(x))φ

0

(x) dx.

Then by the monotone class theorem,

V

contains all bounded Borel functions.

The next problem is differentiation under the integral sign. We want to know

when we can say

d

dt

Z

f(x, t) dx =

Z

∂f

∂t

(x, t) dx.

Theorem

(Differentiation under the integral sign)

.

Let (

E, E, µ

) be a space,

and U ⊆ R be an open set, and f : U × E → R. We assume that

(i) For any t ∈ U fixed, the map x 7→ f(t, x) is integrable;

(ii) For any x ∈ E fixed, the map t 7→ f (t, x) is differentiable;

(iii) There exists an integrable function g such that

∂f

∂t

(t, x)

≤ g(x)

for all x ∈ E and t ∈ U.

Then the map

x 7→

∂f

∂t

(t, x)

is integrable for all t, and also the function

F (t) =

Z

E

f(t, x)dµ

is differentiable, and

F

0

(t) =

Z

E

∂f

∂t

(t, x) dµ.

The reason why we want the derivative to be bounded is that we want to

apply the dominated convergence theorem.

Proof.

Measurability of the derivative follows from the fact that it is a limit of

measurable functions, and then integrability follows since it is bounded by g.

Suppose (h

n

) is a positive sequence with h

n

→ 0. Then let

g

n

(x) =

f(t + h

n

, x) −f(t, x)

h

n

−

∂f

∂t

(t, x).

Since

f

is differentiable, we know that

g

n

(

x

)

→

0 as

n → ∞

. Moreover, by the

mean value theorem, we know that

|g

n

(x)| ≤ 2g(x).

On the other hand, by definition of F (t), we have

F (t + h

n

) −F(t)

h

n

−

Z

E

∂f

∂t

(t, x) dµ =

Z

g

n

(x) dx.

By dominated convergence, we know the RHS tends to 0. So we know

lim

n→∞

F (t + h

n

) −F(t)

h

n

→

Z

E

∂f

∂t

(t, x) dµ.

Since

h

n

was arbitrary, it follows that

F

0

(

t

) exists and is equal to the integral.