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.
(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.