Monotone Convergence Theorem
Theorem 1.
If
0
6
f
n
↑
f
then
R
f
n
d
μ
↑
R
f
d
μ
.
Proof.
As
f
n
increases to
f
it is plain that lim
n
R
f
n
d
μ
6
R
f
d
μ
. In order to
establish the reverse inequality we retrace the way in which the integral is
defined, considering first
f
simple and then
f
arbitrary.
The case
f
simple is conveniently handled in two stages. Suppose first
that
f
=
a
1
A
for some
a
∈
[0
,
∞
] and
A
∈ F
. If
a
= 0 then there is nothing
to do; so let
a >
0 and choose
α
∈
(0
, a
): from
f
>
f
n
>
f
n
1
{
f
n
>
α
}
>
α
1
{
f
n
>
α
}
it follows by monotonicity of the integral that
Z
f
d
μ
>
Z
f
n
d
μ
>
αμ
(
{
f
n
>
α
}
)
whence from
n
↑ ∞ ⇒ {
f
n
>
α
} ↑
A
it follows that
Z
f
d
μ
>
lim
n
Z
f
n
d
μ
>
αμ
(
A
);
as
α
∈
(0
, a
) was arbitrary, we deduce that
Z
f
d
μ
>
lim
n
Z
f
n
d
μ
>
aμ
(
A
) =
Z
f
d
μ.
Now let
f
be simple: say
f
=
M
X
m
=1
a
m
1
A
m
1
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with
a
1
, . . . , a
M
nonnegative and
A
1
, . . . , A
M
∈ F
disjoint. We may rewrite
this decomposition as
f
=
M
X
m
=1
f
1
A
m
and decompose
f
n
similarly:
f
n
=
M
X
m
=1
f
n
1
A
m
.
As
f
n
1
A
m
↑
f
1
A
m
=
a
m
1
A
m
the case just handled yields
Z
f
n
1
A
m
d
μ
↑
a
m
μ
(
A
m
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 Spring '09
 LARSON
 Calculus, lim, FN, Dominated convergence theorem, Monotone convergence theorem

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