Monotone Convergence Theorem (Measure Theory)

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $u : X \to \overline \R_{\ge 0}$ be a positive $\Sigma$-measurable function. 
Let $\sequence {u_n}_{n \mathop \in \N}$ be an sequence of positive $\Sigma$-measurable functions $u_n : X \to \overline \R_{\ge 0}$ such that: 
:$\map {u_i} x \le \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
hold for $\mu$-almost all $x \in X$. 
Then:
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$
=== Corollary ===
== Proof ==
First suppose that: 
:$\map {u_i} x \le \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for all $x \in X$.
From Integral of Positive Measurable Function is Monotone, we have that: 
:$\ds \int u_i \rd \mu \le \int u_j \rd \mu$ for all $i \le j$.
From Monotone Convergence Theorem (Real Analysis): Increasing Sequence, we have: 
:$u_i \le u$ for each $i$.
So, applying Integral of Positive Measurable Function is Monotone again we have: 
:$\ds \int u_i \rd \mu \le \int u \rd \mu$ for each $i$.
So: 
:the sequence $\ds \sequence {\int u_i \rd \mu}_{i \in \N}$ is increasing.
So, from Monotone Convergence Theorem (Real Analysis): Increasing Sequence and Unbounded Monotone Sequence Diverges to Infinity: Increasing, we have: 
:$\ds \sequence {\int u_i \rd \mu}_{i \in \N}$ converges, possibly to $+\infty$.
From Lower and Upper Bounds for Sequences, we also have: 
:$\ds \lim_{n \mathop \to \infty} \int u_n \rd \mu \le \int u \rd \mu$ 
We now aim to prove: 
:$\ds \int u \rd \mu \le \lim_{n \mathop \to \infty} \int u_n \rd \mu$
From Measurable Function is Pointwise Limit of Simple Functions, for each $n$ there exists a increasing sequence of positive simple functions $\sequence {u_{n, k} }_{k \mathop \in \N}$ such that: 
:$\ds u_n = \lim_{k \mathop \to \infty} u_{n, k}$
From Monotone Convergence Theorem (Real Analysis): Increasing Sequence, this is equivalent to: 
:$\ds u_n = \sup_{k \in \N} u_{n, k}$ 
Let: 
:$\ds g_n = \max \set {u_{1, n}, u_{2, n}, \ldots, u_{n, n} }$ 
for each $n$.
From Pointwise Maximum of Simple Functions is Simple, $g_n$ is a positive simple function for each $n \in \N$.
With a view to apply Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, we show that: 
:$\sequence {g_n}_{n \mathop \in \N}$ is increasing
and:
:$\ds \map u x = \lim_{n \to \infty} \map {g_n} x$ 
for each $x \in X$.
We also show that: 
:$\map {g_n} x \le \map {u_n} x$
for each $n \in \N$ and $x \in X$. 
=== Lemma ===
So, from Integral of Positive Measurable Function as Limit of Integrals of Positive Simple Functions, we have: 
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int g_n \rd \mu$ 
Since:
:$g_n \le u_n$ for each $n \in \N$
we have: 
:$\ds \int g_n \rd \mu \le \int u_n \rd \mu$ for each $n \in \N$
from Integral of Positive Measurable Function is Monotone. 
So, from Inequality Rule for Real Sequences: 
:$\ds \lim_{n \mathop \to \infty} \int g_n \rd \mu \le \lim_{n \mathop \to \infty} \int h_n \rd \mu$
So, we have: 
:$\ds \int u \rd \mu \le \lim_{n \mathop \to \infty} \int u_n \rd \mu$
We therefore obtain: 
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$
as required.
Now suppose that: 
:$\map {u_i} x \le \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for $\mu$-almost all $x \in X$.
That is, there exists a $\mu$-null set $N \subseteq X$ such that whenever $x \in X$ has: 
:$\map {u_i} x > \map {u_j} x$ for some $i < j$
and:
:either $\ds \lim_{n \mathop \to \infty} \map {u_n} x$ does not exist or $\ds \map u x \ne \lim_{n \mathop \to \infty} \map {u_n} x$ 
we have $x \in N$.
For each $n \in \N$, define $v_n : X \to \overline \R_{\ge 0}$ by: 
:$\map {v_n} x = \map {u_n} x \times \map {\chi_{X \setminus N} } x$
for each $x \in X$. 
Also define $v : X \to \overline \R_{\ge 0}$ by:
:$\map v x = \map u x \times \map {\chi_{X \setminus N} } x$ 
for each $x \in X$.
Clearly, if $x \in X \setminus N$, we have: 
:$\map {v_n} x = \map {u_n} x$ for each $n$
and:
:$\map v x = \map u x$ 
From the definition of $N$, we have: 
:$\map {u_i} x \le \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$ 
for $x \in X \setminus N$.
So:
:$\map {v_i} x \le \map {v_j} x$ for all $i \le j$
and:
:$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$
If $x \in N$, we have: 
:$\map {v_n} x = 0$ for each $n \in \N$
and:
:$\map v x = 0$ 
So we have: 
:$\map {v_i} x \le \map {v_j} x$ for $i \le j$ 
and:
:$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$
for $x \in N$. 
So, we have: 
:$\map {v_i} x \le \map {v_j} x$ for $i \le j$ 
and:
:$\ds \map v x = \lim_{n \mathop \to \infty} \map {v_n} x$
for all $x \in X$. 
So, from our previous work, we have: 
:$\ds \int v \rd \mu = \lim_{n \mathop \to \infty} \int v_n \rd \mu$
From Characteristic Function of Null Set is A.E. Equal to Zero: Corollary, we have: 
:$\chi_{X \setminus N} = 1$ $\mu$-almost everywhere.
So, from Pointwise Multiplication preserves A.E. Equality, we have: 
:$u \times \chi_{X \setminus N} = u$ $\mu$-almost everywhere
and:
:$u_n \times \chi_{X \setminus N} = u_n$ $\mu$-almost everywhere for each $n \in \N$.
So, we have: 
:$u = v$ $\mu$-almost everywhere
and:
:$u_n = v_n$ $\mu$-almost everywhere for each $n \in \N$.
From A.E. Equal Positive Measurable Functions have Equal Integrals, we therefore have: 
:$\ds \int u \rd \mu = \int v \rd \mu$ 
and:
:$\ds \int u_n \rd \mu = \int v_n \rd \mu$ for each $n \in \N$
giving: 
:$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$
hence the demand.
== Proof 2 ==
Let $Y \subseteq X$ be the set of $x \in X$ such that: 
:$\map {u_i} x \le \map {u_j} x$ for all $i \le j$
and:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
Then :
:$\map \mu {X \setminus Y} = 0$
We define an increasing sequence $\sequence {f_n}_{n \mathop \in \N}$ by: 
:$\map {f_n} x = \begin{cases}
\map {u_n} x & x \in Y \\
0 & x \not \in Y
\end{cases}$
Then:
== Sources ==
* : $11.1 \ \text{(i)}$
Category:Named Theorems
Category:Measure Theory
Category:Monotone Convergence Theorem
Category:Monotone Convergence Theorem (Measure Theory)