Monotone Convergence Theorem (Measure Theory)/Lemma

== Lemma ==
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:
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$
for all $x \in X$. 
Let $\sequence {u_{n, k} }_{k \mathop \in \N}$ be an increasing sequence of positive simple functions such that: 
:$\ds u_n = \lim_{k \mathop \to \infty} u_{n, k}$ 
Let: 
:$\ds g_n = \max \set {u_{1, n}, u_{2, n}, \ldots, u_{n, n} }$
for each $n \in \N$.
Then: 
:$(1) \quad$ $\sequence {g_n}_{n \mathop \in \N}$ is increasing
:$(2) \quad$ $\ds \map u x = \lim_{n \mathop \to \infty} \map {g_n} x$
:$(3) \quad$ $g_n \le u_n$ for each $n \in \N$.
== Proof ==
=== Proof of $(1)$ ===
Let $x \in X$ and $n \in \N$, we can write: 
so: 
:$\sequence {g_n}_{n \mathop \in \N}$ is increasing.
=== Proof of $(2)$ ===
We now show that: 
:$\ds \map u x = \lim_{n \to \infty} \map {g_n} x$ 
for each $x \in X$.
Suppose that $\map u x$ is finite. 
Since: 
:$\sequence {g_n}_{n \mathop \in \N}$ is increasing
we have: 
:$\map {g_n} x \le \map u x$ 
for all $n \in \N$ and $x \in X$, from Monotone Convergence Theorem (Real Analysis): Increasing Sequence.
So it suffices to show that for all $\epsilon > 0$ there exists $N \in \N$ such that: 
:$\map {g_n} x > \map u x - \epsilon$
for all $n > N$.
Fix $\epsilon > 0$. 
We have: 
:$\ds \map u x = \lim_{n \mathop \to \infty} \paren {\map {u_n} x}$
From the definition of a convergent sequence, there exists $N_1 \in \N$ such that: 
:$\ds \size {\map {u_n} x - \map u x} < \frac \epsilon 2$ 
So we have: 
:$\ds \map {u_n} x > \map u x - \frac \epsilon 2$
for $n \ge N_1$.
Noting that: 
:$\ds \map {u_{N_1} } x = \lim_{k \mathop \to \infty} \map {u_{N_1, k} } x$ 
We can pick $N_2 \in \N$ such that:
:$\ds \map {u_{N_1, k} } x > \map {u_n} x - \frac \epsilon 2 > \map u x - \epsilon$
for $k \ge N_2$.
We then have: 
Since $\sequence {g_n}_{n \mathop \in \N}$ is increasing, we therefore have: 
:$\map {g_n} x > \map u x - \epsilon$
for all $n \ge N_1 + N_2 = N$.
Showing that: 
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {g_n} x$ 
in the case $\map u x < \infty$. 
Now suppose that $\map u x = \infty$. 
We aim to show that for each $M > 0$ there exists $N \in \N$ such that: 
:$\map {g_n} x \ge M$ 
for each $n \ge N$. 
Since: 
:$\ds \infty = \lim_{n \mathop \to \infty} \map {u_n} x$
there exists an $N_1 \in \N$ such that: 
:$\ds \map {u_n} x \ge M + \frac 1 2$ 
for each $n \ge N_1$. 
Since: 
:$\ds \map {u_n} x = \lim_{k \mathop \to \infty} \map {u_{n, k} } x$
there exists an $N_2 \in \N$ such that: 
:$\ds \map {u_{n, k} } x \ge \map {u_n} x - \frac 1 2 \ge M$ 
for all $k \ge N_2$. 
As before, we have: 
Again, since $\sequence {g_n}_{n \mathop \in \N}$ is increasing, we therefore have: 
:$\map {g_n} x > M$
for all $n \ge N_1 + N_2 = N$.
Showing that: 
:$\ds \map u x = \lim_{n \mathop \to \infty} \map {g_n} x$ 
in the case $\map u x = \infty$.
=== Proof of $(3)$ ===
Note from the definition of pointwise supremum, we have that:
:$\ds u_{i, n} \le \sup_{k \mathop \in \N} u_{i, k} = u_i$
for each $i \in \N$. 
So:
Category:Monotone Convergence Theorem (Measure Theory)