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 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: