# Monotone Convergence Theorem (Measure Theory)

This proof is about Monotone Convergence Theorem in the context of Measure Theory. For other uses, see Monotone Convergence Theorem.

## 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

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:

$\map {u_i} x \ge \map {u_j} x$ for all $i \le j$

and:

$\ds \map u x = \lim_{n \mathop \to \infty} \map {u_n} x$

for $\mu$-almost all $x \in X$.

Suppose also that $u_1$ is $\mu$-integrable function.

Then $u_n$ is $\mu$-integrable for each $n \in \N$ and $u$ is $\mu$-integrable with:

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int u_n \rd \mu$

## 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 $i \le j$.
$u_i \le u$ 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.
$\ds \sequence {\int u_i \rd \mu}_{i \in \N}$ 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

$(1) \quad$ $\sequence {g_n}_{n \mathop \in \N}$ 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 $n \in \N$.

$\Box$

$\ds \int u \rd \mu = \lim_{n \mathop \to \infty} \int g_n \rd \mu$

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 $n \in \N$

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:

$\ds \int v \rd \mu = \lim_{n \mathop \to \infty} \int v_n \rd \mu$
$\chi_{X \setminus N} = 1$ $\mu$-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.

$\blacksquare$

## 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 by hypothesis:

$\map \mu {X \setminus Y} = 0$

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:

 $\ds \lim_{n \mathop \to \infty} \int u_n \rd \mu$ $=$ $\ds \lim_{n \mathop \to \infty} \int_Y u_n \rd \mu$ Measurable Function Zero A.E. iff Absolute Value has Zero Integral $\ds$ $=$ $\ds \lim_{n \mathop \to \infty} \int f_n \rd \mu$ $\ds$ $=$ $\ds \int \sup_{n \in \N} f_n \rd \mu$ Beppo Levi's Theorem $\ds$ $=$ $\ds \int_Y \sup_{n \in \N} u_n \rd \mu$ Measurable Function Zero A.E. iff Absolute Value has Zero Integral $\ds$ $=$ $\ds \int_Y u \rd \mu$ $\ds$ $=$ $\ds \int u \rd \mu$ Measurable Function Zero A.E. iff Absolute Value has Zero Integral

$\blacksquare$