monotone convergence theorem (integral version)

This is a way to interchange the limit of an integral with the integral of the limit (under certain conditions, since that doesn’t always work).

Theorem

For a measure space $(X,\mathcal{S},\mu )$ and $0\le f_1\le f_2\cdots$ is an increasing sequence of measurable functions, let $f:X\to [0,\infty ]$ such that

$f(x)={\lim }_{k\to \infty }{f}_k(x)$

Then

${\lim }_{k\to \infty }\int f_{k}d\mu =\int fd\mu =\int {\lim }_{k\to \infty }{f}_k(x)d\mu$

Proof

(Before anything else, we know that the function $f$ is measurable)

First, prove ${\lim }_{k\to \infty }\int f_{k}d\mu \le \int fd\mu$. Each $f_k\le f$ thus since Lebesgue integral preserves order ${\lim }_{k\to \infty }\int f_{k}d\mu \le \int fd\mu$.

Next, prove ${\lim }_{k\to \infty }\int f_{k}d\mu \ge \int fd\mu$.

We need a little space, so first let $0\le \varphi \le f$ be a simple measurable function. $\varphi =\sum c_i{\chi }_{A_i}{A}_i\in \mathcal{S}\text{disjoint}$

Choose $\alpha \in (0,1)$ and consider the set $E_k=\{x\in X|f_k(x)\ge \alpha \varphi (x)\}$

Since $f_k\le f_{k+1}$ then $E_k\subseteq E_{k+1}$ so it’s an increasing sequence of sets and $\cup E_k=X$ since we “shrank” the function so $\alpha \varphi (x)\le f$.

So we use these constructed sets $E_k$ along with the sets $A_i$ from $\varphi$.

$A_i=X\cap A_i={\bigcup }_{k=1}^{\infty }(E_k\cap A_i)$

This lets us use these in $\varphi$

$$\begin{array}{rl}{f}_k\ge f_k{\chi }_{E_k}\ge \alpha \varphi {\chi }_{E_k}& =\alpha \sum c_i{\chi }_{A_i}{\chi }_{E_k}\\ & =\alpha \sum c_i{\chi }_{A_i}\cap E_k\end{array}$$

We need these $E_k$ which is an increasing sequence of sets to move the limit inside the measure. Integrating the above gives \begin{equation} \int f_k \geq \int \alpha \sum c_i\chi_{A_i \cap E_k} = \alpha \sum c_i\mu(A_i \cap E_k) \end{equation}

By the continuity of measure, along with the sets $E_k$

$\mu (A_i)=\mu ({\lim }_{k\to \infty }(E_k\cap A_i))={\lim }_{k\to \infty }\mu (E_k\cap A_i)$

So the limit plays nice in equation (1), so we take the limit, \begin{equation} \lim_{k\to\infty} \int f_k \geq \lim_{k\to\infty} \alpha \sum c_i\mu(A_i \cap E_k) = \alpha \sum c_i\mu(A_i) \end{equation}

Now take the supremum over all the partitions (aka let the simple function become a better and better approximation)

${\lim }_{k\to \infty }\int f_k\ge \alpha {\sup }_{0\le \varphi \le f}\int \varphi d\mu =\alpha \int fd\mu$

Then just let $\alpha$ get arbitrarily close to 1. And with that we have both sides so ${\lim }_{k\to \infty }\int f_{k}d\mu =\int {\lim }_{k\to \infty }{f}_{k}d\mu$