# Monotone Convergence Theorem

/Given a sequence of functions $\{f_n\}$ which converges pointwise to some limit function $f$, it is

*not*always true that $$\int \lim_{n\to\infty}f_n = \lim_{n\to\infty}\int f_n.$$ (Take this sequence for example.) The**Monotone Convergence Theorem**(MCT), the**Dominated Convergence Theorem**(DCT), and**Fatou's Lemma**are three major results in the theory of Lebesgue integration which answer the question "When do $\displaystyle{ \lim_{n\to\infty} }$ and $\int$ commute?" The MCT and DCT tell us that*if*you place certain restrictions on both the $f_n$ and $f$, then you can go ahead and interchange the limit and integral. Fatou's Lemma, on the other hand, says "Here's the best you can do if you*don't*make any extra assumptions about the functions."Last week we discussed Fatou's Lemma. Today we'll look at an example which uses the MCT. And next week we'll cover the DCT.

**Monotone Converegence Theorem:**If $\{f_n:X\to[0,\infty)\}$ is a sequence of measurable functions on a measurable set $X$ such that $f_n\to f$ pointwise almost everywhere and $f_1\leq f_2\leq \cdots$, then $$\lim_{n\to\infty}\int_X f_n=\int_X f.$$

In this statement the $f_n$ are nondecreasing, but the theorem holds for a nonincreasing sequence as well. Let's look at an example which, on the surface, looks quite nasty. But thanks to the MCT, it's not bad at all.

## Example

Let $X$ be a measure space with a positive measure $\mu$ and let $f:X\to[0,\infty]$ be a measurable function. Prove that
$$\lim_{n\to\infty}\int_X n\log\left(1+\frac{f}{n}\right)d\mu \;=\;\int_X f\;d\mu.$$

*Proof.*Begin by defining \begin{align*} f_n&=n\log\left(1+\frac{f}{n}\right)\\ &=\log\left(1+\frac{f}{n}\right)^n \end{align*} and note that each $f_n$ is nonnegative (since both $\log$ and $f$ are nonnegative) and measurable (since the composition of a continuous function with a measurable function is measurable). Further $f_1\leq f_2\leq\cdots$. Indeed, $\log$ is an increasing function and for a fixed $x\in X$ the sequence $\left(1+\frac{f(x)}{n}\right)^n$ is increasing. In fact*, it increases to $e^{f(x)}$. In other words, $$\lim_{n\to\infty}f_n(x)=\lim_{n\to\infty}\log\left(1+\frac{f}{n}\right)^n=\log e^{f(x)}=f(x).$$ Hence, by the Monotone Convergence Theorem $$\lim_{n\to\infty}\int_X f_n\;d\mu=\int_x f\;d\mu$$ as desired.

*Not so bad, huh?*

## Remark

Did you know that the MCT has a "continuous cousin"? (Well, maybe it's more like a second cousin.) Have you come across Dini's Theorem before?

**Dini's Theorem:**If $\{f_n:X\to\mathbb{R}\}$ is a nondecreasing sequence of continuous functions on a compact metric space $X$ such that $f_n\to f$ pointwise to a continuous function $f:X\to\mathbb{R}$, then the convergence is uniform.

Here we have a monotone sequence of

*continuous*- instead of measurable - functions which converge pointwise to a limit function $f$ on a compact metric space. By Dini's Theorem, the convergence is actually uniform. So IF the $f_n$ are also Riemann integrable, then we can conclude** $$\lim_{n\to\infty}\int_Xf_n=\int_Xf.$$ Perhaps this doesn't surprise us too much: we've seen before that continuity and measurability are analogous notions (to a certain extent)!*Footnotes** Recall from elementary calculus: $\displaystyle{\lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n=e^x}$ for any $x\in\mathbb{R}$. ** See Rudin's

*Principles of Mathematical Analysis*(3ed.), Theorem 7.16.