Monotone Convergence Theorem

Given a sequence of functions $\{f_n\}$ that 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 that 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 that uses the MCT. And next week we'll cover the DCT.

Monotone Convergence 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 $$\lim_{n\to\infty}\int_X f_n=\int_X f.$$  

Let's look at an example that, on the surface, looks quite nasty. But thanks to the MCT, it's not bad at all.


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
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?


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 that 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)!   


*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.

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