Dominated 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."

Today we're discussing the Dominated Convergence Theorem. First we'll look at a counterexample to see why "domination" is a necessary condition, and we'll close by using the DCT to compute $$\lim_{n\to\infty}\int_{\mathbb{R}}\frac{n\sin(x/n)}{x(x^2+1)}.$$

 
 
 

The Dominated Convergence Theorem: If $\{f_n:\mathbb{R}\to\mathbb{R}\}$ is a sequence of measurable functions which converge pointwise almost everywhere to $f$, and if there exists an integrable function $g$ such that $|f_n(x)|\leq g(x)$ for all $n$ and for all $x$, then $f$ is integrable and $$\int_{\mathbb{R}}f=\lim_{n\to\infty}\int_\mathbb{R} f_n.$$

 
 

Why is domination necessary?

Let's see where things can go wrong if a sequence $\{f_n\}$ is not dominated by any function. Take, for instance, the sequence of functions $\{f_n\}$ where for each $n\in\mathbb{N}$ we define $$f_n(x)=n\chi_{(0,1/n]}(x)=\begin{cases} n, &\text{if $0< x\leq \frac{1}{n} $}\\ 0, &\text{else.} \end{cases}$$
 
Then $f_n\to 0$ pointwise as is evident by looking at graphs of the first few $f_n$. (We've also discussed this sequence before.) But notice there is no function $g$ such that $|f_n(x)|\leq g(x)$ for $x\in(0,1]$ and for ALL $n$. This is because for large values of $n$, the height of $f_n$ is tending towards infinity, or in other words, the $f_n$ are unbounded. Since the area of each rectangle is 1, we see that the integral and the limit do not commute in this example. Explicitly: $$1=\lim_{n\to\infty}\int_0^1 f_n(x)\;dx\quad \neq\quad \int_0^1\lim_{n\to\infty}f_n(x)\;dx=0 $$ where we have use the fact that $\displaystyle{\lim_{n\to\infty}f_n(x)=0}$ and that the Riemann and Lebesgue integral coincide in this case.
 

An example using the DCT

Compute the following integral: $$\lim_{n\to\infty}\int_{\mathbb{R}} \frac{n\sin(x/n)}{x(x^2+1)}\;dx.$$

Solution. Let $x\in\mathbb{R}$ and begin by defining $$ f_n(x)=\frac{n\sin(x/n)}{x(x^2+1)}\qquad \text{for each $n\in\mathbb{N}$.}$$ Observe that each $f_n$ is measurable* and the sequence $\{f_n\}$ converges pointwise to $\frac{1}{1+x^2}$ for all $x\neq 0$: \begin{align*} \lim_{n\to\infty} f_n(x) &=\lim_{n\to\infty} \left(\frac{\sin(x/n)}{x/n}\right)\frac{1}{1+x^2}\\ &=\frac{1}{1+x^2} \end{align*} since $\displaystyle{ \lim_{n\to\infty} \frac{\sin(x/n)}{x/n} }=1$ for a fixed $x$.

From this we also see that $g(x)=\frac{1}{1+x^2}$ works as a dominating function. Indeed, $g$ is integrable on $\mathbb{R}$ (as we'll verify below) and \begin{align*} |f_n(x)|&=\bigg|\frac{\sin(x/n)}{x/n}\cdot\frac{1}{1+x^2} \bigg|\\ &=\frac{|\sin(x/n)|}{|x/n|}\cdot\frac{1}{1+x^2}\\ &\leq \frac{1}{1+x^2}\\ &=g(x) \end{align*} since $|\sin(x)|\leq |x|$ for all $x$. Thus we apply the DCT to conclude \begin{align*} \lim_{n\to\infty}\int_{\mathbb{R}} \frac{n\sin(x/n)}{x(x^2+1)}\;dx &= \int_{-\infty}^{\infty} \frac{n\sin(x/n)}{x(x^2+1)}\;dx\\[4pt] &=\int_{-\infty}^{\infty}\frac{1}{1+x^2}\;dx\\[4pt] &=\tan^{-1}(x)\bigg|_{-\infty}^{\infty}\\[4pt] &=\pi. \end{align*}
 

 
Footnote:

* If $g$ is a continuous function and $f$ is a measurable function, then their composition $g\circ f$ is measurable. And if $g$ and $f$ are both measurable, then so is $fg$.