# Dominated Convergence Theorem

## The Basic Idea

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 $$\lim_{n\to\infty}\int_{\mathbb{R}}\frac{n\sin(x/n)}{x(x^2+1)}.$$

## From English to Math

**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* integrable function $g$ such that $|f_n(x)|\leq g(x)$ for all $x\in(0,1]$ 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 used 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 &= \lim_{n\to\infty}\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 $fg$.