# Dominated Convergence Theorem

/*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?

*not*dominated by

*any*function. Take, for instance, the sequence of functions $\{f_n\}$ where for each $n\in\mathbb{N}$ we define $$f_n(x)=n\chi_{(0,1/n]}(x)=\begin{cases} n, &\text{if $0< x\leq \frac{1}{n} $}\\ 0, &\text{else.} \end{cases}$$

*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

*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 $x$.

*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 $fg$.