The Unapologetic Mathematician

Lebesgue’s Dominated Convergence Theorem

The dominated convergence theorem provides a nice tool to make sure certain sequences of integrable functions converge (in the mean) to integrable limits. Yes, we have the definition and the characterization in terms of convergence in measure, but this theorem is often easier to apply.

If $\{f_n\}$ is a sequence of integrable functions converging in measure or converging a.e. to a function $f$, and if $g$ is an integrable function which “dominates” the sequence $\{f_n\}$ — we have $\lvert f_n(x)\rvert\leq g(x)$ for almost all $x$ — then $f$ is integrable and the sequence $\{f_n\}$ converges to $f$ in the mean. It’s important to note that we do not assume that $f$ is integrable; that’s part of the conclusion.

If we start by assuming that $\{f_n\}$ converges in measure to $f$, then yesterday’s result immediately tells us that $\{f_n\}$ converges in the mean to $f$; the uniformity assumptions from that theorem are consequences of the inequalities

$\displaystyle\int\limits_E\lvert f\rvert\,d\mu\leq\int g\,d\mu$

Now, if we only assume that $\{f_n\}$ converges a.e. to $f$, then we can reduce to convergence in measure by using $g$. By throwing out a set of measure zero, we will assume that $\lvert f_n(x)\rvert$ and $\lvert f(x)\rvert$ are all no more than $g(x)$ everywhere. Then for every fixed positive $\epsilon$ we can write

$\displaystyle E_n=\bigcup\limits_{i=n}^\infty\left\{x\in X\big\vert\lvert f_i(x)-f(x)\rvert\geq\epsilon\right\}\subseteq\left\{x\in X\bigg\vert\lvert g(x)\rvert\geq\frac{\epsilon}{2}\right\}$

but the measure of this latter set is finite, and so $\mu(E_n)<\infty$ for all $n$. A.e. convergence tells us that the measure of the intersection of all the $E_n$ is $0$. By continuity, we conclude that

$\displaystyle\limsup\limits_{n\to\infty}\mu\left(\left\{x\in X\big\vert\lvert f_n(x)-f(x)\rvert\geq\epsilon\right\}\right)\leq\lim\limits_{n\to\infty}\mu(E_n)=\mu\left(\lim\limits_{n\to\infty}E_n\right)=0$

That is, in the presence of a dominating function $g$, convergence a.e. implies convergence in measure, and thus implies convergence in mean.

Incidentally, since we know that

$\displaystyle\left\lvert\int f_n\,d\mu-\int f\,d\mu\right\rvert=\left\lvert\int f_n-f\,d\mu\right\rvert\leq\int\lvert f_n-f\rvert\,d\mu$

we can use convergence in the mean to control the right hand side, and thus get control over the left hand side. That is, we find that

$\displaystyle\lim\limits_{n\to\infty}\int f_n\,d\mu=\int\lim\limits_{n\to\infty}f_n\,d\mu$

The dominated convergence theorem shows that the integral and the limit commute so long as the sequence is dominated by some integrable function.

It should be noted that the dominating function is essential. Indeed, let $X$ be the closed unit interval with Lebesgue measure, and let $E_n=\left(0,\frac{1}{n}\right)$. Now we consider the sequence $\{n\chi_{E_n}\}$ which converges in measure to $0$. However, there is no integrable dominating function; we find that

$\displaystyle\int\lvert n\chi_{E_n}\rvert\,d\mu=1$

and so the sequence cannot converge in the mean to $0$.

June 10, 2010 - Posted by | Analysis, Measure Theory

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3. Gosh, I passed my analysis prelim in 1986 and haven’t looked at this since then! :)

Still, it makes much more sense to me now that I have some experience to put things into context.

Comment by blueollie | June 15, 2010 | Reply

4. Doesn’t it, though?

Yeah, I just scraped by my analysis quals. That’s part of why I’m using this series as an excuse to go back through and work through it all now that I might understand it better.

Comment by John Armstrong | June 15, 2010 | Reply

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9. if there exist a fns say m(x) in lebesgue dominated thm. Is fns greater than the measurable sequence. Try work it out

Comment by jackson akhigbe | December 16, 2010 | Reply