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 is a sequence of integrable functions converging in measure or converging a.e. to a function , and if is an integrable function which “dominates” the sequence — we have for almost all — then is integrable and the sequence converges to in the mean. It’s important to note that we do not assume that is integrable; that’s part of the conclusion.
If we start by assuming that converges in measure to , then yesterday’s result immediately tells us that converges in the mean to ; the uniformity assumptions from that theorem are consequences of the inequalities
Now, if we only assume that converges a.e. to , then we can reduce to convergence in measure by using . By throwing out a set of measure zero, we will assume that and are all no more than everywhere. Then for every fixed positive we can write
but the measure of this latter set is finite, and so for all . A.e. convergence tells us that the measure of the intersection of all the is . By continuity, we conclude that
That is, in the presence of a dominating function , convergence a.e. implies convergence in measure, and thus implies convergence in mean.
Incidentally, since we know that
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
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 be the closed unit interval with Lebesgue measure, and let . Now we consider the sequence which converges in measure to . However, there is no integrable dominating function; we find that
and so the sequence cannot converge in the mean to .