## Egoroff’s Theorem

Let’s look back at what goes wrong when a sequence of functions doesn’t converge uniformly. Let be the closed unit interval , and let . Pointwise, this converges to a function with for , and . This convergence can’t be uniform, because the uniform limit of a sequence of continuous functions is continuous.

But things only go wrong at the one point, and the singleton has measure zero. That is, the sequence converges almost everywhere to the function with constant value . The convergence still isn’t uniform, though, because we still have a problem at . But if we cut out any open patch and only look at the interval , the convergence *is* uniform. We might think that this is “uniform a.e.”, but we have to cut out a set of positive measure to make it work. The set can be as small as we want, but we can’t get uniformity by just cutting out .

However, what we’ve seen is a general phenomenon expressed in Egoroff’s Theorem: If is a measurable set of finite measure, and if is a sequence of a.e. finite-valued measurable functions converging a.e. on to a finite-valued measurable function , then for every there is a measurable subset with so that converges uniformly to on . That is, if we have a.e. convergence we can get to uniform convergence by cutting out an arbitrarily small part of our domain.

First off, we cut out a set of measure zero from so that converges pointwise to . Now we define the measurable sets

As gets bigger, we’re taking the intersection of fewer and fewer sets, and so . Since converges pointwise to , eventually the difference gets down below every , and so for every . Thus we conclude that . And so for every there is an so that

Now let’s define

This is a measurable set contained in , and monotonicity tells us that

We can calculate

And so given any we take . Then for any we have , and thus . Since we can pick this independently of , the convergence on is uniform.