The Unapologetic Mathematician

Mathematics for the interested outsider

Egoroff’s Theorem

Let’s look back at what goes wrong when a sequence of functions doesn’t converge uniformly. Let X be the closed unit interval \left[0,1\right], and let f_n(x)=x^n. Pointwise, this converges to a function f with f(x)=0 for 0\leq x<1, and f(1)=1. 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 \{1\} has measure zero. That is, the sequence f_n converges almost everywhere to the function with constant value 0. The convergence still isn’t uniform, though, because we still have a problem at \{1\}. But if we cut out any open patch and only look at the interval \left[0,1-\epsilon\right], 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 \{1\}.

However, what we’ve seen is a general phenomenon expressed in Egoroff’s Theorem: If E\subseteq X is a measurable set of finite measure, and if \{f_n\} is a sequence of a.e. finite-valued measurable functions converging a.e. on E to a finite-valued measurable function f, then for every \epsilon>0 there is a measurable subset F with \mu(F)<\epsilon so that \{f_n\} converges uniformly to f on E\setminus F. 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 E so that \{f_n\} converges pointwise to f. Now we define the measurable sets

\displaystyle E_n^m=\bigcap\limits_{i=n}^\infty\left\{x\in X\bigg\vert\lvert f_i(x)-f(x)\rvert<\frac{1}{m}\right\}

As n gets bigger, we’re taking the intersection of fewer and fewer sets, and so E_1^m\subseteq E_2^m\subseteq\dots. Since \{f_n\} converges pointwise to f, eventually the difference \lvert f_i(x)-f(x)\rvert gets down below every \frac{1}{m}, and so \lim_nE_n^m\supseteq E for every m. Thus we conclude that \lim_n\mu(E\setminus E_n^m)=0. And so for every m there is an N(m) so that

\displaystyle\mu(E\setminus E_{N(m)}^m)<\frac{\epsilon}{2^m}

Now let’s define

\displaystyle F=\bigcup\limits_{m=1}^\infty\left(E\setminus E_{N(m)}^n\right)

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

\displaystyle\mu(F)=\mu\left(\bigcup\limits_{m=1}^\infty\left(E\setminus E_{N(m)}^n\right)\right)\leq\sum\limits_{m=1}^\infty\mu\left(E\setminus E_{N(m)}^n\right)<\sum\limits_{m=1}^\infty\frac{\epsilon}{2^m}=\epsilon

We can calculate

\displaystyle E\setminus F=E\cap\bigcap\limits_{m=1}^\infty E_{N(m)}^m

And so given any m we take n\geq N(m). Then for any x\in E\setminus F we have x\in E_n^m, and thus \lvert f_n(x)-f(x)\rvert<\frac{1}{m}. Since we can pick this n independently of x, the convergence on E\setminus F is uniform.

May 17, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

  1. […] Uniform Convergence From the conclusion of Egoroff’s Theorem we draw a new kind of convergence. We say that a sequence of a.e. finite-valued measurable […]

    Pingback by Almost Uniform Convergence « The Unapologetic Mathematician | May 18, 2010 | Reply

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