The Unapologetic Mathematician

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.