The Unapologetic Mathematician

Almost Uniform Convergence

From the conclusion of Egoroff’s Theorem we draw a new kind of convergence. We say that a sequence $\{f_n\}$ of a.e. finite-valued measurable functions converges to the measurable function $f$ “almost uniformly” if for every $\epsilon>0$ there is a measurable set $F$ with $\mu(F)<\epsilon$ so that the sequence $\{f_n\}$ converges uniformly to $f$ on $E\setminus F$.

We have to caution ourselves that this is not almost everywhere uniform convergence, which would be a sequence that converges uniformly once we cut out a subset of measure zero. Our example sequence $f_n(x)=x^n$ yesterday converges almost uniformly to the constant zero function, but it doesn’t converge almost everywhere uniformly. Maybe “nearly uniform” would be better, but “almost uniformly” has become standard.

Now we can restate Egoroff’s Theorem to say that on a set of finite measure, a.e. convergence implies almost uniform convergence. Conversely, if $\{f_n\}$ is a sequence of functions (on any measure space) that converges to $f$ almost uniformly, then it converges pointwise almost everywhere.

We start by picking a set $F_m$ for every positive integer $m$ so that $\mu(F_m)<\frac{1}{m}$, and so that $\{f_n\}$ converges uniformly to $f$ on ${F_m}^c$. We set

$\displaystyle F=\bigcap\limits_{m=1}^\infty F_m$

and conclude that $\mu(F)=0$, since $\mu(F)\leq\mu(F_m)<\frac{1}{m}$ for all $m$. If $x\in F^c$, then there must be some $m$ for which $x\in{F_m}^c$. Since $\{f_n\}$ converges to $f$ uniformly on ${F_m}^c$, we conclude that $\{f_n(x)\}$ converges to $f(x)$. Thus $\{f_n\}$ converges pointwise to $f$ except on the set $F$ of measure zero.

May 18, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

1. […] strengthen it slightly by removing the finiteness assumption, but changing from a.e. convergence to almost uniform convergence: almost uniform convergence implies convergence in measure. Indeed, if converges to almost […]

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