The Unapologetic Mathematician

Mathematics for the interested outsider

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

   

Follow

Get every new post delivered to your Inbox.

Join 393 other followers