# The Unapologetic Mathematician

## Convergence Almost Everywhere

Okay, so let’s take our idea of almost everywhere and apply it to convergence of sequences of measurable functions.

Given a sequence $\{f_n\}_{n=1}^\infty$ of extended real-valued functions on a measure space $X$, we say that $f_n$ converges a.e. to the function $f$ if there is a set $E_0\subseteq X$ with $\mu(E_0)=0$ so that $\lim\limits_{n\to\infty}f_n(x)=f(x)$ for all $x\in{E_0}^c$. Similarly, we say that the sequence $f_n$ is Cauchy a.e. if there exists a set $E_0$ of measure zero so that $\{f_n(x)\}$ is a Cauchy sequence of real numbers for all $x\in{E_0}^c$. That is, given $x\notin E_0$ and $\epsilon>0$ there is some natural number $N$ depending on $x$ and $\epsilon$ so that whenever $m,n\geq N$ we have $\lvert f_m(x)-f_n(x)\rvert<\epsilon$

Because the real numbers $\mathbb{R}$ form a complete metric space, being Cauchy and being convergent are equivalent — a sequence of finite real numbers is convergent if and only if it is Cauchy, and a similar thing happens here. If a sequence of finite-valued functions is convergent a.e., then $\{f_n(x)\}$ converges to $f(x)$ away from a set of measure zero. Each of these sequences $\{f_n(x)\}$ is thus Cauchy, and so $\{f_n\}$ is Cauchy almost everywhere. On the other hand, if $\{f_n\}$ is Cauchy a.e. then the sequences $\{f_n(x)\}$ are Cauchy away from a set of measure zero, and these sequences then converge.

We can also define what it means for a sequence of functions to converge uniformly almost everywhere. That is, there is some set $E_0$ of measure zero so that for every $\epsilon>0$ we can find a natural number $N$ so that for all $n\geq N$ and $x\notin E_0$ we have $\lvert f_n(x)-f(x)\rvert<\epsilon$. The uniformity means that $N$ is independent of $x\in{E_0}^c$, but if we choose a different negligible $E_0$ we may have to choose different values of $N$ to get the desired control on the sequence.

As it happens, the topology defined by uniform a.e. convergence comes from a norm: the essential supremum; using this notion of convergence makes the algebra of essentially bounded measurable functions on a measure space $X$ into a normed vector space. Indeed, we can check what it means for a sequence of functions $\{f_n\}$ to converge to $f$ under the essential supremum norm — for any $\epsilon>0$ there is some $N$ so that for all $n\geq N$ we have $\text{ess sup}(f_n-f)<\epsilon$. Unpacking the definition of the essential supremum, this means that there is some measurable set $E_0$ with measure zero so that $\lvert f_n(x)-f(x)\rvert<\epsilon$ for all $x\notin E_0$, which is exactly what we said for uniform a.e. convergence above.

We can also turn around and define what it means for a sequence to be uniformly Cauchy almost everywhere — for any $\epsilon>0$ there is some $N$ so that for all $m,n\geq N$ we have $\text{ess sup}(f_m-f_n)<\epsilon$. Unpacking again, there is some measurable set $E_0$ so that $\lvert f_m(x)-f_n(x)\rvert<\epsilon$ for all $x\notin E_0$. It’s straightforward to check that a sequence that converges uniformly a.e. is uniformly Cauchy a.e., and vice versa. That is, the topology defined by the essential supremum norm is complete, and the algebra of essentially bounded measurable functions on a measure space $X$ is a Banach space.

May 14, 2010