# The limit of an almost uniformly Cauchy sequence of measurable functions

by Fredrik
Tags: functions, limit, measurable, sequence
 Emeritus Sci Advisor PF Gold P: 9,356 I'm trying to understand the proof of theorem 2.4.3 in Friedman. I don't understand why f must be measurable. The "first part" of the corollary he's referring to says nothing more than that a pointwise limit of a sequence of measurable functions is measurable. But it seems to me that f is just an almost uniform limit, not a pointwise limit. Almost uniform convergence implies convergence a.e, but not (everywhere) pointwise convergence. This is my (possibly incorrect) interpretation of what we're doing. We find that there's a subsequence that's "almost uniformly Cauchy" in the sense I'm about to describe, so now the goal is to use that sequence to define a new function, and then we prove that it's measurable. Let's try this for an arbitrary sequence ##\langle g_n\rangle## of measurable functions that 's "almost uniformly Cauchy" in the sense that for all ##\varepsilon>0##, there's a measurable set ##E## such that ##\mu(E)<\varepsilon## and ##\langle g_n(x)\rangle## is uniformly Cauchy on ##E^c##. What I mean by that is that there's an ##N\in\mathbb Z^+## such that for all ##n,m\in\mathbb Z^+## and all ##x\in E^c##, $$n,m\geq N\ \Rightarrow\ |g_n(x)-g_m(x)|<\varepsilon.$$ The assumption that ##\langle g_n\rangle## is almost uniformly Cauchy implies that for all ##k\in\mathbb Z^+##, there's a measurable set ##E_k## such that ##\mu(E_k)<\frac{1}{k}## and ##\langle g_n(x)\rangle## is uniformly Cauchy on ##E^c##. This allows us to define a function ##g:X\to\mathbb R## by $$g(x)= \begin{cases} \lim_n g_n(x) &\text{ if }x\in\bigcup_{k=1}^\infty E_k{}^c\\ 0 &\text{ if }x\notin\bigcup_{k=1}^\infty E_k{}^c. \end{cases}$$ (How else can we define g?) Then we're supposed to be able to show that this g is measurable. Is this what Friedman has in mind? Since $$\mu\bigg(\bigcap_{k=1}^\infty E_k\bigg) \leq\mu(E_k)<\frac{1}{k}$$ for all k, we have almost uniform convergence, but not (everywhere) pointwise convergence.
 Mentor P: 18,244 The limit of an almost uniformly Cauchy sequence of measurable functions So you wish to prove that if $f_n$ converges on A, and if we define [/itex]\lim f_n=f[/itex] on A and f=0 elsewhere, that we have a measurable function? This can be easily done by looking at $f_nI_A$ with $I_A(x)=1$ if x in A and 0 if x notin A. Then we find that $f_nI_A$ converges pointswise on A to f, and thus f is measurable as pointswise limit.