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


by Fredrik
Tags: functions, limit, measurable, sequence
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#1
Feb15-12, 10:47 AM
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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.
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morphism
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Feb15-12, 07:29 PM
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You do have pointwise convergence on A. This is because for each x in A, (f_{n_j}(x)) is a Cauchy sequence of real numbers, hence converges to some real number - call it f(x).
Fredrik
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Feb15-12, 08:12 PM
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Yes, I understand that we have pointwise convergence on A. That enables us to define a function ##f:A\to\mathbb R## in the way you're suggesting. But we're supposed to end up with a measurable function ##f:X\to\mathbb R##, so we must extend that function to X somehow and then prove that the extended function is measurable. Friedman doesn't say how to extend it. I guessed that we just define f(x)=0 for all x not in A. But then we don't have pointwise convergence on X, and we can't use "the first part of corollary 2.2.4" to argue that the function is measurable. (That part says that if ##f_n\to f## pointwise on X, then f is measurable).

I'm thinking that maybe the argument should be that since the complement of A has measure 0, we have ##f_n\to f## a.e., and now the second part of that corollary says that if the measure is complete (in the sense that every subset of a set of measure 0 is measurable), then f is measurable. But I don't see any indication that this theorem only applies to complete measure spaces.

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Feb15-12, 08:17 PM
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The limit of an "almost uniformly Cauchy" sequence of measurable functions


So you wish to prove that if [itex]f_n[/itex] 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 [itex]f_nI_A[/itex] with [itex]I_A(x)=1[/itex] if x in A and 0 if x notin A.
Then we find that [itex]f_nI_A[/itex] converges pointswise on A to f, and thus f is measurable as pointswise limit.
Fredrik
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Feb15-12, 08:26 PM
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Aaaah...##f_n\to f## pointwise on A implies that ##f_n\chi_A\to f\chi_A=f## pointwise on X, and since the product of two measurable functions is measurable, f is a pointwise limit of measurable functions. That sort of trick didn't occur to me at all. Thank you very much.


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