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I read the same proof that almost uniform convergence implies convergence almost everywhere on several sources (Friedman's Foundations of Modern Analysis and online sources), and they all seem to use the same proof:

Proof taken from Proofwiki, https://proofwiki.org/wiki/Convergence_a.u._Implies_Convergence_a.e.

However, I have a problem with this proof. While the set on which $f_n$ does not converge to $f$ is definitely a subset of $B$, I do not see why the reverse inclusion (that $f_n$ does not converge to $f$ for every element of $B$) is true. Hence, might it not be possible that $f_n$ converges to $f$only on a proper subset of $B$ that just happens to be non-measurable? (Which is possible since the measure space is not specified to be complete.) Then the proof would be false.

I would really appreciate help in understanding why my critique of the proof does not hold. Thanks in advance.

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  • $\begingroup$ The proof doesn't give any information about where the sequence of functions doesn't converge, except that this bad behavior lies inside a set of measure zero. Sure, there can be $x\in B$ with $f_n(x)\to f(x)$, but that information isn't needed to show that $f\to f$ pointwise a.e. $\endgroup$ – Aweygan Jan 5 '17 at 18:16
  • $\begingroup$ @Aweygan, yes I agree that that is exactly what the proof tells us. However, I don't understand your statement that the information isn't needed to show that $f_n→f$ pointwise a.e. Isn't the definition of pointwise a.e. precisely that $f_n→f$ except on a set of measure 0? $\endgroup$ – Wayne Lin Jan 5 '17 at 18:19
  • $\begingroup$ I think I understand. So the question can summarized as: Is the set of points $x$ where $f_n(x)\to f(x)$ measurable? If so, your critique would be unnecessary. If not, then your critique is certainly valid. $\endgroup$ – Aweygan Jan 5 '17 at 18:25
  • $\begingroup$ Yes, @Aweygan, that is exactly my critique. (: I believe the proof has a gap due to its assumption that the set of points $x$ where $f_n$ does not converge to $f$ is measurable. $\endgroup$ – Wayne Lin Jan 5 '17 at 18:26
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Actually if $(f_n)_{n \in \mathbf{N}}$ is a sequence of measurable functions to a complete space, then the set of convergence points of $(f_n)$ is always measurable. In your example, $f_n : D \to \mathbf{R}$ and $\mathbf{R}$ is a complete space.

$$\{x \in D \; \vert \; (f_n(x)) \text{ converges}\}=\bigcap_{\epsilon \in \mathbf{Q_+^*}}\bigcup_{N \in \mathbf{N}}\bigcap_{n,m\ge N}\{x \in D \; \vert \; |f_n(x)-f_m(x)| \le \epsilon\}$$

And $\{x \in D \; \vert \; |f_n(x)-f_m(x)| \le \epsilon\}=|f_n-f_m|^{-1}([0,\epsilon])$ is measurable because $|f_n-f_m|$ is still measurable.

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  • $\begingroup$ Thank you for your reply. I do not understand the significance of the (Cauchy?) completion of the codomain though: do you have a source or proof for the theorem you mentioned? I was interested in the completion of the measure, because that would mean any subset of a measure 0 set is measurable, and hence the proof would be true if the measure were complete. $\endgroup$ – Wayne Lin Jan 5 '17 at 18:35
  • $\begingroup$ @WayneLin: Oh ok, I understand what you meant. Yes, the completion of the measure was enough, but here I mean indeed the Cauchy completion of the codomain. $\endgroup$ – md5 Jan 5 '17 at 18:36
  • $\begingroup$ @WayneLin: I included a short proof in my answer. For more detailed explanations you may look here for instance. $\endgroup$ – md5 Jan 5 '17 at 18:45
  • $\begingroup$ Oh okay, I understand the proof in your answer. Thanks so much! I never knew about this property, and wouldn't have known without your answer. $\endgroup$ – Wayne Lin Jan 5 '17 at 18:57
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This answer seeks to address your confusion over the definition of "almost everywhere" (a.e.). According to Rudin - Real and Complex Analysis, something occurs a.e. if the set where it doesn't occur is a subset of a set of measure 0. If the measure isn't complete the actual set on which it doesn't occur could be non-measurable. This fact is used, for example, in Rudin's book in Chapter 8 when proving that (under certain conditions) if the projection of a subset of $U \subset X \times Y$ onto $X$ has measure 0 for almost all $y \in Y$, $U$ has measure 0.

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  • $\begingroup$ Thank you for pointing me to that definition. It is unlike the definition in Friedman - Foundations of Modern Analysis, however, which says in Ch 2.2: "A property $P$ concerning the points $x$ of the measure space $X$ is said to be true a.e. if the set $E$ for which $P$ is not true has measure zero." Rudin's does seems a more useful definition of a.e., though. Do you know if one definition is more widely accepted than the other? $\endgroup$ – Wayne Lin Jan 5 '17 at 19:01
  • $\begingroup$ @WayneLin I really have no idea. However, Wikipedia agrees with Rudin: "In cases where the measure is not complete, it is sufficient that the set is contained within a set of measure zero." en.wikipedia.org/wiki/Almost_everywhere $\endgroup$ – Reinstate Monica Jan 5 '17 at 19:23
  • $\begingroup$ Okay I will use that definition going forward: it makes much more sense. Thank you for looking it up! $\endgroup$ – Wayne Lin Jan 5 '17 at 19:32

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