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Let $(X,\mu)$ be a measure space and $E$ a measurable subset of $X$. We say that, for real valued functions, $\{f_n\}$ converges in measure to $f$ on $E$ so long as $\{f_n\}$ and $f$ are measurable functions and $\forall\ \eta>0$, $\ \lim_{n\to\infty}\mu(\{x\in E\ :\ |f_n(x)-f(x)|>\eta\})=0$. It can be shown that if $\mu(E)<\infty$, then convergence in measure on $E$ is equivalent to the following statement: $$\textrm{Every subsequence of }\{f_n\}\textrm{ has a further subsequence that converges pointwise a.e. to } f\textrm{ on } E$$ However, I have recently discovered that the following statements are equivalent: $$\{x_n\}\textrm{ is a sequence of real numbers that converges to } x$$ and $$\textrm{Every subsequence of }\{x_n\}\textrm{ has a further subsequence which converges to }x.$$

I'm afraid I must be missing something obvious, but, from what I have above, it seems that convergence in measure and pointwise a.e. convergence are the same thing. I know this is not the case since I also have a counter example that works on the set $[0,1]$ with Lebesgue measure. I reach the conclusion by recalling that pointwise a.e. convergence on $E$ is, for every $x\in E$ except for a set of measure 0, convergence of the sequence of real numbers $\{x_n\}$, defined by $x_n=f_n(x)$, to the number $f(x)$. Where is my logic flawed?

Edit: I believe I have figured out where my logic is flawed, but I'd like to see if another finds the same flaw before I answer my own question.

Here is a link to show the second equivalency. The first equivalency is an exercise in Royden and Fitzpatrick's Real analysis.

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    $\begingroup$ It is worth mentioning that in any Hausrorff topological space, it is true that $x_n \to x$ if and only if every subsequence of $x_n$ has a further subsequence converging to $x$. Needless to say, convergence almost everywhere is not topologizable in general. $\endgroup$ – Shalop Aug 3 '16 at 1:31
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    $\begingroup$ Here is where your logic is flawed: The problem lies in the fact that there are uncountably many subsequences of $f_n$, and an uncountable union of measure zero sets need not have measure zero. $\endgroup$ – Shalop Aug 3 '16 at 1:53
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One direction is indeed true. In general convergence almost everywhere of $f_n \to f$ implies that every subsequence of $f_n$ has a further subsequence converging a.e. to $f$.

It is the converse that is false. Indeed, the problem lies in the fact that there are uncountably many subsequences of $f_n$, and an uncountable union of measure zero sets need not have measure zero! Or equivalently, an uncountable intersection of sets of full measure need not have full measure.

Your logic is presumably that if each subsequence of $f_n$ has a further subsequence converging to $f$ almost everywhere, then there exists a set $A$ whose complement has measure zero and such that for each $x \in A$, every subsequence of $f_n(x)$ has a further subsequence converging to $f(x)$. However, this is entirely false.

Indeed, we can try to build a "false proof" of the above fact as follows. Let $\Sigma$ denote the set of all increasing functions $\sigma: \Bbb N \to \Bbb N$, so that $\Sigma$ indexes all of the possible subsequences of $f_n$. Given a subsequence $(f_{\sigma(k)})$ there exists a set $A_{\sigma}$ whose complement has measure zero and such that $f_{\sigma(k)} \to f$ on $A_{\sigma}$, after passing to a further subsequence. Let $A = \bigcap_{\sigma \in \Sigma} A_{\sigma}$. If $\mu(A^c) = 0$ then the above fact is true. However, since the set $\Sigma$ is uncountable, we do not know whether $\mu(A^c)=0$ and therefore we cannot deduce anything.

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