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I came across the following problem in my study of measurable functions:

If $\{f_n\}$ is a sequence of measurable functions on $X$, then $\{x : \lim_{n \to \infty}f_n(x) \mbox{ exists}\}$ is measurable.

I know how to prove this result if we assume the $f_n$ are $\overline{\mathbb{R}}$-valued, since there are particularly nice results about the limit inferior and the limit superior of such functions in this case. I suspect these (lim sup and lim inf) will factor into the more general case, but I was wondering how the proof would proceed!

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HINT (may not be the best approach): A sequence of reals converges to a point in $\mathbb{R}$ iff it is Cauchy. –  Srivatsan Feb 1 '12 at 5:37
    
Another way: If $E$ is the set where the extended limit exists, you just have to intersect it with the set of points $x$ where $(f_n(x))$ is bounded: $B=\bigcup_N\bigcap_n f_n^{-1}([-N,N])$. –  Michael Greinecker Feb 1 '12 at 7:09

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up vote 3 down vote accepted

Expanding on Srivastans comment: Let $C$ be the set of points $x$ where $(f_n(x))$ converges. This is the case if and only if $(f_n(x))$ is Cauchy: $$C=\big\{x:\forall\epsilon>0 \exists N\forall m> N\forall n>N:|f_n(x)-f_m(x)|<\epsilon\big\}$$ $$C=\big\{x:\forall k \exists N\forall m>N\forall n>N:|f_n(x)-f_m(x)|<1/k\big\}$$ $$C=\bigcap_{k=1}^\infty\bigcup_{N=1}^\infty\bigcap_{n=N+1}^\infty\bigcap_{m=N+1}^\infty\{x:|f_n(x)-f_m(x)|<1/k\}$$

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You seem to be comfortable with the idea that E = {x: $\lim_{n \to \infty} f_n (x)$ exists in the extended sense} is measurable so let's bootstrap off of that. Define a function $g(x) = \lim_{n\to \infty} f_n (x)$ where that limit exists in the extended sense and $g(x) = -\infty$ otherwise. Write $g_n$ = $f_n \chi_E - \infty \chi_{E^c}$. By construction $g = \limsup g_n$ and thus $g$ is measurable.

The set {x: $\lim_n f_n (x)$ exists and is real valued} is by construction $g^{-1}(\mathbb{R})$ which is measurable by the measurability of g.

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