Set of convergence is measurable. [duplicate]

Possible Duplicate:
pointwise convergence in $\sigma$-algebra

Problem: Prove that the set of points at which a sequence of measurable real functions converges is a measurable set. (I believe the problem means functions from the reals to the reals.)

Source: W. Rudin, Real and Complex Analysis, Chapter 1, exercise 5.

I have posted a proposed solution in the answers.

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marked as duplicate by t.b., Potato, Jonas Meyer, Did, Asaf KaragilaJul 1 '12 at 12:35

Let the sequence of functions be $\{f_n(x)\}$. The $\lim \inf$ and $\lim\sup$ of this sequence of functions are measurable (extended-valued) functions. Denote them $h(x)$ and $g(x)$. The set $A$ where $g$ and $h$ are both positive infinity or both negative infinity is measurable, as they are each measurable functions.
Consider the function $p(x)=h\chi_{\mathbb{R}-A}-g\chi_{\mathbb{R}-A}$. It is zero precisely where the original sequence of functions has a limit. Then $E=p^{-1}(\{0\})$ is measurable, so and $E\cup A$ is measurable, and it is the set of points where the sequence has a limit, so we are done.