Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

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

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

add comment

1 Answer 1

up vote 4 down vote accepted

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.

share|improve this answer
    
You're not really using that the domain is the set of reals you can do this on any measurable space. You can also replace the range by any complete metric space by observing that a sequence converges if and only if it is Cauchy using the argument I gave in the answer in the thread I marked as a duplicate. –  t.b. Jul 1 '12 at 4:23
    
Thanks for the advice! –  Potato Jul 1 '12 at 4:24
add comment

Not the answer you're looking for? Browse other questions tagged or ask your own question.