It's a common analysis problem to show that if $m(E) < \infty$ to show that convergence in $L^P(E)$ implies convergence in measure. However, I don't see where the necessity of $m(E) < \infty$ comes up in the proof. The proofs I have seen all look like:

Let $A_n = \{x \in E: |f_n(x) - f(x)| > \epsilon \}$. Then,

$\int_E |f_n(x) - f(x)|^p \ge \int_{A_n} |f_n(x) - f(x)|^p \ge \int_{A_n} \epsilon^p = \epsilon^p m(A_n) \ge 0$

Thus $||f_n - f||_p \rightarrow 0$ implies that lim $m(A_n) = 0$ and hence it converges in measure.

Where is the finite measure of $E$ required?

EDIT* Based on the one answer so far, and the comment, I suspect that this was an incorrect question. It came up from a Qualifying exam I was going over. Which would make me want to believe that it is not mistake, but it looks like it must have been a typo of some sort (that would suck to get an impossible question on your exam...). So, the "it's common" was just me making that up. The original question was to prove this for a finite E, and I guess there are just so many similar type questions that specify a finite E I assumed I had always seen it like that for this particular problem. Should I delete this question?

  • $\begingroup$ I am curious where you learned it is "common" to add the unneeded hypothesis that $m(E)<\infty$. $\endgroup$ – Jonas Meyer Aug 15 '13 at 19:31
  • $\begingroup$ Touche, I guess I just made that up! $\endgroup$ – Fractal20 Aug 15 '13 at 20:05
  • $\begingroup$ Fractal20, thanks for clarifying, but it is really nothing to worry about. Regardless of the source, you had a valid question about verifying that finiteness of $m(E)$ is not required for the fact and the presented proof to be correct. $\endgroup$ – Jonas Meyer Aug 15 '13 at 20:14
  • $\begingroup$ I realize this is an old post, but if $p=\infty$ then it is necessary, but otherwise not, so it depends on how you define $p$ $\endgroup$ – f00d Dec 23 '13 at 21:17

It is not required. The argument you presented is correct regardless of whether or not $m(E)$ is finite. I don't know whether adding the unneeded hypothesis is common, but I believe based on Google searching just now that not adding the unneeded hypothesis is common.

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    $\begingroup$ I think the hypothesis that $m(E)<\infty$ is sometimes added because the result is important in probability theory where it is interpreted as $L^1$-convergence implies convergence in probability. In the usual proof found in many probability courses $m(E) < \infty$ is important because the argument usually goes like $L^p$ convergence implies $L^1$-convergence and now use Markov's inequality. $\endgroup$ – user38355 Aug 16 '13 at 10:12

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