Suppose that $f_n$ is a sequence of measurable functions that are all bounded by M, supported on a set E of finite measure, and $f_n(x)\to f(x)$ a.e. x as $n\to \infty$. Then f is measurable, bounded, supported on E for a.e. x, and $\int |f_n-f|\to 0 $ as $n\to\infty$

At here, I understand the restriction of E to be finite measure is because we need to use Egorov theroem. But what is the reason to assume $f_n$ to be bounded? I check the Egorov theorem, it does not require the squence that converge to $f$ to be bounded in order to find a smaller set differs from E by $\epsilon$ and converges uniformly to $f$ on that set.

  • 1
    $\begingroup$ In my opinion, it is needed to assume that to ensure that $f$ is a bounded measurable function and hence we can conclude that $\int_{E}f$ exist. In fact, $|f(x)|\le M$ for all $x\in E$. $\endgroup$ – Juniven Dec 17 '12 at 4:26
  • $\begingroup$ That make sense! $\endgroup$ – Zhixia Zhang Dec 17 '12 at 4:33

Take $X = [0,1]$ with Lebesgue measure. Then let $$f_n = n 1_{[0,\frac{1}{n})}.$$ Then $f_n \rightarrow 0$ a.e. However for all $n$, $$ \int \lvert f_n - 0\rvert = \int \lvert f_n\rvert = 1$$

  • $\begingroup$ Also note that if $f_n \leq M$ for all $n$ and $E$ is of finite measure, then $M1_{E}$ is a dominating function for $f_n$. $\endgroup$ – Deven Ware Dec 17 '12 at 4:41
  • $\begingroup$ Nice post! I like it. $\endgroup$ – Juniven Dec 17 '12 at 5:09

Deven Ware's answer is somewhat along the lines of saying "the reason for assuming uniform boundedness is that otherwise there are counterexamples" (which is a standard argument in mathematics). Here is another reason, which is rather philosophical (or heuristic), due to the proof of the Bounded Convergence Theorem using Egorov's Theorem:

In order to bound the integral of a function, we need to bound either the measure of the domain of the integral, or the function itself. By using Egorov's Theorem we can ensure the existence of a measurable subset $F$ of the domain on which we have uniform convergence whose complement can be made to have arbitrarily small measure. On $F$ we can bound $|f_n-f|$ precisely because we have uniform convergence there. However if we don't have a uniform bound on $\{|f_n|\}_n$, and hence on $\{|f_n-f|\}_n$, $|f_n-f|$ may grow in such a way that nullifies the smallness of the measure of $F^c$. This would be a problem, since even though we can make the measure of $F^c$ arbitrarily small, we may be unable to make it zero (e.g. see this post).


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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