1
$\begingroup$

I have a question about one particular step in the proof of Lebesgue's monotone convergence theorem in Rudin's Principles of Mathematical Analysis.

Theorem $\quad$ Suppose $E \in \mathfrak M, \{f_n\}$ is a sequence of measurable functions such that $0\le f_1(x)\le f_2(x)\le ...$ for $x\in E,$ and $f_n\to f$ as $n\to \infty$ on $E$. Then $\int_E f_nd\mu \to \int_E fd\mu.$

Now, in an intermediate step of the proof (page 319), the author chooses $c$ such that $0<c<1$, lets $s$ be a simple function such that $0\le s\le f$, and then defines $E_n=\{x\mid f_n(x)\ge cs(x)\}$. Then he states that, since $f_n\to f$, we have $E=\cup_{n=1}^\infty E_n,$ which is my question.

It's not immediately evident to me why $f_n \to f$ would imply $E=\cup_{n=1}^\infty E_n$? Is it because of the following reason? First of all, $E_n \subset E,$ if the domain of $f_n$ is assumed to be $E$. Hence $\cup_{n=1}^\infty E_n\subset E.$ On the other hand, for any $x \in E,$ there must exist an $n_0$ such that $f_{n_0}(x)\ge cf(x)\ge cs(x)$, because $0<c<1$ and $f_n \to f.$ So $x\in E_{n_0}\subset \cup_{n=1}^\infty E_n,$ implying $E\subset \cup_{n=1}^\infty E_n.$ This proves the claim.

Is this argument correct? Is there a simpler way to see it? BTW, it appears that the domain of $f_n$ should be assumed to be $E$. Otherwise, $E_n$ may not be contained in $E.$ (This assumption doesn't weaken the theorem though, does it?)

$\endgroup$
  • 1
    $\begingroup$ Yes, your line of reasoning is correct. $\endgroup$ – Juanito May 12 '17 at 8:25
  • 1
    $\begingroup$ And to avoid complications you can also do it with $E_n:=\{x\in E\mid f_n(x)\geq cs(x)\}\subseteq E$. Caution: $f_{n_0}(x)>cf(x)$ asks for the extra condition that $f(x)>0$. Fortunately it is evident that $x\in E_1$ if $f(x)=0$. $\endgroup$ – drhab May 12 '17 at 8:29
  • $\begingroup$ @drhab Indeed. Thanks a lot for the tip! To cover the case when $f(x)=0$, I changed it to $f_{n_0}(x) \ge cf(x)$. $\endgroup$ – syeh_106 May 12 '17 at 8:40

Your Answer

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

Browse other questions tagged or ask your own question.