I'm trying to solve the following problem:

Let $f \ge 0$ be a measurable function on a measurable subset $E \subset \mathbb{R}^d$. Suppose there exits a sequence of measurable subset $\{E_k\}_{k=1}^{\infty}$ of $E$ such that $$m(E\setminus E_k)<\frac{1}{k}\quad \textrm{ and }\quad \lim_{k\to \infty}\int_{E_k} f(x)dx<\infty.$$ Show that $f$ is integrable on $E$.


Consider $$ f_k:=\chi_{E_k}f \Longrightarrow \lim_{k\to \infty}f_k=f \textrm{ a.e on }E \quad \textrm{and}\quad f_k\ge 0. $$ We can see \begin{align*} \int_{E}f dm&=\int_{E}\left(\liminf_{k \to \infty} f_k \right)dm \leq \liminf_{k \to \infty}\int_{E}f_k dm \quad \textrm{by Fatou's Lemma} \\  =&\liminf_{k \to \infty}\int_E \chi_{E_k}f dm\leq \lim_{k \to \infty} \int_{E_k}f dm <\infty \quad \textrm{by hypothesis} \end{align*}

Thanks for any hint.

  • $\begingroup$ Should that be $>0$ in the second inequality? $\endgroup$
    – Ben
    Aug 6 '19 at 17:58
  • $\begingroup$ Sorry, you're accurate. $\endgroup$ Aug 6 '19 at 18:10

Hint: apply Fatou's lemma to the sequence of functions $f_k = f 1_{E_k}$.

  • $\begingroup$ Thanks so much! I just used your hint for the right integral, hope it's okay. I just wonder if the same trick might be applied for the left integral. $\endgroup$ Aug 6 '19 at 23:29
  • $\begingroup$ @CharlesSeife: You don't need to do anything with the integral over $E^c$ at all and it is much better if you just do all your analysis over $E$ itself. But one of the key steps is your claim that $\lim f_k = f$. This is only true almost everywhere on $E$, and you should carefully prove this step. Once you have this, you should use Fatou to show that $$\int_E f \le \liminf \int_E f_k = \liminf \int_{E_k} f.$$ $\endgroup$ Aug 6 '19 at 23:56
  • $\begingroup$ Got it, I guess when we prove that $m(\{x \in E: f_k(x) \nrightarrow f(x)\})=0$, we must use the hypothesis that $m(E\setminus E_k)<1/k$. Sorry, I'm lost to prove it. $\endgroup$ Aug 7 '19 at 21:26
  • $\begingroup$ Actually, I take it back. I don't think it's necessarily true that that $f_k \to f$ almost everywhere on $E$. However, it is true for some subsequence of the $f_k$. One approach is to note that $f_k \to f$ in measure, and then use the theorem that some subsequence therefore converges to $f$ almost everywhere. Another is to choose a subsequence $E_{k_j}$ for which $\sum m(E \setminus E_{k_j}) < \infty$ and use the Borel-Cantelli lemma (which is basically how you prove the theorem previously mentioned). $\endgroup$ Aug 7 '19 at 22:39
  • $\begingroup$ It makes sense. It'll be the first time I'll use Borel-Cantelli Lemma. Thanks a lot for your time. $\endgroup$ Aug 7 '19 at 23:15

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