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Given $(X,\mu)$ a measure space, $\{f_n\}$ is a sequence of measurable and non-negative functions defined on $X$ such that $f_n\to f$ pointwise and $$\int_X f\mathrm d \mu=\lim_{n\to\infty}\int_X f_n\mathrm d\mu.$$ Prove that for all measurable subset $E$, $$\int_E f\mathrm d \mu=\lim_{n\to\infty}\int_E f_n\mathrm d\mu.$$

Tools available to me: Lebesgue monotone/dominated convergence theorem and Fatou's lemma, and I believe it isn't convenient to apply dominated convergence theorem here.

If I should start with simple function case, then should I let $f$ or $f_n$ or both be simple? Unfortunately, all these tricks seem to mess up.

I don't think this problem is hard. But anyhow I'm struggling with it for more than an hour and have hardly made any progress (forgive my stupidity, but it just happens every now and then). So, frankly, I need help. Maybe even a tiny hint can shed enough light on it. Thanks in advance.

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  • $\begingroup$ @user251257 These functions are not assumed to be integrable. Their integrals may diverge. $\endgroup$ Commented Mar 26, 2016 at 5:20
  • $\begingroup$ @user251257 thanks for the source. This is of great help if the assignment question were stated correctly. $\endgroup$
    – Vim
    Commented Mar 26, 2016 at 5:21

2 Answers 2

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The claim is not necessarily true if $\int_Xf=\infty$. For example, let $X=\mathbb{R}$ with Lebesgue measure, and $f_n(x)=n^2\chi_{(0,\frac{1}{n}]}(x)+\chi_{[1,n]}(x)$. Then $f_n\to f$ pointwise, where $f(x)=0$ if $x<1$, and $f(x)=1$ for $x\geq 1$.

Also, $$ \int_{\mathbb{R}}f(x)\;dx=2n-1\to\infty=\int_{\mathbb{R}}f(x)\;dx $$ but if $E=[0,1]$ then $$ \int_Ef_n(x)\;dx=n\to\infty$$ while $$ \int_Ef(x)\;dx=0 $$

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  • $\begingroup$ Oh my gosh! Thanks for pointing out and preventing me from wasting more time on this darned assignment! $\endgroup$
    – Vim
    Commented Mar 26, 2016 at 5:15
  • $\begingroup$ No problem, I imagine the problem meant to include the condition $\int_Xf<\infty$. $\endgroup$ Commented Mar 26, 2016 at 5:16
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Tip: Using \, inbetween $f$ and $d\mu$ enhances visual effect.

By Fatou's Lemma, $\int_E f\,d\mu \leq \liminf_{n \to \infty} \int_E f_n\,d\mu$, and likewise for $E^c$. Observe that $$\begin{align*} \int f\,d\mu - \int_E f\,d\mu &= \int_{E^c} f\,d\mu \leq \liminf_{n \to \infty} \int_{E^c} f_n\,d\mu = \liminf_{n \to \infty} \left(\int f_n\,d\mu - \int_E f_n\,d\mu\right) \\ &= \int f\,d\mu - \limsup_{n \to \infty} \int_E f_n\,d\mu \end{align*}$$ Therefore $\limsup_{n \to \infty} \int_E f_n\,d\mu \leq \int_E f\,d\mu \leq \liminf_{n \to \infty} \int_E f_n\,d\mu$, as desired. Also, the above equation makes sense only if $\int f\,d\mu < \infty$.

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    $\begingroup$ Be careful canceling $\int f\;d\mu$! $\endgroup$ Commented Mar 26, 2016 at 5:08
  • $\begingroup$ Thanks. This I have considered, but what if $\int_E f\, \mathrm d\mu=\int f\, \mathrm d\mu=\infty$? $\endgroup$
    – Vim
    Commented Mar 26, 2016 at 5:09
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    $\begingroup$ @Vim Then the equation does not hold. $\endgroup$ Commented Mar 26, 2016 at 5:09
  • $\begingroup$ @HenryW. Yea you're right. Thanks. It seems that the assignment itself has some problems. $\endgroup$
    – Vim
    Commented Mar 26, 2016 at 5:17

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