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Suppose $\{f_n\}_{n=1}^{\infty}\subset L^{+}$, $\lim_{n\rightarrow \infty} = f$ pointwise, and $\int f d\mu = \lim_{n\rightarrow \infty}\int f_n d\mu < \infty$. Then $\int_{E}f d\mu = \lim_{n\rightarrow \infty}\int_{E}f_n d\mu$ for $E\in M$

Proof: Let $\{f_n\}\subset L^{+}$ and $f_n\rightarrow f$ pointwise and $\int f d\mu = \lim_{n\rightarrow \infty}\int f_n d\mu < \infty$. By, the monotone convergence theorem, since $\{f_n\}\subset L^{+}$ then $f_j\leq f_{j+1}$, and $f = \lim_{n\rightarrow\infty} f_n(= \sup_{n} f_n)$ then we have $\int f = \lim_{n\rightarrow\infty}\int f_n$.

Now, let $E_n = \{x:f(x) > 1/n\}$ and define $$\int f d\mu = \sup\{\int \phi d\mu: 0\leq\phi\leq f, \phi \ \ \text{simple}\}$$ then since $f_n\rightarrow f$ point wise there exists an $n\in\mathbb{N}$ such that $$\int_{E_n}f d\mu = \lim_{n\rightarrow \infty}\int_{E_n}f_n d\mu$$

I am not sure if this is correct, any suggestions is greatly appreciated.

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1 Answer 1

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The set $E$ is any measurable set, it does not depend on $n$. Moreover, neither the monotone convergence theorem nor the dominated convergence theorem can be used. In these situations, Fatou's lemma is your friend.

Let $\chi_E$ be its characteristic function of $E$. Then $$ \int_E f_n\,d\mu=\int \chi_E\,f_n\,d\mu. $$ The functions $\chi_E\,f_n$ are non-negative and converge pointwise to $\chi_E\,f$. By Fatou's lemma $$ \int_Ef\,d\mu\le\limsup_{n\to\infty}\int_E f_n\,d\mu.\tag{1} $$ Let $E^c$ be the complement of $E$. Then $$ \int_{E^c}f\,d\mu\le\limsup_{n\to\infty}\int_{E^c} f_n\,d\mu=\limsup_{n\to\infty}\int(1-\chi_E)\, f_n\,d\mu=\int f\,d\mu-\liminf_{n\to\infty}\int_E f_n\,d\mu.\tag{2} $$ Adding (1) and (2) we get $$ \int f\,d\mu=\int f\,d\mu+\limsup_{n\to\infty}\int_E f_n\,d\mu-\liminf_{n\to\infty}\int_E f_n\,d\mu. $$ This implies that $\lim_{n\to\infty}\int_Ef_n\,\mu$ exists. Moreover, from (2) we see that $$ \lim_{n\to\infty}\int_Ef_n\,d\mu\le\int f\,d\mu-\int_{E^c} f\,d\mu=\int_E f\,d\mu. $$ This together with (1) proves the claim.

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  • $\begingroup$ Doesn't Fatou's Lemma say that $\int f \leq \liminf \int f_n$? $\endgroup$
    – Wolfy
    Commented Dec 5, 2015 at 21:13
  • $\begingroup$ @MorganWeiss: $\liminf\leq\limsup$, always. $\endgroup$ Commented Dec 5, 2015 at 21:30

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