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Let $f_n \in L_0(X), |f_n|\leqslant\phi \in L_1(X), n \in N$ and $f_n \rightarrow f$ in measure. Prove that $f \in L_1(X)$ and $$\lim_{n\to\infty}\int_{X}{f_n}d\mu = \int_{X} fd\mu$$

Here $L_0(X)$ stands for Lebesgue measurability, and $L_1(X)$ — for Lebesgue integrability on $X$.

As far as I understand, we should use the fact that $f_n$ is bounded by a Lebesgue integrable function being itself measurable, so $f$ is also Lebesgue integrable. Is that correct? I remember similar reasoning being used in the classroom. If the first part is correct, how do I prove the equality?

Don Draper
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1 Answers1

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Let $\{f_{n_k}\}_{k=1}^\infty$ be an arbitrary subsequence of $\{f_n\}_{n=1}^\infty$. Then $f_{n_k} \to f$ in measure, so one may extract a further subsequence $\{f_{n_{k_j}}\}_{j=1}^\infty$ of $\{f_{n_k}\}_{k=1}^\infty$ such that $f_{n_{k_j}}\to f$ pointwise a.e. in $X$. By Fatou's lemma $f\in L_1(X)$; by the dominated convergence theorem $\int_X f\, d\mu = \lim\limits_{j\to \infty} \int_X f_{n_{k_j}}\, d\mu$. Since $\{f_{n_k}\}_{k=1}^\infty$ was arbitrary, the result follows.

kobe
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  • Why does the result follow from $f_{n_k}$ being arbitrary? – GuPe Dec 25 '18 at 19:29
  • @Gaucho If every subsequence of a sequence of real numbers converges to some number $A$, then the sequence converges to $A$. In this case, the sequence under consideration is $\int_X f_n, d\mu$. – kobe Dec 25 '18 at 19:30
  • But it is only proved that a subsequence of every subsequence converges to $\int f$. – GuPe Dec 25 '18 at 19:45
  • @GuachoPerez no, it has been shown that every subsequence of $\int_X f_n, d\mu$ has a further subsequence which converges to $\int_X f, d\mu$, so $\int_X f_n, d\mu \to \int_X f, d\mu$. Indeed, since $\int_X f_{n_j}, d\mu$ has a subsequence which converges to $\int_X f, d\mu$, then $\int_X f_{n_j}, d\mu$ to converges to the same number. But since $\int_X f_{n_j}, d\mu$ is an arbitrary subsequence of $\int_X f_n, d\mu$, $\lim_n \int_X f_n, d\mu = \int_X f, d\mu$. – kobe Dec 25 '18 at 19:53
  • @Gaucho take a look here: https://math.stackexchange.com/questions/397978/every-subsequence-of-x-n-has-a-further-subsequence-which-converges-to-x-the. – kobe Dec 25 '18 at 19:55
  • Of course. Nice solution, (+1) – GuPe Dec 25 '18 at 20:15