Given a sequence $\{f_n\}$ of measurable functions, why does there exist a subsequence $\{f_{n_k}\}$ such that $\lim_{k \to \infty} \int_E f_{n_k} = \liminf \int_E f_n$? I need to use this in a theorem I am proving, but I don't see how to justify it. Just for your information, I am trying to prove the convergence-in-measure version of Fatou's lemma.
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maybe this helps https://math.stackexchange.com/questions/912749/subsequence-of-a-sequence-converging-to-its-lim-sup-and-lim-inf?utm_medium=organic&utm_source=google_rich_qa&utm_campaign=google_rich_qa – Uskebasi Apr 09 '18 at 16:51
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Hint: This is just a statement about real sequences, namely that for a real sequence $(x_n)$, there exists a subsequence $(x_{n_k})$ such that $\lim_{k \to \infty} x_{n_k} = \liminf x_n$. Can you show this instead?
B. Mehta
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Hmm...Let me try. By definition, $\liminf x_n = \lim_{n \to \infty} \inf_{i \ge n} x_i$, so given $k=1$, there exists a $n_1 \ge 1$ such that $0 \le x_{n_1} - \inf_{i \ge 1} x_i$. Given $k=2$, there exists $n_2 \ge 2$ such that $0 \le x_{n_2} - \inf_{i \ge 2} x_i < \frac{1}{2}$. If $n_2 = n_1$, then replace $n_2$ with $n_2 + 1$. The only problem is I don't see why $0 \le x_{n_2} - \inf_{i \ge 2} x_i < \frac{1}{2}$ still holds... – user193319 Apr 09 '18 at 23:35
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...I've never really been good at these proofs requiring constructions. Any that I've read in books were remarkably handwazy and therefore unconvincing, so that I've never come across a good "constructive" proof which could serve as a model for my own. – user193319 Apr 09 '18 at 23:35
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