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Suppose $\{f_n\}:[0,1] \to \mathbb{R}$ is uniformly integrable and $f_n \to g$ in measure. Show that $\lim\limits_{n\to \infty}\int_{[0,1]}|f_n-g|=0$

Attempt. I am not sure whether I am supposed to use the definition of convergence in measure to do this, or to use the fact that some subsequence $\{f_{n_k}\}$ converges pointwise to $g$. I will try to use the pointwise convergence. Since $\{f_n\}$ is uniformly integrable, for $\epsilon>0$, there exists $\delta>0$ such that $\mu(E)<\delta \implies \int_E|f_n|<\epsilon$ where $\mu$ is Lebesgue measure. Assume $\{f_{n_k}\}$ is a subsequence that converges pointwise to $g$. By Egoroff's theorem, there exists a measurable set $E$ such that $\mu(E)<\delta$ and $\{f_{n_k}\} \to g$ uniformly on $E^c$. So there exists $N \in \mathbb{N}$ such that $k \geq N \implies |f_{n_k}-g|<\frac{\epsilon}{\mu(E^c)}$. Then $$\int_{[0,1]}|f_{n_k}-g|=\int_E|f_{n_k}-g|+\int_{E^c}|f_{n_k}-g|$$

Now I am stuck. Is $g$ uniformly integrable, and can I canclude the left term in the sum on the right is $\int_{E}|f_{n_k}|f_{n_k}-g|\leq \int_E|f_n|+\int_E|g|<2\epsilon$? What do I do here? Also if I can finally show theat $\lim\limits_{k \to \infty}\int_{[0,1]}|f_{n_k}-g|=0$, how do I show that for the sequence?

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Hint: You are on the right track. Use Fatou's Lemma to conclude that $\int_E |g| \leq \lim \inf \int_E|f_{n_k}|$ and you can finish the proof.

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  • $\begingroup$ So $\int_{E}|g|\leq \liminf\int_{E}|f_{n_k}|<\epsilon$ so that $\int_{[0,1]}|f_{n_k}-g|\leq \int_{E}|f_{n_k}|+\int_E|g|+\int_{E^c}|f_{n_k}-g|<3\epsilon$ for $k \geq N$ for some $N$. Then why does this imply $\int_E|f_n-g|\to 0$? $\endgroup$
    – Jolie
    Dec 16, 2021 at 23:59
  • $\begingroup$ You have to complete the proof using the following basic theorem: A sequence of real num bers $(x_n)$ converges to $x$ if and only if every subsequence of $(x_n)$ has a further subsequence which converges to $x$. This allows you to reduce the proof to the case of a.e. convergence. @Jolie $\endgroup$ Dec 17, 2021 at 0:19

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