# Uniform integrability and convergence in measure

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?

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.
• 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 $\int_E|f_n-g|\to 0$? Dec 16, 2021 at 23:59
• 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 Dec 17, 2021 at 0:19