# A problem about Lebesgue integral convergence

Suppose that $$\{E_n\}_n$$ is a sequence of Lebesgue measurable subset in $$\mathbb R$$ satisfies $$\displaystyle\lim_{n\to \infty}\mu(E_n)=0$$. $$\{f_n\}_n$$ is a Lebesgue measurable function sequence on $$\mathbb R$$ satisfies

$$\displaystyle\lim_{n\to \infty}\int_{\mathbb R-E_n}|f_n-f|d\mu=0$$

Here $$\mu$$ is the one-dimensional lebesgue measure.

(1) Prove the following proposition or give an counterexample:

There exist a subsequence $$\{f_{n_{j}}\}_j$$of $$\{f_n\}_n$$ that, the real number sequence $$\{f_{n_{j}}(x)\}_j$$ converges to $$f(x)$$ a.e. $$\mathbb R$$.

(2) $$\{E_n\}_n$$, $$\{f_n\}_n$$ and $$f$$ satisfies the above conditions, with the following additional condition:

$$\displaystyle\limsup_{n\to \infty}\int_{\mathbb R}|f_n(x)|^2d\mu\lt\infty$$

Prove the following proposition or give an counterexample:

$$\displaystyle\lim_{n\to \infty}\int_{\mathbb R}|f_n-f|d\mu=0$$

Here are my confusion:

a. I'm aware that, $$L_1$$ convergence lead to convergence in measure by Chebyshev inequality, and convergence in measure implies that there exist a subsequence that converges a.e., however I don't know how to prove this with the condition in this problem.

b. For (2), I don't know how to apply this uniformly bounded condition.

Any help would be appreciated, thanks a lot!

• For (2) it is not clear what "the above conditions" refer to. Commented Aug 20, 2023 at 18:20

For the first question you should be able to show that $$f_n$$ converges in measure to $$f$$ and apply what you know. Define $$A_n(\epsilon):= \{ x: \vert f_n(x)-f(x)\vert\geq \epsilon \}$$ and $$\hat{A_n}(\epsilon):=A_n(\epsilon) \setminus E_n$$.
Then $$A_n(\epsilon)\subseteq \hat{A_n}(\epsilon) \cup E_n$$, and it is enough to show $$\mu\big(\hat{A_n}(\epsilon)\big)\to 0$$. But since
$$\int_{\mathbb R-E_n}|f_n-f|d\mu\geq \int_{\hat{A_n}(\epsilon)}|f_n-f|d\mu \geq \epsilon\cdot\mu\big(\hat{A_n}(\epsilon)\big),$$
you can conclude that $$\mu\big(\hat{A_n}(\epsilon)\big)\to 0$$. So you do have convergence in measure.
For the second question, you should see that you are given a bound on $$L^2$$ norms and asked to show $$L^1$$ convergence. Since a.e convergence does not imply $$L^1$$ convergence, you can find a sequence of relatively simple functions such that $$f_n\to0$$ a.e and $$\int\vert f_n(x)\vert^2 dx\to 0$$, but $$\int\vert f_n(x)\vert dx\to \infty$$.