A problem about Lebesgue integral convergence

83 Views Asked by At

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!

1

There are 1 best solutions below

0
On BEST ANSWER

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 $\int\vert f_n(x)\vert dx\to \infty$.