Show $\lim_{n\rightarrow\infty} \int_{E_n}gdm=0\mbox{.}$

Question

Show that if $g$ is summable on $A$ and $A\supset E_1 \supset E_2,\ldots$ and $\cap_{n=1}^{\infty} E_n= \emptyset$ then

$$\lim_{n\rightarrow\infty} \int_{E_n}gdm=0\mbox{.}$$

Proof: Assume the contrary. So there is $\epsilon>0$ such that for every $N\in \mathbb{N}$ there is $n>N$ such that $$\Big |\int_{E_n}gdm \Big| \ge \epsilon \mbox{.}$$ But that also means that for every $N=1,2,\ldots$ there is $n>N$ such that $E_n$ is not empty. Contradiction with $\cap_{n=1}^{\infty} E_n= \emptyset$.

Is this a correct "proof"? Would you add anything?

Answer 1

In your attempt, you did not use the integrability of $g$ (and the sets $E_n$ may have all positive measure).

Here are some hints:

• if $g$ is the indicator function of a measurable set, then the result is a consequence of properties of a measure.
• When $g$ is a linear combination of indicator functions of a measurable sets, it is just linearity of the limit.
• If $g$ is non-negative and integrable, approximate it by functions like in the second bullet.
• Write $g$ as the difference of two non-negative integrable functions.

Alternatively, if you know the monotone convergence theorem, you can use it with $f_n:=|g|-|g|\mathbf 1(E_n)$.