Showing Convergence

Question

Let $(X,M,\mu)$ be a measurable space and $f$ be a real valued integrable function on $X$. Let $E_n=\{x\in X: f(x)\geq nq\}$ for every $n\in \mathbb{N}$ and fixed $q>0$ . Show that

• $\lim_{n \to \infty} \begin{equation*} \int_{E_n} f d\mu \end{equation*}=0$
• $f\chi_{E_n}\to 0$ in measure $\mu$
• $\lim_{n \to \infty} \mu^*(E_n)=0$ ($\mu^*$ is outer measure corresponding $\mu$)

My attempt:

if $f\geq0$ then $\begin{equation*} \int_{E_n} f d\mu \end{equation*}= \int_{X} f d\mu-\int_{X} f\chi_{E_n} d\mu $ now because $f\chi_{E_n}$ is accending sequence of integrable functions that coverge to f almost every where by monotonic covergence theorem we have $$\lim_{n \to \infty} \begin{equation*} \int_{E_n} f d\mu \end{equation*}=0$$ so $f\chi_{E_n}\to 0$ in $L^1$ and hence $f\chi_{E_n}\to0$ in measure $\mu$

$E_n=f^{-1}([nq,\infty))$ and f is measurable so $E_n$ is measurable and hence $$\mu^*(E_n)=\mu(E_n)$$

Answer 1

Your attempt for the first question is good, or more directly, since $|f\chi_{E_n}| \leq |f|$ and $\int_X |f|d\mu < +\infty$, we can apply dominated convergence to say

$$\lim \int_{E_n}fd\mu = \lim \int_X f\chi_{E_n}d\mu = \int_X \lim f\chi_{E_n}d\mu = \int_X 0d\mu = 0$$

For the second one, you didn't prove it. You need to show for any $\epsilon >0$, we have

$$\mu\{x\in X : |f\chi_{E_n}|\geq \epsilon\} \to 0$$

Remark that $$\{x\in X : |f\chi_{E_n}|\geq \epsilon\} \subset E_n$$ so if we have $\mu(E_n) \to 0$, the second conclusion is proved.

Now we will prove the third one, then the third one implies the second one.

Since $E_n$ is measurable, replace the outer measure by measure to simplify, then remark that:

$$\mu(E_n)nq \leq \int_{E_n}fd\mu \leq \int_X |f|d\mu$$

Then it's easy to conclude