Let $f$ be a measurable function on $E$ that is finite a.e. on $E$ and $m^* (E) < \infty$. For each $\epsilon > 0$, show that there is a measurable set $F$ contained in $E$ such that $f$ is bounded on $F$ and $m^* (E\setminus F) < \epsilon$. What is $F$? I know $F \subset E$ do I have to say that in my proof?
Let $E\setminus F = E \cap F^c$ which implies $m^*(E\setminus F) = m^*(E \cap F^c)$.
I know $m^*(F) = m^*( F\cap E) + m^*(F\setminus E^c)$ but I have
$m^*(F) = m^*(E\cap F^c ) + m^*(F\cap E)$. I don’t see where this came from.
Since $m^*(E) < \infty$ that implies $m^*( E\cap F^c) = m^*(F) – m^*( F\cap E)$.
I will create $I_n$ a countable cover for $E$, where $I_n=(a_n,b_n)$.
I’m going to create a closed set for F so the function f will be bounded on it.
$$ F_n=[a_n,b_n- \epsilon /2^n ] $$ so let $$ F = \bigcup F_n. $$
Using the definition of outer measure I get $$ m^*(F) = \sum l(F_n )=\sum (b)_n- a_n-\epsilon/2^n )= \sum(b_n-a_n-\epsilon).$$
Since $F\subset E$, it is easy to see that $m^*(E\cap F)=m^*(F)$ so $$m^*(E\cap F) = m^*(F) \leq m^*(I_n )=\sum (b_n-a_n)$$ which implies $$m^*(E\cap F) \leq \sum l(I_n )- \sum l(F_n )=\sum (b_n-a_n )-\sum (b_n-a_n-\epsilon)=\epsilon.$$
Thus there is an $ F$ such that $F\subset E$ and $F$ is closed so $f$ is bounded on it and $m^*(E\setminus F) \leq \epsilon$.
Nothing in your work has included f. I do not see why suddenly f should be bounded.
Best Answer
You need to construct such an $F$ with the properties outlined. The set $A_n=\{x\in E:|f(x)|>n\}$ is an open set with $A_n\subseteq E$. Since f finite a.e. and $m^*(E)<\infty$, there exists a $s$ such that $m^*(A_s)<\varepsilon$. Take $F=E\setminus A_s$.
$F$ is closed and $F\subseteq E$ with $m^*(E\setminus F)< \varepsilon$. Also, $|f(x)|\leq s$ on $F$ by definition of $A_s$.