Let $f$ be measurable with $f$>0 a.e. If $ \int_E f$=0 for some measurable set E, show that m(E)=0.

Question

Let $f$ be measurable with $f$>0 a.e. If $ \int_E f$=0 for some measurable set E, show that m(E)=0.

My solution: suppose $\int_E f$=0. This implies $f$=0 a.e on E. m(E)=$m(E \cap \{f=0\}) + m(E \cap \{f\gt 0\})$. Since $f$=0 a.e on E this implies {$f \gt 0$} measure zero. $m(E \cap \{f\gt 0\})$=0. Since $f$>0 a.e, m{$f$=0}=0 and $E \cap \{f=0\}$ is a subset of a set with measure 0 so m($E \cap \{f=0\}$)=0 also. Thus m(E)=0.

my quesion if $f$=0 a.e does that imply $f$>0 at countably many points? If $f$ is measurable When you are told $f$=0 a.e or $f$>0 a.e what can you conclude? Can someone please identify the flaw in my solution or give me a hint at how to approach this problem?

Answer 1

Your real question has allready been answered by Starfall. This shows you how to do it correctly.

For convenience assume that $f(x)>0$ for every $x\in E$.

For $n=1,2,\dots$ define: $$E_n:=E\cap\{f\geq\frac1{n}\}$$

Observe that: $$1_{E_n}\leq nf1_{E}$$ so that: $$0\leq m(E_n)\leq n\int_E f=0$$ and consequently: $$m(E_n)=0\text{ for all }n$$

Also it is evident that: $$E_1\subseteq E_2\subseteq\cdots$$ and: $$E=\bigcup_{n=1}^{\infty}E_n$$ so that: $$E_n\uparrow E$$ and consequently: $$m(E_n)\uparrow m(E)$$

This together proves that $m(E)=0$ and under the weaker condition "$f>0$ a.e." only a small adjustment is needed.

Answer 2

No, it doesn't. The characteristic function of the Cantor set is a counterexample for $ E = [0, 1] $.