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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?

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    What does "countably finite" mean? If it means either "countable" or "finite", the claim that the set on which $ f > 0 $ is "countably finite" need not be true - it may just be Lebesgue null. – 2017-02-19
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    @Starfall hi Starfall I edited my post I meant just countable. Does my solution hold? what can we conclude from f=0 a.e? – 2017-02-19

2 Answers 2

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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.

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    to adjust for f>0 a.e can we say $E$=$\cup_{n=1}^\infty E_n \cap {$f=0}$ since m($E_n$)=0, m($\cup_{n=1}^\infty E_n) \le 0$ so m($\cup_{n=1}^\infty E_n$)=0 and m{$f$=0}=0 so we have m($E$)=0. Also another question if f>0 a.e that means m{f=0}=0 correct? – 2017-02-20
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    $f>0$ a.e. $\iff m\{f\leq0\}=0\implies m\{f=0\}=0$ – 2017-02-20
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No, it doesn't. The characteristic function of the Cantor set is a counterexample for $ E = [0, 1] $.