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I have been working on reproducing the following proof for this theorem and have come up with a few questions I cannot seem to answer:

$(1)$ Given that $f$ is a measurable function such that $f >0$ a.e. Prove that, if $\displaystyle\int_{E}fdm=0$ for some measurable set $E$, then $m(E) = 0$. (where $m$ is the Lebesgue measure on $\Bbb{R}$)

My original attempt depended on a problem I did before this one in which I showed that for a Lebesgue measurable function $f$ on $\mathbb{R}$, $f$ is finite almost everywhere. In the proof of this Here, for $n \in \Bbb{N}$ define the set $E := \{x \in \Bbb{R}: f(x) \geq n\}$ such that $E := \bigcap_{n=1}^{\infty}E_n$, I came up with the following inequaility $$\int_{\mathbb{R}}fdm \geq \int_{E_n}fdm \geq n\cdot m(E_n) \geq n \cdot m(E).$$ I figured for $(1)$ I could use this inequality (and the set $E$) to my advantage and say: since $m(E) \leq \frac{1}{n} \int_{E}fdm$, we are given that $\int_{E}fdm=0$ so therefore $m(E) =0$. But, it seems there is an obvious counterexample here to my work. With that said, I started reading this proof and had some questions (I would comment and ask the author, but it seems they haven't been on this site for awhile):

$\bullet$ I'm really not seeing how $\mathbb R^n \setminus F$ has measure zero and why we need to know this for the rest of the proof?

$\bullet$ Why is this line correct: "if $m(E) \neq 0$ then we must have $m(F_n \cap E) >0$ for some $n$ and this would immediately imply that $\int_E f(x)dx>0$"? I don't think I understand why $m(F_n \cap E) >0$ implies $\int_E f(x)dx>0$.

StubbornAtom
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Taylor Rendon
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1 Answers1

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Let $A=\{x\in\mathbb{R}\mid f(x)>0\}$, then $m(A^{c})=0$ because $f>0$ a.e.. For each $n$, define $A_{n}=\{x\in\mathbb{R}\mid f(x)>\frac{1}{n}\}$. We have that $\frac{1}{n}m(A_{n}\cap E)\leq\int_{A_{n}\cap E}f\leq\int_{E}f=0$ and hence $m(A_{n}\cap E)=0$. Note that $A=\cup_{n}A_{n}$, so $A\cap E=\cup_{n}(A_{n}\cap E)$. Therefore, $m(A\cap E)\leq\sum_{n=1}^{\infty}m(A_{n}\cap E)=0$. Finally, $m(E)=m(E\cap A)+m(E\cap A^{c})=0$.

Danny Pak-Keung Chan
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