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Let $m$ denote the Lebesgue measure on the Lebesgue sigma algebra $\mathcal{M}$. If $f:\mathbb{R} \to [0,\infty)$ is Lebesgue measurable and $\int_{(n,n+1]}f dm = 0$ for all $n \in \mathbb{Z}$, then $\int_E fdm =0$ for all $E \in \mathcal{M}$.

I think the statement is true. I tried to prove it below. However, the question was given as "Prove/Disprove". If it is actually false, please let me know why.

Proof. Let $E \in \mathcal{M}$. Note that $f$ is measurable and $\chi_{_{(n,n+1]}} $ is measurable for all $n \in \mathbb{Z}$. Hence, $f\chi_{_{(n,n+1]}} $ is measurable for all $n \in \mathbb{Z}$. Then notice that

\begin{align} \int\sum_{n\in \mathbb{Z}}f\chi_{_{(n,n+1]}}dm & = \int \left( \sum_{n\in \mathbb{N}}f\chi_{_{(n,n+1]}} + \sum_{n\in \mathbb{N}}f\chi_{_{(-n,-n+1]}}\right)dm\\ & = \sum_{n\in \mathbb{N}}\int f\chi_{_{(n,n+1]}}dm + \sum_{n\in \mathbb{N}}\int f\chi_{_{(-n,-n+1]}}dm\\ & = \sum_{n\in \mathbb{N}}\int_{(n,n+1]} fdm + \sum_{n\in \mathbb{N}}\int_{(-n,-n+1]} fdm\\ & = 0 & (\text{by assumption}). \end{align}

Now, we have \begin{align} \int_E fdm & = \int_{E \cap (\cup_{n \in \mathbb{Z}}(n,n+1])}fdm\\ & = \int f \chi_{_{E \cap (\cup_{n \in \mathbb{Z}}(n,n+1])}}dm\\ & = \int f \chi_{_{(\cup_{n \in \mathbb{Z}}E\cap(n,n+1])}}dm\\ & = \int \sum_{n\in \mathbb{Z}} f\chi_{_{E\cap(n,n+1]}}dm\\ & \leq \int \sum_{n\in \mathbb{Z}} f\chi_{_{(n,n+1]}}dm\\ & = 0 & (\text{by the above}). \end{align}

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  • $\begingroup$ There is missing $E$ in the second display (or $\le$). $\endgroup$
    – d.k.o.
    Mar 10, 2019 at 4:15
  • $\begingroup$ d.k.o. It is fixed. Is it correct now? $\endgroup$
    – user506873
    Mar 10, 2019 at 4:25
  • $\begingroup$ What if $f(x) = \sin (2 \pi x)$? $\endgroup$
    – copper.hat
    Mar 10, 2019 at 4:36
  • $\begingroup$ @copper.hat $f$ is nonnegative $\endgroup$
    – d.k.o.
    Mar 10, 2019 at 4:40
  • $\begingroup$ @d.k.o.: Thanks, I missed that. $\endgroup$
    – copper.hat
    Mar 10, 2019 at 4:41

1 Answer 1

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If $f\ge 0$ and $\int_A f = 0$ then $f=0$ ae. on $A$. Hence $f$ is zero ae on every $(n,n+1]$, hence $f=0$ a.e It follows that $\int_E f =0$.

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  • $\begingroup$ Nice. That is a much simpler proof. $\endgroup$
    – user506873
    Mar 10, 2019 at 4:48

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