# 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 $E$.

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}

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

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 $$\int_E f =0$$.