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Let $([0,1], \mathcal{B}([0,1]), \mu)$ be a measure space with Lebesgue measure $\mu$. Let $f\in L_1([0,1])$ and a fixed constant $b\in (0,1)$ so that $$ \int_B fd\mu=0 $$ for all $B\in \mathcal{B}([0,1])$ with $\mu(B)=b$. Show that $f=0$ $\mu$ a.e.

My proof is as follows: (that is different from Want to show that $f=0$ a.e.).

W.L.O.G, assume that for any open set $G$ with $\mu(G)=b$ we have $ \int_G fd\mu=0$.

Then for every $a\in \mathbb{R}$, we have $$\int_a^{a+b}f(x)dx=0$$

Then for every $c>0$ we have $$ \frac{1}{c}\int_a^{a+c}f(x)dx=\frac{1}{c}\int_{a+b}^{a+b+c}f(x)dx $$

As $c\to 0$, we get $f(a)=f(a+b)$ a.e. for $a\in \mathbb{R}$. Define $$E_n:=\{a\in\mathbb{R}: f(a)=f(a+b)=\dots=f(a+nb)\neq f(a+(n+1)b)\}$$ with $\mu(E_n)=0$ for $n\in\mathbb{N}$.

Let $S:=\mathbb{R}\setminus \cup_{n=1}^\infty E_n$. we have $$ \lim_{n\to\infty}\frac{1}{b/n}\int_a^{a+b/n}f(x)dx=f(a) $$ for all $a\in S\setminus\mathbb{Z}$.

Let $G=(a,a+b/n)\cup(a+b/n, a+2b/n)\cup\dots\cup (a+n-1, a+n-1+b/n)$. $G$ is open and $m(G)=1$. So $$ \int_G fd\mu=0=\frac{1}{b/n}\int_a^{a+b/n}f(x)dx\to f(a) $$

So $f(a)=0$ a.e. for every $a\in\mathbb{R}$.

Hermi
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    You need more hypotheses on $\mu$. For example, this is not true if $\mu$ is the probability measure which is $1$ at ${0}$ and $0$ at every other point. One situation I know that this is true is if $\mu$ is the Lebesgue measure, so is that what you meant? –  Apr 16 '24 at 23:13
  • @WhileIAm You are right. We need to add $\mu$ is Lebesgue measure. I have added it. Thank you! – Hermi Apr 17 '24 at 00:07
  • In that case, hint: use the Lebesgue differentiation theorem. Try seeing what happens when the shrinking set for the theorem is the overlap of two sets of measure $b$. –  Apr 17 '24 at 00:17
  • @WhileIAm We do not teach this theorem. Can we use another result to show that? – Hermi Apr 17 '24 at 00:29
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    https://math.stackexchange.com/questions/4437750/want-to-show-that-f-0-a-e – user23571113 Apr 17 '24 at 03:04
  • @user23571113 I just update a proof. Can you please check that? Does it make sense? – Hermi Apr 18 '24 at 22:02

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