If $\int_A f\,dm = 0$ for all $A$ having some fixed measure $C$, then $f = 0$ almost everywhere

Let $ f \in L^1[0,1]$. Assume that there is a constant C, with $0 < C < 1$, such that for every measurable set $A \subset [0,1] $ with $m(A)=C$, we have $ \int_{A} f dm = 0 $. Prove that $f = 0$ almost everywhere.

I tried to do my contradiction but I could not get my head around it. Any hints or ideas are appreciated.

Solution 1:

If $f$ is not $0$ almost everywhere, then clearly $m(\{f>0\})>0$ and $m(\{f<0\})>0$ (since on sets of measure $C$ where $f$ is not $0$ almost everywhere, there must be positive parts and negative parts to cancel out). Now suppose $C\leq 1/2$. Either $m(\{f\geq 0\})\geq 1/2$ or $m(\{f\leq 0\})\geq 1/2$; suppose WLOG $m(\{f\geq 0\})\geq 1/2$. Then we can find a subset $A\subseteq\{f\geq 0\}$ of measure $C$ such that $m(A\cap\{f>0\})>0$. This implies $\int_A f>0$, which is a contradiction.

Now suppose $C>1/2$, and write $I=\int_{[0,1]}f$. Note that if $m(B)=1-C$, we have $\int_B f=I$. Let $g=f-I/(1-C)$; then $g$ has the property that for any $B$ of measure $1-C$, $\int_B g=0$. By the previous paragraph, this implies $g$ is $0$ almost everywhere. Thus $f$ is constant almost everywhere (with value $I/(1-C)$), and this clearly implies $f$ is $0$ almost everywhere.