This is an exercise from Natanson's Theory of Functions of a Real Variable.
Suppose $f\in L^1[a,b]$, and $\alpha\in(0,b-a)$ is a constant. If for all $E\subset[a,b]$ such that $m(E)=\alpha$, we have $\int_Ef(x)\mathrm dx=0$. Show that $f=0$ a.e. on $[a,b]$.
I start by suppose the contrary: $\exists\epsilon>0$, $m\{x\in[a,b]:f(x)>\epsilon\}=\beta>0$ (without loss of generality; otherwise we can consider $-f$). If $\beta\ge\alpha$, then the contradiction is obvious. If $\beta<\alpha$, I can reach no further result.
Any hints are appreciated.
Let $\epsilon=((b-a)-\alpha)/3$. Suppose that $m(E_1)<\epsilon$, $m(E_2)=m(E_1)$, and $E_1\cap E_2=\emptyset$. There exists a set $F$ so that $F\cap E_j=\emptyset$ and $m(F\cup E_j)=\alpha$ for $j=1,2$. Hence $$\int_{E_1} f=-\int_F f=\int_{E_2}f.$$
Now if $m(E_1)=m(E_2)<\epsilon$ (but not assuming that $E_1\cap E_2=\emptyset$) there exists $E$ with $m(E)=m(E_j)$ and $E\cap E_j=\emptyset$ for $j=1,2$ (because $\epsilon<(b-a)/3$). Hence $\int_{E_1}=\int_E=\int_{E_2}$.
So: If $m(E)<\epsilon$ then $\int_Ef$ depends only on $m(E)$. Now we're done, for any number of reasons. For example it follows that $$\lim_{h\to0}\frac1h\int_x^{x+h}f(t)\, dt$$is independent of $x$; hence $f$ is constant (or rather equal to some constant almost everywhere) and that constant must be $0$.
(If we don't know the Lebesgue Differentiation Theorem yet we can give a more elementary argument for the punchline...)