Suppose $f \in L^1([0,1],\lambda)$ and let $\alpha \in (0,1)$. Assume that $\int_E f=0$ for all$E$ s.t $m(E)=\alpha$ show $f=0$ a.e.
I am not sure how to tackle this problem when $\alpha>1/2$, it is easy when $\alpha<1/2$ by considering the sets where $f>0$ or $f<0$ and integrating over them and adding elements of $f=0$ / subtracting to get the measure to be $\alpha=1/2$
Choose $n$ large enough such that $\alpha+{ 1\over n} \alpha < 1$. Choose any collection of $n+1$ pairwise disjoint sets $E_1,..,E_{n+1}$ such that $\lambda E_k = {\alpha \over n}$. Note that the measure of the union of any $n$ of these is $\alpha$ and hence the integral of $f$ over any $n$ is zero.
In particular, the integral over $E_1,...,E_n$ is zero. Now swap any of these $E_k$ with $E_{n+1}$ and we still have a set of measure $\alpha$ and the integral over this collection is zero. It follows that the integral of $f$ on any of the $E_k$ is the same as the integral over $E_{n+1}$ and hence it follows that the integral over any of the $E_k$ is zero.
In particular, by choosing large enough $n$ we can assume that $\alpha$ is arbitrarily small (but not arbitrary, of course).
Let $A = \{x | f(x) >0 \}$. If $\lambda A >0$, we can choose $\alpha \in (0,\lambda A)$ and find some $B \subset A$ such that $\lambda B = \alpha$ from which we get $\int_B f = 0$, a contradiction. Hence $\lambda A = 0$. The same reasoning applies to $\{x | f(x) <0 \}$.
