Let $f$ measurable and almost everywhere finite on $[0,1]$. Let $\int_{E} f\,d\lambda = 0$ for any measurable $E\subset[0,1]$ with $\lambda(E) = \frac{1}{2}$. (Where $\lambda$ is the Lebesgue measure and the function is Lebesgue measurable).
I would like to show that $f$ is almost everywhere $0$. That is the last part in a many part question but it would kill every other part. I believe my proof is correct.
Put $A=f^{-1}((0,\infty])$, $B=f^{-1}([-\infty,0))$, $C=f^{-1}(\{0 \})$. Notice that $A\cup B \cup C = [0,1]$.
Claim: We have that $\lambda(A) < \frac{1}{2}$ and $\lambda(B) < \frac{1}{2}$ (and so $\lambda(C) > 0$).
Proof: Suppose that $\lambda(A) \geq \frac{1}{2}$. Put $A' = \{x\in[0,1]:\lambda(A\cap[0,x]) \geq \frac{1}{2} \}$. This set is nonempty as it contains $1$. Put $\alpha = \inf A'$. The continuity of the measure provides that $\lambda(A\cap[0,\alpha]) = \frac{1}{2}$. By hypothesis we then have $$\int_{A\cap[0,\alpha]}f\,d\lambda = 0. $$ Since $f$ is nonnegative on the domain of integration, it must be that this $f$ is almost everywhere $0$ on it. This contradicts that $f$ is strictly positive on $A$. A similar proof gives the result for $B$.
Since $\lambda(B) < \frac{1}{2}$ it must be that $\lambda(A\cup C) = \lambda(A)+\lambda(C) \geq \frac{1}{2}$. But then we may do the same trick.
Claim: We have that $\lambda(A) = \lambda(B) = 0$ and so $\lambda(C) = 1$ (and $f$ is 0 almost everywhere)
Proof:Put $A'' = \{x\in[0,1] : \lambda(A\cup(C\cap[0,x]))\geq\frac{1}{2} \}$. This set is also nonempty because it contains $1$. Put $\beta = \inf A''$. The continuity of the measure provides that $\lambda(A\cup(C\cap[0,\beta])) = \frac{1}{2}$. By hypothesis we have $$\int_{A\cup(C\cap[0,\beta])}f\,d\lambda = 0 .$$ Since $f$ is nonnegative on the domain of integration, it must be that $f$ is almost everywhere $0$ on it. Since $f$ is strictly positive on $A$, this can only be so $\lambda(A) = 0$. A similar proof gives the result for $B$.