If integral is 0 on any set of measure 1/pi, then the function is 0 a.e. This is a problem in my Qualifying Exam.
"Suppose $f:[0,1]\to \mathbb{R}$ is in $L^1$ (Lebesgue measure) and for
   every measurable $A\subset [0,1]$ with $m(A)=\frac 1{\pi}$ we have
   $\int_A f dm=0$. Prove that $f=0$ a.e."
I could not do it back then. I did my research and we have a similar problem here Integral vanishes on all intervals implies the function is a.e. zero. But the same method cannot be applied.
Anyway, I cannot think of anything except for let $B$ be a set of measure $1/4$ and try to make the integral 0. However, I forgot that this is on the real line, so there is no monotonicity here. Anyone can help?

Let $E = \{f > 0\}$.  If $m(E) \ge \dfrac 1\pi$ then $E$ contains a subset $A$ with $m(A) = \dfrac 1\pi$ and necessarily $\displaystyle \int_A f > 0$. Thus $m(E) < \dfrac 1\pi$. Likewise, if $F = \{f < 0\}$, then $m(F) < \dfrac 1\pi$.
Define $G = \{f = 0\}$ and note that $m(G) = 1 - \dfrac 2\pi > \dfrac 1\pi$.
Suppose that $m(E) > 0$. Select $H \subset G$ with $m(H) = \dfrac 1\pi - m(E)$ and observe that $\displaystyle \int_{E \cup H} f > 0$, contrary to hypothesis. Thus $m(E) = 0$. Likewise $m(F) = 0$.
