Show an integrable function a.e. 0 $f\in L^1[0,1]$, there is a constant $0<c<1$, suppose for any measurable set $A\in [0,1]$ with $m(A)=c$, we have 
$$\int_A f =0$$
Prove that $f=0$ a.e.
I know how to prove this when $f$ is nonnegative, but I don't know how to deal with this version. Any suggestion?
Thanks!
 A: Let $S_R = \{x | f(x)\, R \, 0 \}$ where $R \in \{<,=,>\}$. If $mS_< =0$ then its easy to see that we must have $mS_> = 0$, and similarly if $m S_> = 0$, we must have $mS_< = 0$.
So, suppose $m S_< >0, m S_> >0$.
The idea of the rest of the proof is to show that if we have two subsets $A,B$ such that $mA=mB$ and $mA$ is small enough, then $\int_A f = \int_B f$. Then by choosing $A \subset S_<, B \subset S_>$ we obtain a contradiction.
Note that if $mA >0$, then by considering the continuous function
$t \mapsto m(A \cap[0,t])$ we can see that for any $\lambda \le mA$, we can find
a subset $B\subset A$ such that $mB = \lambda$.
Now choose two sets $A,B$ such that $mA=mB$ and $mA \le \min(1-c,{1 \over 2} c)$. Note that $m([0,1]\setminus (A\cup B)) \ge 1-2mA$, so we can find some subset $C$
such that $C \subset [0,1]\setminus (A\cup B)$ and $mC = c-mA$ (because
$mA \le 1-c$ implies $1 -2 mA \ge c-mA$). Then $m (A \cup C) = m(B \cup C) = c$, and so
$\int_A f + \int_C f = 0 = \int_Bf + \int_C f$, from which we get
$\int_A f = \int_B f$.
Now let $\lambda = \min(m S_<,m S_>, {1 \over 2} c)>0$, and choose $A \subset S_<, B \subset S_>$ such that $mA = mB = \lambda$. Then we have $0>\int_A f = \int_B f > 0$
which is a contradiction.
Hence we must have $m S_< = m S_> = 0$.
