Integration theory Any help with this problem is appreciated. 
Given the $f$ is measurable and finite a.e. on $[0,1]$. Then prove the following statements
$$ \int_E f = 0 \text{ for all measurable $E \subset [0,1]$ with $\mu(E) = 1/2$ }\Rightarrow f = 0 \text{ a.e. on } [0,1]$$
$$ f > 0 \text{ a.e. } \Rightarrow \inf ~ \left\{\int_E f : \mu(E) \geq 1/2\right\} > 0 $$
 A: For $(1)$, we can define the sets $ P:= \{x:f(x)\ge 0\}$ and $ N:=\{x:f(x)\le 0\}$. Then either $\mu(P)\ge \frac{1}{2}$ or $\mu(N)\ge \frac{1}{2}$. Suppose $\mu(P)\ge \frac{1}{2}$, define $SP:= \{x:f(x)> 0\},$ then $\ SP\subset P$. If  $ \mu(SP)<\frac{1}{2}$, we can choose a set $E$ such that $$SP \subset E,\ f(x)\ge 0\  on \ E,\  and\  \mu(E)=\frac{1}{2}$$ According to the hypothesis, $\int_Ef=0$ which implies $\mu(SP)=0$($f$ is non-negative on $E\ \Rightarrow \ f=0\ a.e.\ on\ E$). I think you are able to show the case when $\mu(SP)>\frac{1}{2}$. Similarly, if we define $ SN:=\{x: f(x)<0\} $, we can show $\mu(SN)=0$.
For $(2)$, forst define $A_n:=\{x:f(x)>\frac{1}{n}\} $, then we know $A_n$ is incresing and $\lim_{n\to \infty}\mu(A_n)=1$ since $f>0\ a.e.$. Fix sufficiently large $n_0$ so that $\mu(A_{n_0})>1-\epsilon_0$, then for any $E$ with $\mu(E)\ge \frac{1}{2}$, we have 
$$ \int_Ef\,\mathrm{d}\mu=\int_{E\cap A_{n_0}^c} f\,\mathrm{d}\mu+\int_{E\cap A_{n_0}} f\,\mathrm{d}\mu\ge \int_{E\cap A_{n_0}} f\,\mathrm{d}\mu\ge \frac{1}{n_o}\cdot \mu(E\cap A_{n_0})$$
Note that $\mu(E\cap A_{n_0})\ge \mu (E)+\mu (A_{n_0})-1> \frac{1}{2}+(1-\epsilon_0)-1=\frac{1}{2}-\epsilon_0$, hence $\int_E f\,\mathrm{d}\mu>\frac{1}{n_0}\cdot (\frac{1}{2}-\epsilon_0)>0$.
