Show that $\int _E f=0$ for each subset $E $ of $\mathbb R $ of finite Lebesgue measure Let $ f : \mathbb R \rightarrow \mathbb R$ be a bounded Lebesgue measurable function such that $\int_a^b f =0$ for all real $a,b.$ 
Show that $\int _E f=0$ for each subset $E $ of $\mathbb R $ of finite Lebesgue
measure
Actually I am new to measure theory.So maybe above is simple I can't proceed

Given any open set $O \subset \mathbb{R}$, $\exists \{I_k\}_{k=1}^{\infty}$ disjoint collection of open intervals. Thus, $\int_{O}{f}=\sum\limits_{k=1}^{\infty}\int_{I_k}{f}=\sum\limits_{k=1}^{\infty}{0}=0$. Thus, from our assumption we get that the integral over any open set is zero. Similarly, we get for any $K \subset \mathbb{R}$ closed $\int_{K}{f}=\int_{\mathbb{R}}{f}-\int_{\mathbb{R}-K}{f}=0$. Define $F^+=\{x:f(x)>0\}$ and $F^-=\{x:f(x)<0\}$. Using inner regularity of Lebesgue measure it is not hard to see that $m(F^+)=m(F^-)=0$ and thus $f=0$ a.e. Thus, we get for any measurable set E we get that$\int_{E}{f}=0$. 
