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
 A: Let $F(x)=\int_0^x f.$ Then $F\equiv 0$ from the given hypothesis. Therefore $F'\equiv 0.$ But $F'(x) = f(x)$ for a.e. $x$ by the Legesgue differentiation theorem. Thus $f=0$ a.e., hence $\int_E f = 0$ for any measurable set $E.$ (Using a big gun there, but thought I'd toss this in.)
A: Hint: We have, $$f = f^+-f^-$$ where $f^+$ and $f^-$ denote the positive and negative part of $f$, respectively. By assumption, the ($\sigma$-finite) measures
$$\nu(dx) := f^+(x) \, dx \qquad \mu(dx) := f^-(x) \, dx$$
satisfy
$$\mu((a,b)) = \nu((a,b)).$$
Conclude from the uniqueness of measure theorem that $\mu = \nu$ on $\mathcal{B}(\mathbb{R})$.
A: 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$\int_{E}{f}=0$. 
A: Hint: We can find a countable collection of open intervals $I_k$ such that
$$
E \subset U = \bigcup_{k \in \Bbb N} I_k, \qquad
\left(\sum_{k=1}^\infty m(I_k) \right) - m(E) < \epsilon
$$
Now, note that 
$$
\left| 
\int_U f\,dx - \int_E f\,dx 
\right| = 
\left|
\int_{U \setminus E} f\,dx
\right|
$$

Alternative: Show that $\int_U f\,dx = 0$ whenever $U$ is a $G_\delta$ set.  Then, note that there is a $U$ such that $E \subset U$ with $m(U \setminus E) = 0$.
