Show $\int_Af(x)\,dx=0$ for every measurable subset $A$ of $[0,1]$. Let $f$ be Lebesgue-integrable on $[0,1]$. Suppose $\int_a^bf(x)\,dx=0$ for all $0\leq a\leq b\leq 1$. Show $\int_Af(x)\,dx=0$ for every measurable subset $A$ of $[0,1]$. 
*Let $A$ be a measurable subset of $[0,1]$. Then $A$ can be written as the union of disjoint, countable? intervals. Since $\int_a^bf(x)\,dx=0$ for all $0\leq a\leq b\leq 1$, each integral of $f$ over each interval is $0$ so $\int_Af(x)\,dx=0$. 
I'm not sure if I did it right...
 A: The Lebesgue measure is regular, and in particular outer regular. Therefore there exists a sequence $(O_n)$ of open measurable sets containing $A$ such that $\lambda(A)=\inf \lambda(O_n)$.
Then $\displaystyle \int_A f(x)\, \mathrm{d}x=\int_{O_n} f(x)\, \mathrm{d}x - \int_{O_n - A} f(x)\, \mathrm{d}x$.
Since $O_n$ is an open set of $\mathbb{R}$, it is a countable union of open intervals and $\int_{O_n} f(x)\, \mathrm{d}x=0$. By dominated convergence you can show that $\int_{O_n - A} f(x)\, \mathrm{d}x \to 0$: use the sequence of functions $f1_{O_n-A}$ dominated by $|f|$.
A: This is a standard application of Dynkin's $\pi$-$\lambda$ Theorem.
Define $\mathcal{P}=\{(a,b]\mid0\leq a<b\leq1\}\cup\{\emptyset\}$
and $\mathcal{L}=\{A\in\mathcal{B}([0,1])\mid\int_{A}f(x)\,dx=0\}$.
It is routine to verify that $\mathcal{P}$ is a $\pi$-class and
$\mathcal{L}$ is a $\lambda$-class. That is:


*

*For any $A,B\in\mathcal{P}$, we have $A\cap\mathcal{B\in\mathcal{P}}$,

*$\emptyset\in\mathcal{L}$,

*$A^{c}\in\mathcal{L}$ whenever $A\in\mathcal{L}$, where the complement
is taken with respect to $[0,1]$,

*$\cup_{i=1}^{\infty}A_{i}\in\mathcal{L}$ whenever $A_{i}\in\mathcal{L}$
and $A_{1},A_{2}\ldots$ are pairwisely disjoint.


By the given condition, we have $\mathcal{P}\subseteq\mathcal{L}$.
Therefore, by Dynkin's $\pi$-$\lambda$ Theorem, we have $\sigma(\mathcal{P})\subseteq\mathcal{L}$.
Note that $\sigma(\mathcal{P})=\mathcal{B}([0,1])$. Therefore, $\mathcal{L=B}([0,1])$.
Q.E.D.
