Prove that the Lebesgue integral of $f\chi$ equals to $0$ indicates that $f=0$ a.e. 
Suppose $f: [0,1] \to \mathbb{R}$ is bounded, measurable, and $$\int_{[0,1]}f \chi_{[0,a)}\, d\mu = 0$$ for all $a \in [0,1]$.
Prove that $f=0$ a.e.

I know that if $\int_{[0,1]}f\, d\mu = 0$, then $f=0$ a.e., but I don't really understand why there is a simple function here.
 A: This might be elaborate, but it uses your hint and very basic theorems.
We work on the hint you were given, to show $\int_{(a,b)} f \, d\mu = 0$ for any $a,b \in [0,1]$
Suppose $f: [0,1] \to \mathbb{R}$ is bounded, measurable, and $\int_{[0,1]}f \chi_{[0,a)}\, d\mu = 0$ for all $a \in [0,1]$. 
This means that $\int_{[0,a)} f \, d\mu =0$ for any $a$. Likewise, 
$\int_{[0,b)} f \, d\mu =0$ for any $b \in [0,1], b>a$. Then $[a,b) \cup [0,a) = [0,b)$, and this is a disjoint union. So $$0+\int_{[a,b)} f \, d\mu=\int_{[0,a)} f \, d\mu +\int_{[a,b)} f \, d\mu =\int_{[0,b)} f \, d\mu =0$$
Hence $\int_{[a,b)} f \, d\mu = 0$, for any $a,b$. 
Now $[a,b) = (a, b) \cup \{a\}$. This is a disjoint union. Therefore, $$\int_{(a,b)} f \, d\mu = \int_{[a,b)} f \, d\mu - \int_{\{a\}} f \, d\mu = \int_{[a,b)} f \, d\mu - 0=\int_{[a,b)} f \, d\mu =0$$ 
$\blacksquare$
Now suppose for contradiction that $f=0$ a.e. on $(a,b)$ for any $a, b \in [0,1], a\neq b$ is false. So there exists some $(a,b)$ (necessarily non-empty interval) such that $f\neq 0$ on some subset $E$ of $(a,b)$, of positive measure. 
Now if $f\neq 0$ on some subset $E$, with $\mu(E)>0$, then $f>0$ or $f<0$ on some $B \subseteq E$, with $\mu(B) >0$.  WLOG assume $f>0$ on $B$. 
Take a closed subset of $C$ of $B$ of positive measure. $C$ exists, by If $E$ is Lebesgue measurable, show that there exists a closed set $F$ with $F \subset E$ and $m(E\setminus F)<\epsilon$ (if $\mu(B) =\epsilon>0$, you can always find a closed $C\subseteq B$ such that $\mu(C) > \frac{\epsilon}{2}$ for example).
We then have $\int_{C} f\, d\mu >0$. Since $C$ is closed, $\mathbb{R}\setminus C$ is  open, and hence $(a,b)\setminus C = (\mathbb{R}\setminus C) \cap (a,b)$ is open.
Since $(a,b)\setminus C$ is open, $(a,b)\setminus C$ is a countable disjoint union of open intervals, i.e. for some $x_k, y_k \in [0,1]$, $(a,b)\setminus C=\bigcup_{k=1}^\infty (x_k,y_k)$. Any open subset of $\Bbb R$ is a at most countable union of disjoint open intervals. [Collecting Proofs]
Therefore $0=\int_{(a,b)}f=\sum_{k=1}^\infty\int_{(x_k,y_k)}f\, d\mu+\int_{C} f\, d\mu$ since these are all disjoint sets and the Lebesgue integral is countably additive: Prove the Countable additivity of Lebesgue Integral..
As $\int_{C} f \, d\mu>0$, $\sum\limits_{k=1}^\infty\int_{(x_k,y_k)}f\, d\mu<0$. So for some $k$, $$\int_{(x_k, y_k)} f\, d\mu <0$$  
But this is a contradiction, since $\int_{(a,b)} f\, d\mu  = 0$ for all $a,b$.
We can do the same if we had supposed $f <0$ on $B$ earlier.
Therefore, $f = 0$ a.e. on all open intervals $(a,b)$ in $[0,1]$, particularly $(0,1)$. Therefore $f=0$ a.e. on $[0,1]$.
