Why is f bounded? I am looking at the exercise: Let $f:[0,1] \to \mathbb{R}$ integrable such that $|f(x)| \leq \int_0^x |f(t)|dt, \forall x \in [0,1]$.
Show that $f=0$.
At the solution,it is taken that $f$ is bounded. Why??Because of the fact that $f$ is integrable?

$f$ is bounded because $f$ is integrable and satisfies your inequality. Notice that the integral is monotonically increasing in $x$. In other words:
$$|f(x)|\leq\int_{0}^{x}|f(t)|dt\leq\int_{0}^{1}|f(t)|dt=C,$$
For some finite $C$. 
As a hint for the exercise, try integrating both sides of the inequality from $0$ to $x$ and notice that if $f$ is bounded, then $\int_0^y\int_0^xf(t)dtdx\leq yC$. On the other hand, $|f(y)|\leq\int_0^y|f(x)|dx$. What happens if you do three integrals...four integrals...
