If $\displaystyle f \in L^1(E)$ is bounded, and E is of finite measure, i.e $\displaystyle m(E)<\infty$, (m here is the lebesgue measure but it can be arbitrary).

Is $\displaystyle f \in L^2(E)$?

Is it enough to say:

since $\displaystyle |f|<M$ for some real M.

$\displaystyle \int _E \left|f\right|^2dm<\int _EM^2dm=M^2\int _Edm=M^2m(E)<\infty $

But what I dont get is that this argument only uses the fact f is bounded, not the fact that f is integrable.

I know that $\displaystyle L^2(E) \subseteq L^1(E)$ since E is of finite measure. I wonder if this has something to do with the answer?