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?

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?

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