Question on the integrability of the product of functions If $f \in L^2([0,1]$ and $g$ continuous on [0,1] , then why the product $fg$ is in $L^2([0,1])$?
By definition, we have $\left(\int_{[0,1]} |f|^2 d\mu \right)^{1/2} < + \infty$, why do we have $\left( \int_{[0,1]} |fg|^2 d\mu \right)^{1/2} < + \infty$?

[0,1] is compact set in $R^1$, continuous function on compact set is bound that
for some M, $|g| < M$,
$$
\left( \int_{[0,1]} |fg|^2 d\mu \right)^{1/2} <  \left( M^2\int_{[0,1]} |f|^2 d\mu \right)^{1/2} < \infty
$$
