Below, assume that all integrals/expectations are finite.
If the measure is a probability measure, then your question reads $ {\rm E}|XY| \le {\rm E}|X|{\rm E}|Y|? $ (Here $X$ and $Y$ are random variables.) It does hold $ |{\rm E}(XY)|^2 \le {\rm E}(X^2 ){\rm E}(Y^2 ). $
Note that $X$ and $Y$ (which are measurable functions from $\Omega$ to $\mathbb{R}$) correspond to $f$ and $g$. That is, the correct inequality is $ \bigg|\int_\Omega {fg\,d\mu } \bigg|^2 \le \bigg(\int_\Omega {f^2 \,d\mu } \bigg)\bigg(\int_\Omega {g^2 \,d\mu } \bigg) $ (generalized below), where $\mu$ is the probability measure.
Your inequality, however, is evidently false (in general), for if $X$ is nonnegative, it gives ${\rm E}(X^2 ) \le {\rm E}^2 (X)$, or ${\rm Var}(X) \leq 0$.
EDIT: More (but not most) generally, if $p,q \in (1,\infty)$ satisfy $1/p + 1/q = 1$, then $ {\rm E}|XY| \le ({\rm E}|X|^p )^{1/p} ({\rm E}|Y|^q)^{1/q}, $ or $ \int_\Omega {|fg|\,d\mu } \le \bigg(\int_\Omega {|f|^p \,d\mu } \bigg)^{1/p} \bigg(\int_\Omega {|g|^q \,d\mu } \bigg)^{1/q}. $ For further details, see here (note the section "Generalization for probability measures").