Essential supremum product inequality. Suppose that $f,g\in\mathcal{L}_{\infty}(X,\Sigma,\mu)$ where $(X,\Sigma, \mu)$ is a measure space. 
I am trying to show that $||fg||_{\infty} \leq ||f||_{\infty}||g||_{\infty}$.
I know that $|f(x)|\leq ||f||_\infty$ $\mu$-almost everywhere but I'm not even sure if this is useful since I'm not sure if we can say that $|f(x)||g(x)|\leq||f||_{\infty}||g||_{\infty}$ almost everywhere. 
How should I be approaching this?

Let 
$$
A := \{x\in X:\ |f(x)| \leq \|f\|_\infty\},
\qquad
B := \{x\in X:\ |g(x)| \leq \|g\|_\infty\}.
$$
Since $\mu(X\setminus A) = \mu(X\setminus B) = 0$, you have that $\mu(X\setminus (A\cup B)) = 0$.
Hence
$$
|f(x) g(x)| \leq \|f\|_\infty \|g\|_\infty
\qquad \forall x\in A\cup B,
$$
i.e. $\mu$-a.e.
By definition, this fact implies that $\|fg\|_\infty \leq \|f\|_\infty \|g\|_\infty$.
