Product of measurable functions is measurable

Solution 1:

Hint: Try looking at $f^2$, $g^2$ and $(f+g)^2$.

Solution 2:

I preassume that $f,g:A\to\mathbb R$ where $\langle A,\mathcal A\rangle$ is a measurable space and $\mathbb R$ and $\mathbb R^2$ are both equipped with their usual Borel-$\sigma$-algebras.

If $\mathcal B$ denotes the Borel-$\sigma$-algebra on $\mathbb R$ then the Borel-$\sigma$-algebra on $\mathbb R^2$ equals $\mathcal B^2$.

Now let $A^2$ be equipped with the $\mathcal A^2$. Then:

• Function $\delta:A\to A^2$ prescribed by $a\mapsto\langle a,a\rangle$ is measurable.

• If $f,g$ are measurable then so is $f\times g:A^2\to\mathbb R^2$ prescribed by $\langle a,b\rangle\mapsto\langle f(a),g(b)\rangle$.

• Function $\times:\mathbb R^2\to\mathbb R$ prescribed by $\langle x,y\rangle\mapsto xy$ is continuous, hence measurable.

Then the composition: $$\times\circ(f\times g)\circ\delta:A\to\mathbb R$$ is measurable, and it is prescribed by $a\mapsto f(a)g(a)$.