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Here's the question I'm having a go at:

"Prove that if $f$ and $g$ are measurable then $fg$ is also measurable (express the product using sums and powers of functions)"

I've had a look for the proof that the pointwise sum of measurable functions is measurable and this is relatively simple, however I'm not sure how to do this question using the tip in brackets. Any help or tips are much appreciated, thanks.

user416936
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2 Answers2

27

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

Demophilus
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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 $a\mapsto f(a)g(a)$.

drhab
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