Let $f, g : \mathbb{R} \rightarrow \mathbb{R}$ be measurable functions. Show that: $(x, y) \rightarrow f(x) + g(y)$ is measurable in the sense of the product measure.
Ok, I figured out a way to do this by proving that $(x,y) \rightarrow (f(x), g(y))$ is measurable and then move on to prove that $(x,y) \rightarrow x+y$ is measurable and finally apply these results to the compound of the two. However I am unable to prove that $x+y$ is measurable. Would anyone care to give me a tip?
On
A more complicated answer: Let f,g be measurable real functions.Let $a,b\in R$ with $a<b.$ Let $c=(b-a)/2.$ Let $h=f+g.$ Define $$S(a,b)=\cup_{q\in Q}[f^{-1}(a-q,a-q+c)]\cap g^{-1}(q,q+c).$$ Then $S(a,b)=h^{-1}(a,b)$ because:
(1) If $h(x)\in (a,b),$ take $q\in (f(x)-c,f(x))\cap Q,$ so $f(x)\in (a-q,a-q+c),$ and we must have also $g(x)\in (q,q+c)$ to satisfy $f(x)+g(x)\in (a,b)=(a,[a-q+c]+[q+c]).$ So $x\in S(a,b)$.... And
(2) If $x\in S(a,b)$ then for some $q\in Q$ we have $h(x)\in ([a-q]+q,[a-q+c]+[q+c])=(a,b).$
Every $S(a,b)$ is a measurable set by the def'n of $S(a,b).$ So $h^{-1}V$ is a measurable set whenever $V$ is a bounded open interval. So $h^{-1}W$ is a measurable set whenever $W$ is open, because $W=\cup H$ where $H$ is a countable set of bounded open intervals, and $h^{-1}W=\cup_{U\in H}h^{-1}U.$
The map $(x,y) \mapsto x + y$ is measurable because it is continuous. Do you understand why?