Given x->f ,y->g are measurable, show that (x,y)->f(x)g(x) is measurable

Question

So I said that since f and g are measurbale they are the limit of step functions ($\phi$$_n$ and $\theta$$_n$). Then that f(x)g(x) = lim($\phi$$_n$)lim($\theta$$_n$). I don't think that last point is correct though. Some help would be appreciated.

Thanks.

Answer 1

you can prove this without step functions.

$f+g$ and $f-g$ and $f^2$ are measurable thus

$fg=\frac{(f+g)^2-(f-g)^2}{4}$ is measureable.