If $f,g$ are measurable and $\Phi$ is continuous, then $\Phi(f(x),g(x))$ is measurable.

Solution 1:

Sketch: you need to show that $\{x:h(x)>a\}$ is measurable for each $a\in \mathbb R.$ So,

set $U=\{(u,v):\Phi(u,v)>a\}.\ U$ is open because

$\Phi$ is continuous.

so it is a countable union of rectangles $U=\bigcup_n (a_n,b_n)\times (c_n,d_n).$

Now, for each integer $n,\ \{x:(f(x),g(x))\in (a_n,b_n)\times (c_n,d_n)\}$ is measurable because

it is equal to $\{x:a_n<f(x)<b_n\}\cap \{c_n<g(x)<d_n\}$ and $f$ and $g$ are measurable.

But, $\{x:h(x)>a\}=$

$\{x:(f(x),g(x))\in U\}=\bigcup_n\{x:(f(x),g(x))\in (a_n,b_n)\times (c_n,d_n)\}$

from which the claim follows because

countable unions of measurable sets are measurable.