If $f,g$ are measurable and $\Phi$ is continuous, then $\Phi(f(x),g(x))$ is measurable. Let $E\subseteq\mathbb{R}^d$ be a (Lebesgue) measurable set and let $f,g$ be two measurable functions defined on E. I would like to show that if $\Phi$ is a continuous function on $\mathbb{R}^2$, then the function $h:x\mapsto\Phi(f(x),g(x))$ is measurable. The proof remains unknown to me, but I can address the problem if it is only one-dimensional. Specifically, if $\Phi$ is a continuous function on $\mathbb{R}$, then I can show that $\Phi\circ f$ is measurable. Indeed, since $\{\Phi<a\}$ is an open set $G$, we can conclude that $\{\Phi\circ f<a\}=f^{-1}(G)$ is measurable. How about the two-dimensional problem? Does anyone have an idea? Thank you.

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.

