$(f,g)^{-1}(A) $ for $f,g$ Lebesgue measurable functions and $A$ closed set

Question

If $f,g:\mathbb{R} \rightarrow \mathbb{R}$ Lebesgue measurable functions, $φ:\mathbb{R} \times\mathbb{R}\rightarrow \mathbb{R}$ continuous function, prove that $G:\mathbb{R} \rightarrow \mathbb{R}$, $G(x)=φ(f(x),g(x))$ is Lebesgue measurable.

For $b\in \mathbb{R}$ $φ^{-1}((-\infty,b])\in \mathcal{B}(\mathbb{R^2}) \subseteq \mathcal{M}_{{λ_2}^{*}}$, because $φ$ is continuous so $A=φ^{-1}((-\infty,b])$ is closed set, but what can I say about $(f,g)^{-1}(A) $ ? I can't use any theorem for measurable functions that take values in sets different from $\mathbb{R}$.

Answer 1

Pick $U$ open set in $\mathbb{R}$, then as $\phi$ is continuous $\phi^{-1}(U)$ is open in $\mathbb{R}^2$ and as such can be written as a countable union of closed rectangles. It suffices to write down now what $G^{-1}(U)$ is and use measurability of $f$ and $g$ on closed intervals.

Answer 2

The result follows from theorems that hold generally and are elementary to prove:

Let $L$ denote the Lebesgue sigma-algebra on $\mathbb{R}$, and for topological space $X$, let $B_X$ denote the Borel sigma-algebra on $X$. By assumption, $f$ and $g$ are $(L, B_{\mathbb{R}})$ are measurable. Therefore $(f, g)$ is $(L, B_{\mathbb{R}} \otimes B_{\mathbb{R}})$ measurable since if $E, F \in B_{\mathbb{R}}$, then $(f, g)^{-1}(E \times F) = f^{-1}(E) \cap g^{-1}(F) \in L$. Now we can note that $B_{\mathbb{R}} \otimes B_{\mathbb{R}} = B_{\mathbb{R}^2}$ simply because $\mathbb{R}^2$ is second countable (if $X$, $Y$ are second countable topological spaces, then $B_X \otimes B_Y = B_{X \times Y}$). Thus $(f, g)$ is $(L, B_{\mathbb{R}^2})$ measurable. Since $\phi$ is continuous, $\phi$ is $(B_{\mathbb{R}^2}, B_\mathbb{R})$ measurable. Thus the composition $G = \phi \circ (f, g)$ is $(L, B_{\mathbb{R}})$ measurable.