Let $u$ and $v$ be real measurable functions on a measurable space $X$, let $\phi$ be a continuous mapping of the plane into a topological space $Y$, and define $h(x)=\phi(u(x),v(x))$ for $x\in X$. Then $h:X\to Y$ is measurable.
My attempt: It’s suffice to show, $f:X\to \Bbb{R}^2$ defined by $f(x)=(u(x),v(x))$ is measurable. $\Bbb{R}$ is second countable. By exercise 8(a) section 13 of Munkres’ topology, $\mathcal{B}_{\Bbb{R}}=\{(a,b)| a\lt b$ and $a,b \in \Bbb{Q}\}$ is countable basis of $\mathcal{T}_{\Bbb{R}}$. $\Bbb{R}^2$ is second countable. By exercise 6 section 16 of Munkres’ topology, $\mathcal{B}_{\Bbb{R}^2}=\{ P\times Q| P,Q\in \mathcal{B}_\Bbb{R}\}=\{(a,b)\times (c,d)| a\lt b$, $c\lt d$ and $a,b,c,d \in \Bbb{Q}\}$ is countable basis of $\mathcal{T}_{\Bbb{R}^2}$. Let $V\in \mathcal{T}_{\Bbb{R}^2}$. Then $V=\bigcup_{i\in I}B_i$ ; $B_i \in \mathcal{B}_{\Bbb{R}^2}$, $\forall i\in I$. So $B_i =P_i \times Q_i$, where $P_i,Q_i \in \mathcal{B}_\Bbb{R} \subseteq \mathcal{T}_\Bbb{R}$. By elementary set theory, $f^{-1}(B_i)=f^{-1}(P_i \times Q_i)=u^{-1}(P_i)\cap v^{-1}(Q_i)$. Since $u$ and $v$ are measurable functions, we have $u^{-1}(P_i), v^{-1}(Q_i)\in M$ ($\sigma$-algebra of $X$). So $f^{-1}(B_i)=u^{-1}(P_i)\cap v^{-1}(Q_i) \in M$, $\forall i\in I$. By subset of countable set is countable, $\{B_i|i\in I\}\subseteq \mathcal{B}_{\Bbb{R}^2}$ is countable. $f^{-1}(V)=f^{-1}(\bigcup_{i\in I}B_i)=\bigcup_{i\in I}f^{-1}(B_i)$. By definition of $\sigma$-algebra, $f^{-1}(V)=\bigcup_{i\in I}f^{-1}(B_i) \in M$. Hence $f$ is measurable. By theorem 1.7 of Papa Rudin, $h$ is measurable. Is this proof correct?
My solution is inspired by very similar problem, Theorem 18.4 of Munkres’ Topology. IMO, Rudin’ proof is not “precise” and left out most important details in proof.