Continuous Composition of Measurable Functions into Second Countable Space is Measurable

Theorem
Let $\struct {X, \Sigma}$ be a measurable space.

Let $\struct {X_i, \tau_i}$ for $i = 1, \ldots, n$ and $\struct {Y, \tau_Y}$ be topological spaces such that $X_1, \ldots, X_n$ are second countable.

Let $f_i: \struct {X, \Sigma} \to \struct {X_i, \map \BB {X_i, \tau_i} }$, $i = 1, \ldots, n$ be measurable functions where $\map \BB {X_i, \tau_i}$ denotes the Borel-sigma algebra.

Let $F: \struct {\ds \prod_{i \mathop = 1}^n X_i, \tau} \to \struct {Y, \tau_Y}$ be continuous where $\tau$ denotes the product topology on the finite Cartesian product $\ds \prod_{i \mathop = 1}^n X_i$.

Then the composition $h: \struct {X, \Sigma} \to \struct {Y, \map \BB {Y, \tau_Y} }$, $x \mapsto \map F {\map {f_1} x, \ldots, \map {f_n} x}$ is measurable.

Proof
By Mapping Measurable iff Measurable on Generator, it suffices to check that $h$ is measurable on open sets.

Thus let $U \in \tau_Y$ be given.

As $F$ is continuous by assumption, the pre-image $F^{-1} \sqbrk U$ is in $\tau$.

By:
 * definition of basis
 * Countable Product of Second-Countable Spaces is Second-Countable

there exists an at most countable indexing set $J \subseteq \N$ and indexed family of open sets $\family {U^i_j}_{j \mathop \in J} \subseteq \tau_i$, $i = 1, \ldots, n$ such that:
 * $F^{-1} \sqbrk U = \ds \bigcup_{j \mathop \in J} \prod_{i \mathop = 1}^n U^i_j$

Putting things together:

By assumption each $f_i$ is measurable.

By definition of the Borel-sigma algebra:
 * $\tau_i \subseteq \map \BB {X_i, \tau_i}$

Hence it follows that every $f_i^{-1} \sqbrk {U^i_j}$ is in $\Sigma$.

Therefore by Sigma-Algebra Closed under Countable Intersection:


 * $\ds \bigcap_{i \mathop = 1}^n f_i^{-1} \sqbrk {U^i_j}$

is also in $\Sigma$.

Finally, we note that by definition of $\sigma$-algebra, $\Sigma$ is an at most countable union of measurable sets.

Hence it follows that:
 * $h^{-1} \sqbrk U = \ds \bigcup_{j \mathop \in J} \bigcap_{i \mathop = 1}^n f_i^{-1} \sqbrk {U^i_j}$ is in $\Sigma$

This concludes the proof.