# Proving that a function is measurable

Let $(X,\mathcal{A},\mu)$ be a measure space and $(Y,d)$ a separable metric space. Given $f,g:X\to Y$ $\mu$-measurable functions, prove that $h:X\to \mathbb{R}$, defined by $x\mapsto d(f(x),g(x))$ is measurable.

I already know that the function $x\mapsto (f(x),g(x))$ from $X$ to $Y\times Y$ is measurable, but I don't know how to prove that $d$ also is, or find a way to use the separableness of $Y$ to prove that the composition is measurable.

• hint: d is continuous. – Matematleta Jun 17 '15 at 0:43

The mapping $d:Y\times Y \to \mathbb{R}$ is continuous with the product topology. Given any open set $U\subset \mathbb{R}$, $d^{-1}(U)$ is open in $X\times Y$. Hence, $\phi^{-1}(d^{-1}(U))$ is $\mu$-measurable, where $\phi: X\to Y\times Y$ is the map $x\mapsto (f(x),g(x))$. Hence, $(d\circ \phi)^{-1}(U)$ is measurable, giving us that $d\circ \phi$ is a measurable function.

• But that is assuming that the $\sigma$-algebra of the measurable sets of $Y$ contains the open sets. – jiyanez Jun 17 '15 at 0:49
• True, but you did not specify any $\sigma$-algebra on $Y$, in which case I believe the convention is to take the $\sigma$-algebra to be the Borel $\sigma$-algebra generated by open sets. – Andrew Jun 17 '15 at 0:50
• In that case the separableness of $Y$ doesn't matter. – jiyanez Jun 17 '15 at 0:52