Prove composition is a measurable function

Question

Let $(X, \Sigma )$ be a measurable space and $f:X \rightarrow \mathbb R$ a measurable function. If $h: \mathbb R \rightarrow \mathbb R $ is continuous, then prove that $h \circ f$ is a measurable function.

I unfortunately lost my lecture notes so I really need some help with this module (linear analysis).

Please can someone guide me on how to do this question.

Thanks

Answer 1

Since $h$ is continuous, $h^{-1}((-\infty ,a))$ is open and thus borelien. Since $f$ is mesurable and $h^{-1}((-\infty ,a))$ is borelien, $f^{-1}(h^{-1}((-\infty ,a)))$ is measurable. Therefore $h\circ f$ is measurable.

Answer 2

On an abstract level, the composition of two measurable maps $f:X\to Y$ and $g:Y\to Z$ is measurable; this is very easy to verify from the definition of a measurable map.

The only remaining step is to prove, as Rick did, that continuous real maps are measurable with respect to the Borel sigma-algebra.