# Measurable function (Composite function)

Let $(\Omega,\Sigma)$ be a given measurable space and let $f$ be a $\Sigma$-measurable function. If $h:[-\infty,\infty]\rightarrow[-\infty,\infty]$ is a continuous function, then the composite function $hf$ is measurable.

This can be proven easily if every continuous function is measurable as I just need to show that the inverse image of $hf$ is measurable for a given measurable set say $E$.

i.e. $(hf)^{-1}(E)=f^{-1}(h^{-1}(E))\in\Sigma$

But is a continuous function measurable? Or this can be proven in another way?

As the composition $hf$ needs to be well-defined, I assume the following domain and target sets of $f$ and $h$: $f:\Omega \rightarrow \mathbb{R}$ and $h: \mathbb{R}\rightarrow \mathbb{R}$ (as $(-\infty,\infty)=\mathbb{R}$).
Thus we have a measurable function $h$ from $(\Omega, \Sigma) \rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}))$ and a continuous function $f$ from $(\mathbb{R},\mathcal{B}(\mathbb{R})) \rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}))$, where I assumed suitable $\sigma$-Algebras for the function $h$, which I hope are what you are looking for.
Statement: The composition $hf:(\Omega,\Sigma)\rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}))$ is measurable.
The proof comes from the fact that since $h$ is a continuous function from $\mathbb{R}\rightarrow \mathbb{R}$, it is a measurable function from $(\mathbb{R},\mathcal{B}(\mathbb{R}))\rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}))$.
As $\mathcal{B}(\mathbb{R})$ is generated by open subsets of $\mathbb{R}$ it is sufficient to check measurability of $hf$ on an open subset $A\subset \mathbb{R}$, i.e. we have to check that $(hf)^{-1}(A)\subset \Omega$ for an open set $A\subset \mathbb{R}$. As $h$ is continuous and $A$ is open, $h^{-1}(A)$ is open, i.e. $h^{-1}(A)\in \mathcal{B}(\mathbb{A})$. As $f$ is measurable $f^{-1}(h^{-1}(A))\in \Sigma$, thus $(hf)^{-1}(A)= f^{-1}(h^{-1}(A))\in \Omega$ and we conclude.