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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?

LanaDR
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2 Answers2

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Usually with "measurable" on the real line "Borel measurable" is meant... if you do also every continuous function is ofc measurable because the preimage of open sets under continuous functions are open sets as well and open sets are a generator of the Borel Algebra

Gono
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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 $(\mathbb{R},\mathcal{B}(\mathbb{R}))\rightarrow (\mathbb{R},\mathcal{B}(\mathbb{R}))$.
Detailed proof:
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.

H.Vogel
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