Composition of measurable & continuous functions, is it measurable?

Solution 1:

Edit: following the comment of kahen below, I modified my answer: Lebesgue-measurability of a function $h$ is measurability of $h\colon (X,\mathcal{L}_X)\to (Y,\mathcal{B}_{Y})$, not $h\colon (X,\mathcal{L}_X)\to (Y,\mathcal{L}_{Y})$)

You have that $f\colon (X,\mathcal{L}_X)\to (\mathbb{R},\mathcal{B}_\mathbb{R})$ is Lebesgue-measurable (for the $\sigma$-algebras $\mathcal{L}_X$, $\mathcal{B}_\mathbb{R}$). As $g\colon \mathbb{R}\to \mathbb{R}$ is continuous, it is Borel-measurable ($g\colon (\mathbb{R},\mathcal{B}_\mathbb{R})\to (\mathbb{R},\mathcal{B}_\mathbb{R})$ is measurable for the $\sigma$-algebras $\mathcal{B}_\mathbb{R}$, $\mathcal{B}_\mathbb{R}$). You want to show that $g\circ f$ is Lebesgue-measurable. i.e. $g\circ f \colon (X,\mathcal{L}_X)\to (\mathbb{R},\mathcal{B}_\mathbb{R})$ is measurable.

Take any $B\in\mathcal{B}_\mathbb{R}$: you need to show that $(g\circ f)^{-1}(B)\in \mathcal{L}_X$.

By measurability of $g$, you have that for since $B\in\mathcal{B}_\mathbb{R}$, $B^\prime = g^{-1}(B)\in \mathcal{B}_\mathbb{R}$. By measurability of $f$, this implies that $f^{-1}(B^\prime)\in \mathcal{L}_X$, i.e. $(g\circ f)^{-1}(B)\in \mathcal{L}_X$. This shows that $g\circ f$ is measurable for the $\sigma$-algebras $\mathcal{L}_X$, $\mathcal{B}_\mathbb{R}$ (i.e., $g\circ f\colon (X, \mathcal{L}_X)\to (\mathbb{R}, \mathcal{B}_\mathbb{R})$ is measurable), as wanted.

Solution 2:

Note that $ (g \circ f)^{-1}(A)=f^{-1}(g^{-1}(A))$. You need to show this is Lebesgue measurable if $ A $ is open. Now the key is that if $A$ is open, then so is $ g^{-1}(A) $. So $(g \circ f)^{-1}(A)=f^{-1}(B) $ where $ B $ is open. Conclude from there.