Is Borel measurable function composed with Lebesgue measurable function a Lebesgue measurable function?

Question

Let $g$ be a Borel measurable function and $f$ be a Lebesgue measurable function.

Then, is $g(f(x))$ a Lebesgue measurable function?

Answer 1

Yes, because Borel/Lebesgue measurability means that $f^{-1}(A)$ is Borel/Lebesgue measurable for every Borel set $A$.

Hence,

$$ (g \circ f)^{-1}(A) = f^{-1} ( g^{-1}(A)) $$

is Lebesgue-measurable, because $g^{-1}(A)$ is Borel-measurable.

But composition the other way around is in general not measurable.