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Let $f,g:[0,1]\rightarrow [0,1]$ be measurable functions.Is $g\circ f$ measurable or not?

The composition is definitely measurable from the axiom definition of measurable function. But if we want to prove it from the classic definition of measurable function,we have to prove that preimage of an open set of this composition is measurable. In order to do that,we have to prove that the preimage of a mesurable set by a measurable function is a measurable set.Now I don't know what to do with the mesurable sets in $[0,1]$ except for the Borel sets.

Jack
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2 Answers2

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

user288742
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It's not true.

Let $c$ be the Cantor "devil's staircase" function, so that $c$ is continuous and $g(C) = [0,1]$ where $C$ is the Cantor set. Let $f(x) = \sup\{y: c(y) = x\}$ for $x \in [ 0,1]$, $-1$ otherwise. Since $f$ is increasing on $[0,1]$, it is measurable. $f$ maps $[0,1]$ into the Cantor set. Let $E$ be a nonmeasurable subset of $[0,1]$. There is a subset $V$ of the Cantor set with $f^{-1}(V) = E$. Define $g$ by $g(x) = x$ if $x \in V$, $g(x) = -1$ otherwise. Since $g = 0$ a.e., $g$ is measurable. But $(g \circ f)^{-1}((0,\infty)) = E$ is nonmeasurable.

Robert Israel
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