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Given an example of a continuous function $\phi$ and a lebesgue measurable function $F$. both defined on $[0,1]$, such that $F\circ\phi$ is not lebesgue measurable.

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Here is an example, taken from here:

Let $f:[0,1]\to \mathbb R$ be the Cantor function. It has range $[0,1]$. Define the function $g:[0,1]\to \mathbb R$ by $g(x)=x+f(x)$. The function $g$ has range $[0,2]$, is continuous and injective on $[0,1]$ and has a continuous inverse on its range. It also maps the Cantor set $C\subset [0,1]$ to a set of masure $1$, i.e, $m(g(C))=1$. Since $m(g(C))=1$, there exists a nonmeasurable set $D$ contained in $g(C)$. Then $E= g^{-1}(D)$ is contained in $C$ so it has measure zero. Define $h$ to be the characteristic function of $E$. Then $h$ is measurable on $[0,1]$ but $h(g^{-1})$ is the nonmeasurable characteristic function of the non-measurable set $D$.

As pointed out, originally this example was taken from the Dover book "Counterexamples in Analysis" by Gelbaum/Olmsted. Hope this helps.

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