Give an example where composite function is non-measurable Give an example of a continuous function $g$ and a measurable function $h$ such that $h \circ g$ is non-measurable. 
 A: If you are considering the Borel $\sigma$-algebra on the codomain of $g$ you won't find an example: in that case, if $g$ is continuous then it is measurable, and the composition of measurable functions is still measurable. 
If you are considering the $\sigma$-algebra of Lebesgue measurable sets instead, there is an example: see the answer to this post. You can check Wikipedia's article on the Cantor function which the answerer calls the devil's staircase function.
The key to constructing that example lies first in the fact that the Cantor function maps a set of Lebesgue measure $0$ (the Cantor set) to the set of positive measure $[0,1]$ continuously, and secondly that every Lebesgue measurable set of positive measure contains a non-measurable set. 
The cantor function is not injective, because it is almost everywhere constant. 
The addition of the identity $(f(x)=x)$ fixes that issue and makes it a continuous bijection with continuous inverse. From the argument in that answer it follows that its continuous inverse is non-measurable.
Note: I don't really see the role of the composition in your question, but once you find the non-measurable continuous function you could take $h$ to just be the identity.
