Self composition measurable

Question

If f is a lebesgue measurable function, is f $\circ$ f lebesgue measurable?

I know that composition of lebesgue measurable functions need not be lebesgue measurable. However I am not able to give an example when functions we are composing are same.

Answer 1

Let $g,h : [0,1] \to [0,1]$ be two Lebesgue measurable function such that $g \circ h$ is not Lebesgue measurable.

Define $f : [0,3] \to [0,3]$ by $$f(x) = \begin{cases} h(x)+2, & 0 \le x \le 1 \\0, & 1 < x < 2 \\ g(x-2), & 2 \le x \le 3. \end{cases}$$ Note that for $x \in [0,1]$, we have $f(f(x)) = g(h(x))$. If $f \circ f$ were measurable, then its restriction to $[0,1]$ would also be measurable, but $(f \circ f)|_{[0,1]} = g \circ h$ which is not measurable.