A composition of Lebesgue measurable functions need not be measurable

There exist a Lebesgue measurable function \(h\colon\RR\to\RR\) and a continuous function \(\phi\colon[0,2]\to[0,1]\) such that \(h\circ\phi\) is not Lebesgue measurable.

Let \(C\subseteq[0,1]\) be the Cantor set and \(f\colon[0,1]\to[0,1]\) the Cantor–Lebesgue function, which is continuous, nondecreasing, and constant on each connected component of \([0,1]\setminus C\). Put \(g(x)\coloneqq f(x) + x\).

1. The map \(g\colon[0,1]\to[0,2]\) is continuous and strictly increasing with \(g(0) = 0\) and \(g(1) = 2\), so it is a homeomorphism, and \(\phi\coloneqq \inverseof{g}\colon[0,2]\to[0,1]\) is continuous.

2. The set \([0,1]\setminus C\) is a disjoint union of open intervals of total length \(1\). On each such interval \(f\) is constant, so \(g\) maps it onto an interval of the same length. Hence \(m(g([0,1]\setminus C)) = 1\) and \(m(g(C)) = m([0,2]) - 1 = 1 > 0\).

3. Every subset of \(\RR\) of positive Lebesgue measure contains a subset that is not Lebesgue measurable, so there is a non-measurable \(A\subseteq g(C)\).

4. The set \(B\coloneqq \inverseof{g}(A)\) is contained in \(C\), which has measure \(0\), so \(B\) is Lebesgue measurable and \(h\coloneqq\chi_B\) is a Lebesgue measurable function.

5. For \(y\in[0,2]\), \((h\circ\phi)(y) = 1\) if and only if \(\inverseof{g}(y)\in \inverseof{g}(A)\), that is, \(y\in A\). Hence \(\inverseof{(h\circ\phi)}(\theset{1}) = A\) is not Lebesgue measurable, and \(h\circ\phi\) is not Lebesgue measurable.