Past exam of the martingales and financial mathematics course of the University of Oxford / 2007 / A / 1 / b / iii / Solution

The two child intervals partition their parent:Their preimages under $g$ satisfy the same identity. Since all $A$ sets lie inside $f(M)$ and $f$ is onto its image, equality of their preimages under $f$ impliesOn the first child, $h_{n+1}=2k{2}^{-n-1}=k{2}^{-n}=h_n$; on the second it is $h_n+2^{-n-1}$. Outside the image both functions are zero. ThusFor $m>n$, summing these bounds givesThis tends to zero uniformly in $m>n$, proving that each real sequence $h_n(t)$ is Cauchy and has a finite limit.