= Solution
{c}
The two child intervals partition their parent:
$$
B_{n,k}=B_{n+1,2k}\,\dot\cup\,B_{n+1,2k+1}.
$$
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 $f$ implies
$$
A_{n,k}=A_{n+1,2k}\,\dot\cup\,A_{n+1,2k+1}.
$$
On the first child, $h_{n+1}=2k\,2^{-n-1}=k2^{-n}=h_n$; on the second it is $h_n+2^{-n-1}$. Outside the image both functions are zero. Thus
$$
\boxed{0\leq h_{n+1}(t)-h_n(t)\leq2^{-n-1}.}
$$
For $m>n$, summing these bounds gives
$$
|h_m(t)-h_n(t)|\leq\sum_{j=n}^{m-1}2^{-j-1}
\leq2^{-n}.
$$
This tends to zero uniformly in $m>n$, proving that each real sequence $h_n(t)$ is Cauchy and has a finite limit.
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