= Solution
{c}
If $t\notin f(M)$, all <indicator functions> vanish and $h_n(t)=0$. If $t\in f(M)$, the partition in part (i) puts it in exactly one $A_{n,k}$, so exactly one term can be nonzero and the sum is a finite real number at that point. There is no ambiguous infinite sum of positive and negative terms.
For <measurability>, each symmetric partial sum
$$
h_{n,m}(t)=\sum_{k=-m}^m k2^{-n}\mathbf1_{A_{n,k}}(t)
$$
is <Borel measurable> by closure under finite linear combinations. For each $t$, these partial sums eventually include its unique possible term and converge to $h_n(t)$. The measurable-limit result from part (a) therefore proves \b[$h_n$ is <Borel measurable>].
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