 Solution

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 sumis 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 $h_n$ is Borel measurable.