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

measurability with respect to $\sigma (f)$ gives a Borel set $C_{n,k}$ with $g^{-1}(B_{n,k})=f^{-1}(C_{n,k})$. Restrict it to the actual image:This is Borel by the assumed Borel property of $f(M)$, and its preimage under $f$ is unchanged.

The intervals $B_{n,k}$ partition the real line, so their preimages under $g$ partition $M$. If two of the restricted sets overlapped, a point in their intersection would equal $f(x)$ for some $x$, making $g(x)$ lie in two disjoint intervals, a contradiction. Similarly any $t\in f(M)$ has a preimage $x$, whose real value $g(x)$ belongs to one of those intervals; that puts $t$ in the corresponding $A_{n,k}$. Thus the restricted sets are disjoint and their union is exactly $f(M)$. This restriction is what removes arbitrary overlaps of the initially chosen Borel sets outside the image.

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.

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.

For $x\in M$, let $k$ be the unique index such that $g(x)\in B_{n,k}$. Then $f(x)\in A_{n,k}$ and $h_n(f(x))=k{2}^{-n}$. The chosen interval is open on the left and closed on the right, soThis remains correct at dyadic endpoints; changing it to a floor approximation with a non-strict left endpoint would describe a different approximation.

By part (iii), define the finite limit $h(t)={\lim }_n{h}_n(t)$ for every $t\in \mathbb{R}$. Part (a) makes this limit Borel measurable. Letting $n\to \infty$ in the displayed error bound givesThis is the factorization conclusion of the Doob-Dynkin lemma under the specified Borel-image hypothesis.