Past exam of the mathematics course of the University of Cambridge / 2013 / iii / Paper 24 / 4 / c

Relative to a filtration $({\mathcal{F}}_t)$, progressive measurability means that, for every $T<\infty$, the mapis measurable for the product sigma-algebra $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$ and the Borel sigma-algebra on $\mathbb{R}$.

Fix $T>0$ and divide $[0,T]$ into $2^m$ equal subintervals with mesh $h_m=T/2^m$. DefineSince $X$ is adapted, every random variable $X_{k{h}_m}$ is ${\mathcal{F}}_{k{h}_m}$-measurable, hence ${\mathcal{F}}_T$-measurable. Each approximation is consequently $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$-measurable. For $t<T$, its sampling time lies strictly to the right of $t$, tends to $t$, and never exceeds $T$. Right continuity implies $X_t^{(m)}\to X_t$; at $T$ equality is exact. Thus $X$ is the pointwise limit of measurable functions on this product space. As $T$ was arbitrary, $X$ is progressively measurable. This is the theorem that right-continuous adapted processes are progressively measurable.

The right-endpoint approximations need not themselves be adapted at their intermediate times. What the proof requires is their joint measurability with respect to the single terminal sigma-algebra ${\mathcal{F}}_T$. The proof uses the pathwise càdlàg convention. If path regularity is assumed only almost surely, under a completed filtration setting the stochastic process to zero on its common exceptional null event gives an indistinguishable progressively measurable version. Arbitrary values on that null event need not make the original stochastic process progressively measurable: even with a complete filtration, a null sample point may be assigned a non-Borel time function. This is why almost sure path regularity does not ensure progressive measurability.