Predictable step processes are dense for the compensator L² norm

Put $f(s,\omega )=Y_s(\omega ){\lambda }_s(\omega )$. By [F1], $f$ is product-measurable, so [F2] defines $\mu$ and gives $\mu ([0,T]\times \mathrm{\Omega })=\int fd(ds\times P)=\mathbb{E}{\int }_0^{T}f(s,\omega )ds=\mathbb{E}{A}_T<\mathrm{\infty }$. Countable additivity follows from the product measure; [F5] also gives it by monotone convergence on disjoint unions. Lebesgue measure gives $\mu (\{0\}\times B)=0$ for every $B\in {\mathcal{F}}_0$.

Let $\mathcal{G}$ contain the empty set, whole space, time-zero generators, and interval rectangles from [F1]. Their finite intersections are empty or generators, so $\mathcal{G}$ is a $\pi$-system generating ${\mathcal{P}}_T$. Let $\mathcal{D}$ be the sets whose indicators lie in the $L^2(\mu )$-closure of the span of generator indicators. The whole space is in $\mathcal{D}$ because ${\mathbf{1}}_{(0,T]\times \mathrm{\Omega }}$ is an elementary step and the time-zero slice is null. Complements preserve $\mathcal{D}$ by subtraction from $\mathbf{1}$; disjoint countable unions preserve it because finite partial sums converge in $L^2(\mu )$, with squared tail norm the measure of a decreasing tail tending to zero as $\mu$ is finite. Thus $\mathcal{D}$ is a $\lambda$-system containing $\mathcal{G}$, and [F3] gives $\mathcal{D}={\mathcal{P}}_T$.

By [F4], every $L^2(\mu )$ class is an $L^2$ limit of finite simple functions on measurable sets. Each indicator in such a simple function belongs to the closure from step 2.1, so each simple function does as well; hence the span of bounded predictable step indicators is dense in $L^2(\mu )$.