The Kunita-Watanabe Inequality says:
Let $X,Y$ be two continuous locale martingales and $H,G$ two product-measurable functions on $(0,\infty)\times \Omega$, then
$$ \int_0^t|G_s||H_s|d|\langle X,Y \rangle|_s \le \sqrt{\int_0^tG^2_s d\langle X\rangle_s}\sqrt{\int_0^tH^2_s d\langle Y\rangle_s} $$
Suppose I have proved the inequality for the following set of functions:
$$\mathcal{C}:=\{G=\sum_{i=1}^ng_i \mathbf1_{(t_i,t_i+1]}, n\in \mathbb{N}\mbox{ and $g_i$ bounded and measurable}\}$$
Using Monotone Class Theorem I want to extend this first to $G\in \mathcal{C}$ and $H$ bounded and product-measurable. And in a second step to $G,H$ both product-measurable.
In our class we used the following Monotone Class Theorem: Link (Theorem 2).
How do you choose $\mathcal{K}$ and $\mathcal{H}$ in this setting (in both steps)?
In addition, product-measurable means with respect to the product $\sigma$-algebra $\mathcal{B}(0,\infty)\otimes \mathcal{F}$. Can the Kunita-Watanabe Inequality be applied to functions, which are measurable with respect to the predictable sigma field on $(0,\infty)\times \Omega$? The predictable sigma field is generated by all adapted and left continuous processes.
Thanks for your help.
Best Answer
The answer to the second is of course. The funny thing about kunita-watanabe is that it does not require the progressively measurable etc. If you have it, $\int HdY$ is a martingale, and K-W is no different from the fact about quadratic variation/covariation that underlies it.
Once you have established that fact, you are proving a fact about borel measures on $\mathbb R$. Revuz-Yor uses density of the class $\mathcal C$, without proof of the density. I think you want $\mathcal K = \mathcal C$, and $\mathcal H$ to be all bounded $\mathcal B(0,\infty) \times \mathcal F$ functions, leaving you with a brief argument showing that it is enough to prove your inequality for all bounded H and G. I suppose this fills in the density lacuna in Revuz and Yor's argument as well.