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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.

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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.

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@ mike: Thanks for your answer. The second was already clear. About your answer for my second question: If $\mathcal{C}=\mathcal{K}$ and $\mathcal{H}$ the space of all bounded $\mathcal{B}(0,\infty) \times \mathcal{F}$. And suppose, we can apply monotone class theorem. Shouldn't $\mathcal{H}$ be a set of functions, which are bounded, product measurable and satisfy the inequality. Otherwise I do not see how the extension works. A little bit more detail would be appreciated. Then you will get the bounty. Thanks –  math Jun 7 '12 at 15:46
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