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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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This isn't the kind of approximation argument necessary for this problem. The trouble is that if we have (G, H) for which this holds and (G, I), say, then it isn't a priori clear to me that this holds for (G, H + I). So we can't show that the set of functions for which this holds (the candidate $\cal{H}$) is a vector space. (A succinct proof of this inequality is given on George Lowther's blog, Theorem 6.) –  Ben Derrett Dec 17 at 8:56

1 Answer 1

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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