I am solving the exercises in a book I have about Lévy processes ("Lévy Processes and Stochastic Calculus", Applebaum, 2003), and I cannot get my head around an exercise that seems rather simple. I have been trying to use different lemmas to prove the following result, but I'm not convinced by any.
I am trying to prove that the stochastic integral $$ \int_0^T\int_A H(t,x)\,\tilde{N}(dt,dx) = \int_0^T\int_A H(t,x)\,N(dt,dx) - \int_0^T\int_A H(t,x)\,\nu(dx)dt $$ is finite (almost surely), where $H$ satisfies the regular square-integrability condition and $\nu$ is the Poisson random measure.
Does somebody know how to prove this result?
Thanks very much
First of all, note that the stochastic integral
$$\int_0^T \int_A H(t,x) \tilde{N}(dt,dx)$$
is almost surely finite if $H$ is square integrable, but this is in general not true for the integrals
$$\int_0^T \int_A H(t,x) N(dt,dx) \qquad \text{and} \int_0^T \int_A H(t,x) \, \nu(dx) \, dt.$$
This means that we can in general not write
$$\int_0^T \int_A H(t,x) \tilde{N}(dt,dx) = \int_0^T \int_A H(t,x) N(dt,dx)- \int_0^T \int_A H(t,x) \nu(dx) \, dt.$$
Now let $H$ be square-integrable, i.e.
$$\mathbb{E} \left( \int_0^T \int_A H(t,x)^2 \nu(dx) \, dt \right)<\infty. \tag{1}$$
Then it follows from Itô's isometry that
$$\mathbb{E}\left( \left| \int_0^T \int_A H(t,x) \tilde{N}(dt,dx) \right|^2 \right) = \mathbb{E} \left( \int_0^T \int_A H(t,x)^2 \, \nu(dx) \, dt \right) \stackrel{(1)}{<} \infty.$$
This shows that $\int_0^T \int_A H(t,x) \tilde{N}(dt,dx) \in L^2(\mathbb{P})$; hence, in particular, $$\int_0^T \int_A H(t,x) \tilde{N}(dt,dx)<\infty\quad \text{almost surely.}$$