Let $N$ be the Poisson random measure associated to a Levy process $X$, with intensity $\nu$. Furthermore, let $A$ be bounded from below and $f \in L^1(A,\nu)$ be measurable. It is well known that
\begin{align} E \Big[ \int_A f(y) N(dy,t) \Big] = t\int_A f(y) \nu(dy). \end{align}
I need a similar result for functions on the form $f(X_t,y)$. In particular, I need
\begin{align} E \Big[ \int_A f(X_t,y) N(dy,t) \Big] = t\int_A E[f(X_t,y)] \nu(dy), \end{align}
where $f(X_t,\cdot) \in L^1(A,\nu)$. Any ideas as to where I can find such a result, or a way to prove it? Additional assumptions may be put on $X$ and $f$.
Let $g$ be a predictable function such that
$$\mathbb{E} \left( \int_0^t \!\! \int |g(s,y)| \, \nu(dy) \, ds \right)<\infty,$$
then one can show that the integral
$$\int_0^t \!\! \int g(s,y) \, N(dy,ds)$$
is well-defined and
$$\mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, N(dy,ds) \right) = \mathbb{E} \left( \int_0^t \!\! \int g(s,y) \, \nu(dy) \, ds \right).$$
Applying this for $g(s,y) := f(X_s,y) 1_A(y)$ yields
$$\mathbb{E} \left( \int_0^t \! \! \int_A f(X_s,y) \, N(dy,ds) \right) = \mathbb{E} \left( \int_0^t \int_A f(X_s,y) \, \nu(dy) \, ds \right).$$
(I hope that's what you meant; or do you really want to fix $t$ and consider $g(s,y) := f(X_t,y) 1_A(y)$? This works also fine, but you have to be careful about the proper measurability of $g$.)
A standard reference is e.g.
Watanabe/Ikeda: Stochastic Differential Equations and Diffusion Processes, North-Holland.