Ito's integral and Fubini's theorem

Question

For Ito's integral: \begin{equation*} \int_0^T f(t, X_t) dX_t, \end{equation*} where $X_t$ is Brownian motion. Because, the Ito integral is martingale, which means that \begin{equation} \mathbb{E}\left\{\int_0^T f(t, X_t) dX_t\right\}=\lim_{N\rightarrow\infty} \sum_{i=0}^{N-1} f(t_i, X_i) \mathbb{E} \left\{X_{i+1} - X_i\right\} =0 \end{equation} By using the Fubini's Theorem, we have \begin{equation*} \mathbb{E}\left\{\int_0^T f(t, X_t) dX_t\right\}=\int_0^T \mathbb{E}\left\{f(t, X_t)\right\}dX_t=0 \end{equation*} Can anyone suggest how to prove that: \begin{equation*} \int_0^T \mathbb{E}\left\{f(t, X_t)\right\}dX_t=0? \end{equation*}

Answer 1

Fubini's theorem requires you to have two (not necessarily distinct) measure spaces. I think that if you spit this out explicitly, your life will be much easier. Here, they are $([0,T],\lambda)$, where $\lambda$ is the 1-dimensional Lebesgue measure and, $(\Omega, \mathscr F, \mathbb P)$ the probabiity space on which your BM is defined.

The stochastic process $X$ is a function of two variables $X:[0,T]\times\Omega\to\mathbb R,\;\;$ $X:(t,\omega)\mapsto X_t(\omega)$. $\mathbb E$ denotes integration with respect to the $\omega\in \Omega$ variable. So, \begin{align} \mathbb E\int_0^Tf(t,X_t(\omega))dX_t(\omega)& =\int_\Omega\int_0^Tf(t,X_t(\omega))dX_t(\omega)d\mathbb P(\omega)\\ &=\int_0^T\int_\Omega f(t,X_t(\omega))dX_t(\omega)d\mathbb P(\omega)\\ &=\int_0^T\mathbb E\left[f(t,X_t(\omega))dX_t(\omega)\right]. \end{align}

This is all provided the hypotheses of Fubini's theorem hold, of course.

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This answer is formal as $dX_t$ is not actually a measure... but you'll get into those details if you decide to pursue stochastic analysis further.