Assume a filtered probability space $(\Omega,\{\mathcal F\_t\}\_{t\in[0;T)}, \mathbb P)$ with an $\mathbb R^n$-valued Brownian motion $\{W\_t\}\_{t\in[0;T)}$ and the filtration $\{\mathcal F\_t\}\_{t\in[0;T)}$ being the filtration generated by the Brownian motion, augmented by the nullsets.


Assume an $\mathbb R^n$-valued, progressively measurable stochastic process $\{Z\_t\}\_{t\in[0;T)}$ with the property
$$
\mathbb E\bigg( \int\_0^t \Vert Z\_t \Vert^2 \mathrm dt \bigg) < \infty
$$
for all $t \in [0;T)$, but possibly
$$
\mathbb E\bigg( \int\_0^T \Vert Z\_t \Vert^2 \mathrm dt \bigg) = \infty
$$


Assume a stopping time $\tau \colon \Omega \to [0;T)$.


I want to prove (but I don't know whether it's true...)
$$
\mathbb E\bigg( \int\_0^\tau Z\_t \mathrm dW\_t \bigg) = 0.
$$
I think this can be reduced to a proof of the statement
$$
\mathbb E\bigg( \Big| \int\_0^\tau Z\_t \mathrm dW\_t \Big| \bigg) < \infty
$$
because then I could use the dominated convergence theorem with the sequence $1\_{\{\tau \le T - 1/n\}} \cdot \int\_0^\tau Z\_t \mathrm dW\_t$.


But how to obtain this integrability property? Or is it even wrong? Thanks!



Is a stopped Ito-integral integrable if the Ito integrand is only square-integrable on an open interval?
A counterexample should be just the deterministic $$Z\_t = \frac{1}{\sqrt{T-t}}$$ with $$\tau := \inf{\biggl\{s>0 : \int\_0^s Z\_u \, dW\_u =1\biggr\}}$$. You have $\tau < T$ a.s.and thus $$\mathbb{E}\biggl[ \int\_0^\tau Z\_u \, dW\_u\biggr] =1$$.


