Is a stopped Ito-integral integrable if the Ito integrand is only square-integrable on an open interval?

Question

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!

Answer 1

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 $$\mathbb{E}\biggl[ \int_0^\tau Z_u \, dW_u\biggr] =1$$.