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

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!

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 $$\mathbb{E}\biggl[ \int_0^\tau Z_u \, dW_u\biggr] =1$$.
• Thanks. Intuitively, I see why $\mathbb P[\tau < T] = 1$. But is there a way to briefly argue for this mathematically? – Kolodez Feb 15 at 7:59
• Consider a time change $u = Log(T-t)$. Then this is equivalent to that Brownian motion hits 1 in finite time. – Stephan Sturm Feb 15 at 8:14