Let $(\Omega,\mathcal F,P)$ be a probability space and $B_t$ be a Brownian motion on it. If a progressively measurable space $X_t$ satisfies $E\int_0^T|X_t|^2dt<\infty$, then the integral $\int_0^TX_tdB_t$ is a martingale. If we have $E(\int_0^T|X_t|^2dt)^{1/2}<\infty$, can we still construct the integral $\int_0^TX_tdB_t$ and is it still a martingale? Thanks!
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1To clarify, you mean $E\left[\left(\int_0^T |X_t|^2,dt\right)^{1/2}\right] <\infty$? – Nate Eldredge May 31 '19 at 03:49
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What did you mean by"If a progressively measurable space$X_t$ – Zbigniew Jul 16 '19 at 22:21