Given an adapted process which is right continuous with left limits ($X(t)$) and satisfies $\int_0^TX^2(s)ds < \infty$, I want to prove that it's indefinite integral with respect to Brownian motion will be a Martingale. For that I need to prove that the process is integrable.
Given $Y(t) = \int_0^t X(s)dB(s)$, to prove : $E[|Y(t)|] < \infty$ for $t \in [0,T]$ ($B$ stands for Brownian motion). I have shown that $E[Y(t)] = 0 <\infty$ but am unable to show it for $E[|Y(t)|]$.
I will give an answer summarizing our comments for completion.
It is not true in general that given $X_t$ satisfying $\int_0^T X_s^2 ds < \infty$ a.s. for all T, $\int_0^T X_s dB_s$ is a martingale. It is true that the stochastic integral is a local martingale. If we were given that $E\int_0^T X_s^2 ds < \infty$ for all T, then by construction of stochastic integral it is a continuous, square-integrable martingale. Here is one counter-example, which I will not prove, showing that a local martingale need not be a martingale.
(Exercise 3.3.36 Karatzas and Shreve Brownian Motion and Stochastic Calculus) Let R be a Bessel process with dimension $d \geq 3$ starting with $r=0$. Then $M_t := 1/R_t^{d-2}$ is a local martingale but not a martingale.