How to show that this is a martingale?

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Let $H_s$ be a predictable and bounded process. How can I show that $$M_t = \int_0^t H_s \, dW_s$$ is a martingale?

Clearly since $H_s \in L^2_\text{loc} (W)$ we have that $M_t$ is a local martingale, but then how to proceed?

I have tried to use Lebesgue dominated convergence theorem, writing

$$\lim_{k \to \infty} E[M_{t \wedge \tau_k} \mid \mathcal F_s] = M_s$$

where $\tau_k$ is the localising sequence for the local martingale $M_t$

So If I can bring the limit inside I am done. I need to show that exists $Y$ such that

$$|M_{t \wedge \tau_k}| \le Y \in L^1$$

that is

$$\left|\int_0^{t \wedge \tau_k} H_s \, dW_s\right| \le Y \in L^1$$

If this was a normal integral, I would just write

$$\left|\int_0^{t \wedge \tau_k} H_s \, dW_s\right| \le M\left|\int_0^{t \wedge \tau_k} \, dW_s \right|= M |W_{t \wedge \tau_k} | \le M |W_{\max}| \in L^1$$

where $\max(\omega, t) $ is the value where the brownian motion assumes the highest value in the interval $[0,t]$ (of course as a function of $\omega$)

But I am not sure of my argument, especially when I substitute the $M$ there in the inequality seems very suspicious.. Can you provide some help?

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So the thing is that what you are looking for is exactly the content of lemma 3. I report the claim here :

Lemma 3 : Let $X$ be a càdlàg square integrable martingale and ${\xi}$ be a bounded predictable process. Then, $\int\xi\,dX$ is a square integrable martingale.

In your case $W$ is a Brownian motion, so you fit the conditions for the lemma as a Brownian motion is a continuous square integrable martingale.

The content of the proof is a bit too long to report here but G. Lowther's blog is GREAT and really self contained, he provides all the details in a much better way that I would (or more realistically "could"). Just to say a few words on the argument of the proof, it is based in spirit on the approach of "standard machine" used in the classical measure theory as it goes back to elementary process where the result is clear and then extends it thanks to topological stability properties of the limits. Best regards