The question:
Suppose that $(M_t) $ $ : t \in [0,\infty)$ is a bounded continuous martingale with finite variation. Prove that $M_t^2 = M_0^2 + 2 \int_0^t M_s dM_s,$ where the final integral is almost surely well-defined. (Hint: Cite here known results about one-dimensional functions of bounded variation.) Deduce that $M$ is almost surely constant
My attempt
Im fairly sure the last part is quite easy by taking the expected value of both sides and knowing that the stochastic integral is a martingale.
My struggle lies within the first bit. Can I just use Itô formula using $f(x) = \frac{x^2}{2} $?
That is to say: $f(M_t) -f(M_0) = \int_0^t f'(M_s)dM_s + \frac{1}{2}\int_0^tf''(M_s)ds$
This then gives me $\frac{M_t^2}{2} - \frac{M_0^2}{2} = \int_0^tM_sdM_s + t$
This is very nearly what I need but I have the annoying $+t$ on the end. Can I get rid of this somehow? I haven't at all used the continuity, boundedness or finite variation so i think I've done something wrong. What is the "well known result" the hint suggests I use?
EDIT
My textbook states:
I do not understand how any of the equalities follow or hold? Is it that because the process is of finite variation we can just. use normal calculus? What is the formula being used here please
Yes, because the process is of finite variation, we can just use normal calculus. Specifically, using integration by parts we have \begin{align*} \int_0^t M_s dM_s = M_t^2 - M_0^2 - \int_0^t M_s dM_s \end{align*} and the claim follows by simply rearranging that expression.