A specific local martingale Knowing that M and N are two local martingales. Can I do the following simple reasoning for claiming that $N^\tau(M^\tau - M)$ is a local martingale? : Since a stopped local martingale is still a local martingale, and since the product of two local martingale is still a local martingale, then $N^\tau M^\tau$. Applying the same reasoning also at $N^\tau M$ I could conclude that it is also a local martingale. Finally the difference of 2 local martingale is still a local martingale. It follows that $N^\tau(M^\tau - M)$ is a local martingale. If this reasoning does not hold, what is wrong? How could I prove it without using Ito or advanced techniques?

Let $M^{(n)}$ and $N^{(n)}$ denote $M$ and $N$ stopped at $\sigma_n\wedge\tau_n$. These are both bounded martingales. Form $X^{(n)}_t:=N^{(n)}_{t\wedge\tau}(M^{(n)}_{t\wedge\tau}-M^{(n)}_t)$. As previously noted (now corrected), $$ X^{(n)}_t=\cases{0,&$0\le t<\tau$,\cr N^{(n)}_\tau(M^{(n)}_{\tau}-M^{(n)}_t),&$t\ge \tau$.\cr} $$ Now decompose (for $0\le s