This is ex 4.22 from chapter 1 of''Brownian motion, martingales, and stochastic calculus by Jean-Francois Le Gall''.
exercise 4.22: Processes on defined on a probability space $(\Omega,\mathcal{F},P)$ equipped with a complete filtration $(\mathcal{F}_t)_{t \in [0,\infty]}$. Let U be an $\mathcal{F}_0$-measurable real random variable, and let M be a continuous local martingale. Show the process $N_t=UM_t$ is a continuous local martingale.
Definition of continuous local martingale:an adapted process $M=(M_t)_{t \geq 0}$ with continuous sample paths and such that $M_0=0$ a.s. is called a continuous local martingale if there exists a nondecreasing sequence $(T_n)_{n \geq 0}$ of stopping times such that $T_n \uparrow \infty$ and, for every n, the stopped process $M^{T_n}$ is a uniformly integrable martingale.
Property of continuous local martingale: A martingale with continuous sample paths is a continuous local martingale and the sequence $T_n=n$ reduces M.
I am not sure [https-//math.stackexchange.com/questions/4289186/how-to-prove-that-um-t-t-geq-0-is-a-continuous-local-martingale?rq=1][1] this is a feasible way, since ex 4.22 has a stronger condition with a complete filtration.
Bounty: I haven't made any further progress, so an answer is appreciated.
update: I will follow neijimban's hint and try the techniques used in prop 4.7 of the book: We write $M_t=M_0+N_t.$ there exists a sequence $(T_n)$ of stopping times that reduces N. Then if $s \leq t$, we have for every n, $N_{ s \wedge T_n}=E[N_{ t \wedge T_n} | \mathcal{F}_s]$. Adding both sides the random variable $M_0$ (which is $\mathcal{F}_0$-measurable and in $L^1$ by assumption), we get $M_{s \wedge T_n}=E[M_{t \wedge T_n} | \mathcal{F}_s]$. Then I am not sure if we can assume there is a random variable Z such that $|M_{t \wedge T_n}| \leq Z$, and then proceed by dominated convergence to get that $M_{ t \wedge T_n}$ converges to $M_t$ in $L^1$. Then as $n \rightarrow \infty$, we get $M_s=E[M_t | \mathcal{F}_s]$ to get uniformly integrable. Let $\tau_n= \inf \{t \geq 0: |UM_t| \geq n\}$ or $\tau \wedge T_n$ where $(T_n)_{n \geq 0}$ reduces $M$. Since $T_n$ are stopping times and $M^{T_n}$ is a continuous local martingale and $|M^{T_n}| \leq n$. If we assume $M_0 \in L^1$, $M^{T_n}$ is dominated by $n+|M_0|$. So $\lim_{n \rightarrow \infty} \int_{|M^{T_n}|>n} |M^{T_n}| dP(\omega)=0$ and this is the continuous local martingale. I am not sure if this works, so please provide an alternative if this is not correct, thanks!
