Integral of a deterministic function with respect to a martingale is a martingale? Let $(X_t)$ be a sample continuous stochastic process on $[0,T]$ and a martingale with respect to a given filtration $(\mathcal F_t)$. Let $f:[0,T] \rightarrow \mathbb R$ be a smooth function.
Is $ Y_t= \int_0^t f(s) d X_s $ a martingale with respect to $(\mathcal F_t)$ ? If yes, how to prove it? If not, can we add conditions on $f$ to make the statement true?
 A: The sketch of just proving the martingale property is the following. Let $(M_t)_{t \in [0,T]}$ be a continuous $L^2$-martingale on $(\mathscr{F}_t)_{t\in [0,T]}$, define
$$(f\cdot M)_t:=L^2\textrm{-}\lim_{n \to \infty}(f_n\cdot M)_t \ \ \ \color{red}{(!)}$$
where $(f_n)_{n \in \mathbb{N}}$ is a sequence of simple processes approximating $f$ $\color{red}{(!)}$. Then $(f\cdot M)_t$ is a martingale. To see this, we first use $L^2$ convergence:
$$|E[(f_n\cdot M)_t]-E[(f\cdot M)_t]|^2\leq E[|(f_n\cdot M)_t-(f\cdot M)_t|^2]\stackrel{n\to \infty}{\to} 0$$
And the fact that the simple stochastic integrals $(f_n \cdot M)_t$ are martingales $\color{red}{(!)}$: let $F\in \mathscr{F}_s$, $s<t$
$$E[\mathbf{1}_F(f\cdot M)_t]=\lim_{n \to \infty}E[\mathbf{1}_F(f_n\cdot M)_t]=\lim_{n \to \infty}E[\mathbf{1}_F(f_n\cdot M)_s]=E[\mathbf{1}_F(f\cdot M)_s]$$

$\color{red}{(!)}$ Technical details are set aside here. As suggested by @user6247850, look for a complete answer on a standard textbook. In particular, key parts are the existence of $(f\cdot M)_t$ as defined above and how $f_n$ approximates $f$ and how simple stochastic integrals are martingales.
