prove that process $Z_t=(\int_0^tf(s)dW_s)-\int_0^tf(s)g(s)ds$ is a local martingale Hello i have 2 questions. 
$1)$ I need to prove that if $X_t, Y_t$ - Ito processes, then
$X_tY_t=X_0Y_0+\int_0^tX_sdY_s+\int_0^tY_sdX_s+\int_0^tdX_sdY_s, t\ge 0$
it is enought to wright this like that? From Ito formula
$d(X_tY_t)=Y_tdX_t+X_tdY_t+\frac{1}{2}(dX_tdY_t+dY_tdX_t)=Y_tdX_t+X_tdY_t+dX_tdY_t$
so
$X_tY_t=X_0Y_0+\int_0^tX_sdY_s+\int_0^tY_sdX_s+\int_0^tdX_sdY_s$
Is it okay?
$2$. Using formula from point $1$ prove that process
$Z_t=(\int_0^tf(s)dW_s\int_0^tg(s)dW_s)-\int_0^tf(s)g(s)ds$
is a local martingale for any $f,g\in P_{[0,T]}^2$ where
$P_{[0,T]}^2=\{X:\Omega\times[0,+\infty]: X - \text{adapted to }   (F_t), P(\int_0^T|X(t)|^2dt<+\infty)=1\}$
 A: Let $Y_t=\int_0^t g(s)dW_s$ and $X_t=\int_0^t f(s)dW_s$, where $g$ and $f$ satisfy the condition you imposed, hence they ($X$ and $Y$) are continuous local martingales.
Then making use of the expression that you've obtain in the previous step 
$$X_t Y_t=X_0Y_0+\int_0^tX_sdY_s+\int_0^tY_sdX_s+\int_0^tdX_sdY_s$$
you have that $$\int_0^tdX_s\cdot dY_s=\langle X,Y \rangle_t=\bigg\langle\left(\int_0^t g(s)dW_s\right),\left(\int_0^t f(s)dW_s\right)\bigg\rangle(t)=\int_0^tf(s)g(s)ds$$
Hence your $(Z_t)$ is a process defined as
$$Z_t=\int_0^tX_sdY_s+\int_0^tY_sdX_s$$
You can now use some localizing sequence (of stopping times)
$$\tau_n=\inf\{t\geq 0: \int_0^t|X(s)|^2d\langle Y \rangle_s\geq n\}\wedge T$$
$$\sigma_n=\inf\{t\geq 0: \int_0^t|Y(s)|^2d\langle X \rangle_s\geq n\}\wedge T$$
Then we have 
$$E\left(\int_0^{t\wedge\tau_n\wedge \sigma_n}X^2(s)d\langle Y\rangle_s\right)<\infty$$
Same holds for Y.
Hence $$Z_{t\wedge \tau_n\wedge \sigma_n}=\int_0^{t\wedge \tau_n\wedge \sigma_n} X_sdY_s+\int_0^ {t\wedge \tau_n\wedge \sigma_n} Y_sdX_s$$
is a martingale, and this implies that $Z_t$ is a local martingale.
