Partial Derivative of Ito Integral I am curious whether the following is true:

$\frac{\partial}{\partial t} \int_0^t f(s) dW_s = 0$

I think I was able to successfully derive this property by Ito's Lemma:
Let $G = f(t) W(t)$. 
Then applying Ito's Lemma gives $d(f(t) W(t)) = W(t) f'(t) dt + f(t) dW(t)$
Thus,
$f(t) W_t = \int_0^t W_s f'(s) ds + \int_0^t f(s) dW_s$
$\int_0^t f(s) dW_s = \int_0^t W_s f'(s) ds - f(t) W_t$
Which gives:
$\frac{\partial}{\partial t} \int_0^t f(s) dW_s = \frac{\partial}{\partial t} \int_0^t W_s f'(s) ds - f'(t) W_t = f'(t) W_t - f'(t) W_t= 0$
Therefore, $\frac{\partial}{\partial t} \int_0^t f(s) dW_s = 0$
Here are my questions:


*

*Are the above steps valid? Is the conclusion true? In particular I saw this thread (Leibniz Rule applied to Brownian integral)

*If it is true, is this the best way to reason through/justify why. And if it is false, why?


Thank you for your help!!
 A: @Lorenzo already pointed out that your reasoning does not work and that the stochastic integral
$$M_t := \int_0^t f(s) \, dW_s$$
fails in general to be differentiable with respect to $t$ (e.g. if we choose $f:=1$).
In fact, it is possible to show the following stronger statement:

Proposition Let $f$ be a progressively measurable function satisfying the integrability condition
$$\mathbb{E} \left( \int_0^t f(s)^2 \, ds \right)<\infty \quad \text{for all $t \geq 0$}.$$
The following statements are equivalent:

*

*The process $M_t:= \int_0^t f(s) \, dW_s$ has a modification $(\tilde{M}_t)_{t \geq 0}$ whose sample paths $t \mapsto \tilde{M}_t$ are differentiable for all $\omega \in \Omega$.

*$f=0$ almost everywhere


Since $f=0$ implies, by Itô's isometry, $M_t=0$ almost surely, it follows easily that the second statement implies the first one. For the proof of the converse we can use the following result:

Lemma Let $(N_t)_{t \geq 0}$ be a martingale whose sample paths are continuous and of bounded variation. Then $N_t = N_0$ almost surely.

Since the process $(\tilde{M}_t)_{t \geq 0}$ has differentiable paths, they are, in particular, of bounded variation. Moreover, the integrability condition on $f$ ensures that the stochastic integral $M_t = \int_0^t f(s) \, dW_s$ is a martingale, and this implies that the modification $(\tilde{M}_t)_{t \geq 0}$ is a martingale.  Applying the above lemma, we find that $\tilde{M}_t = 0$ almost surely. Thus,
$$\mathbb{E}(M_t^2) = \mathbb{E}(\tilde{M}_t^2) = 0.$$
On the other hand, Itô's isometry shows
$$\mathbb{E}(M_t^2) = \mathbb{E} \left( \int_0^t f(s)^2 \,ds \right).$$
Hence,
$$\mathbb{E} \left( \int_0^t f(s)^2 \,ds \right)=0.$$
As $t >0 $ is arbitrary, this proves $f=0$ almost everywhere.
