# Malliavin derivative of a Lebesgue integral.

Let $X_t$ a random process such that its Malliavin derivative is well defined for all $t$. Then I have read that :

$D_s(\int_0^t \! X_u \, \mathrm{d}u)=\int_s^t \! D_s(X_u) \, \mathrm{d}u.$

What I don't understand is why in the second integral, the domain of integration is now $[s,t]$.

I assume that $$X_t$$ is progressively measurable with respect to the corresponding Wiener process $$W_t$$ (otherwise the formula doesn't hold).
$$X_t$$ being $$\sigma(W_s,s\le t)$$-measurable implies that it does not depend on the "tail" $$\{w(v),t as a functional from $$L_2(C([0,T],\mathbb{R}),\text{Wiener measure})$$, $$t. Thus, $$D_hX_t$$ is zero on the $$\{h: \operatorname{supp}(h)\in(t,T]\}$$ and $$D_s(X_t)=0$$ for $$s>t$$, therefore $$D_s\left(\int_0^t X_udu\right)=\int_0^t D_s(X_u) du=\int_s^t D_s(X_u) du.$$