I've been finding some difficulties in an exercise about the stochastic integral:
Consider the following stochastic process: $$X_t=\int_0^t\sigma_udW_u$$ Where $\sigma_u$ is a cadlag deterministic adapted process, $W_u$ is a standard Brownian motion and $\{\mathcal{F}_t\}_{t\geq 0}$ is the standard augmented filtration. Is it true that $X_t-X_s$ is independent on $\mathcal{F}_s$?
MY ATTEMPT
Note that the process $X_t-X_s=\int_s^t\sigma_udW_u$ and for any partition of the interval $[s,t]$ of the type $\pi^n=\{s=t_0^n<t_1^n<...<t^n_{k_n}=t\}$ with $\max|t^n_i-t^n_{i-1}|\xrightarrow[n\to\infty]{}0$: $$\lim_{n\to\infty}\sum_{i=1}^{k_n}\sigma_{t^n_{i-1}}(W_{t^n_i}-W_{t^n_{i-1}})\xrightarrow[n\to\infty]{\mathbb{P}}\int_s^t\sigma_udW_u$$ We have that for any $n$ the LHS is independent on $\mathcal{F}_s$ thanks to the independece properties of Brownian increments ($\sigma_s$ is deterministic).
Is there any result that allows me to say that also the limit is independent on $\mathcal{F}_s$?