In this question I want to prove the property which is answered in this question: Why can we change the differentials?(Brownian stochastic integrals)
Description of the problem:
Assume you have a filtered probability space $(\Omega, \mathcal{F},\mathcal{F}_t,P)$, assume that $B_t$ is a Brownian motion with respect to this filtered probability space.
Define
$$\text{sign}(x)=1, x \ge 0\\ \text{sign}(x)=-1, x <0.$$
It can then be shown that $X_t=\int_0^t \text{sign}(B_s)dB_s$ is a Brownian motion. It is then stated in the answer that we have
$$\text{sign}(B_s)dX_s=\text{sign}(B_s)\text{sign}(B_s)dB_s=dB_s.$$
This is referred to as associtivity for stochastic integrals. I would like to prove this property for this example.
Attempt:
I would like to prove this by calculating the stochastic integral $\int_0^t \text{sign}(B_s)dX_s$ and showing that it is equal to $B_t$.
Let $\{q_n^{(k)}\}$ be a finite collection of rational numbers in $[0,t]$ for each $k$. Assume that as $k$ increases it increases to $\mathbb{Q}\cap[0,t]$. Assume also that for each $k$ we have $q_n^{(k)}<q_{n+1}^{(k)}$. For each $k$ look at the simple function $s_k$
$$s_k (s) = \sum\limits_{i=0}^{N_k-1} \text{sign}(B_{q_i^{(k)}})1_{[q_i^{(k)}, q_{i+1}^{(k)},)}(s).$$
$N_k$ is the number of rational numbers in our partition for each $k$. Assume always that $q_0^{(k)}=0, q_{N_k}^{(k)}=t$.
Since the lebesgue measure of the zeros of a brownian motion is zero a.s., and that we have a countable number of rational numbers, and that the Brownian motion is continuous, we have that a.s. $s_k$ converges to $\text{sign}(B_s)$ lebesgue a.e.. We have that $\omega$-wise $\int_0^t |s_k(t,\omega)-\text{sign}(B_s(\omega))|^2 dt$ converges to zero by the dominated convergence theorem. Again by the dominated convergence theorem we have that $E[\int_0^t |s_k(t,\omega)-\text{sign}(B_s(\omega))|^2 dt]$ converges to zero.
This means that we can use $s_k$ as the simple processes in the definition of the stochastic integral $\int_0^t \text{sign}(B_s)dX_s$.
So for each $k$ define $I_k$ as
$$\sum\limits_{i=0}^{N_k-1} \text{sign}(B_{q_i^{(k)}})(X_{q_{i+1}^{(k)}}-X_{q_{i}^{(k)}}).$$
From the stocahstic integral theory we have that $I_k$ converges in $L^2(\Omega)$ to $\int_0^t \text{sign}(B_s)dX_s$, and by taking a subsequence we can get a.s. convergence.
Now I come to a problem. We have that $(X_{q_{i+1}^{(k)}}-X_{q_{i}^{(k)}})$ is again a limit. This value is $\int\limits_{q_{i}^{(k)}}^{q_{i+1}^{(k)}} \text{sign}(B_s)dB_s$. If I again use the definition of a stochastic integral, I am working with two limits at the same time, and I am not able to make it work. Do you see how to solve this? Can you please help?