Some thinking in stochastic integral I am new to stochastic calculus and have a question when I read the following from the textbook:
$$X_t=x+\int_a^t f(s,X_s)ds+\int_a^t \sigma(s,X_s)dB(s),\quad a\leq t\leq b$$
Here $X_t$ is of multidimensional case, so the $f$ is vector and $\sigma$ is a matrix valued measurable function. Typicall I see that $f$ and $\sigma$ satisfy Lipschitz and linear growth condition and continuous on the domain.
So my question is that:
(1) I know that $B_i$ and $B_j$ are independent to each other. So are the component of $f$ and $\sigma$ independent to each other? i.e., $f_i$ and $f_j$ are independent to each other? $\sigma_{i,j}$ and $\sigma_{k,l}$ are independent to each other?
(2) How to write out the rigorous proof for $\int_t^{t+\epsilon}\mathbb{E}[|X_s|^2]ds<\infty$ ?
 A: The independence of $f_i$ and $f_j$ (and $\sigma_i$,$\sigma_j$) depends on how they are transformations? For example, $f_i$ could be a function of both $X_{s}^{i}$ and $X_{s}^{j}$.
Assuming that $f(s,X_{s}) = \begin{pmatrix} f_{1}(s,X_{s}^1) & f_{2}(s,X_{s}^2) & \ldots & f_{n}(s,X_{s}^n)\end{pmatrix}$, the only stochasticity governing $f_{i}$ is $B_{i}$, so by independence of the Brownian motions, the $f_i$'s are independent as well.
The integral is finite as long as the expectation is finite. Applying Itô's lemma we have
$$
(X_{t}^{i})^2 = X_{0}^{i} + 2\int_{0}^{t}X_{s}f_{i}(s,X_{s})ds+\int_{0}^{t}\sigma_{i}(s,X_{s})^2ds+2\int_{0}^{t}X_{s}\sigma_{i}(s,X_{s})dB_{s}
$$
As $f$ and $\sigma$ are continuous, these integrals are finite. Taking expectations, we find that
$$
\mathbb{E}\left[ (X_{t}^{i})^2 \right] = 2\mathbb{E}\left[ \int_{0}^{t}X_{s}f_{i}(s,X_{s})ds \right] + \int_{0}^{t}\sigma_{i}(s,X_{s})^2ds
$$
To evaluate the expectation we would need to know more about $f$, but as everything is continuous, the integral is finite.
