Change of variable in Ito integral I am currently reading Oksendale's SDE and got confused at one equation involving a change of variable. 
Here is the equation (with $\{X_t\}$ is Ito diffusion and $X_{t}^{s,x}$ is a Ito diffusion solution to the initial value problem with initial time $s$ and initial position $x$,) 
\begin{align*}
X_{s+h}^{s,x}&= x+\int_s^{s+h}b(X_u^{s,x})du+\int_{s}^{s+h}\sigma(X_u^{s,x})dB_u\\
&=x+\int_0^{h}b(X_{s+v}^{s,x})dv+\int_{0}^{h}\sigma(X_{s+v}^{s,x})d\tilde{B}_v
\end{align*}
which the change of variable $u=s+v$ and where $\tilde{B_{v}}=B_{s+v}-B_s$.
I thought it would be just $dB_v$ instead of $d\tilde{B}_v$ but apparently it seems not. I personally could not find a reference for it. Does anybody know why it is?
 A: This is not a proper answer as it lacks formalism, still you could find it useful in order to understand the intuition! Let me know if it helped you.
$$\int_{s}^{s+h}\sigma(X_u^{s,x})dB_u$$
In the original stochastic integral you can see that the variable $u$ ranges from $s$ to $s+h$, hence your integrator ranges from $B_s$ to $B_{s+h}$.
If you do as you proposed, without changing the Brownian motion you would obtain the following:
$$\int_{0}^{h}\sigma(X_{s+v}^{s,x})d{B}_v$$
Notice that in this expression the variable $v$ ranges from $0$ to $h$ and hence your integrator ranges from $B_0=0$ to $B_h$. 
You can see that this is not the same as before, simply because you are integrating your function against a different portion of the sample path of the BM.
Hence in order to keep things equal you need your integrator to range from $B_s$ to $B_{s+h}$, and you achieve it by building a new Brownian motion,
$\tilde{B}_v=B_{s+v}-B_s$.
Hence your new integrator ranges from $B_s-B_s$ to $B_{s+h}-B_s$. (You still have this "minus" $B_s$ but as a matter of fact the integral gives you the same result as if you consider $B_{s+v}$ alone, the fact is that if you define $\tilde{B}_v=B_{s+v}$ then it is not a Brownian motion since $P(\tilde{B}_0=0)\neq 1$).)
