Discounted price process in Black-Scholes model is a martingale with respect to Q. I have been presented a proof that the discounted price process in the Black and Scholes formula is a martingale, but there is something important omitted, and I am not able to fill in the gap. I will present the presentation of the proof I am given here:
start of proof:
The price process is modelled by $dS_t = \mu dt+\sigma dB_t$, $\mu, \sigma$ constant. We also have a constant interest rate r, and we define $\tilde{S}_t = e^{-rt}S_t $.
Using the Ito's lemma on $e^{-rt}S_t$, we find that $d\tilde{S}_t =\tilde{S}_t(\mu-r)dt+\tilde{S}_t\sigma d B_t$
Then we have this version of Girsanovs theorem:

If $X_t$ is a stochastic process where: $E[\exp(\int_o^
tX_t^2/2)]<\infty, \forall t.$ And B is a Brownian motion under an
  original probability space $(\Omega, \mathcal{F},P)$. Then
  $\tilde{B}_t=B_t-\int_o^tX_s ds$ is a brownian motion on the space
  $(\Omega, \mathcal{F},Q)$, where $Q(A)=E_P[1_A \exp(\int_0^tX_t
dB_t-1/2\int_0^tX_t^2ds)]$.

We then define $X_t = (r-\mu)/\sigma$. It satisfies the novikov condition because it is a constant, and hence. $\tilde{B}_t=B_t-(\mu-r)/\sigma \cdot t$ is a Brownian motion under Q.
But now comes by problem, in order to finish the proof they say that:
since we have $\tilde{S}_t=\tilde{S}_0+\int_0^t\tilde{S}_s(\mu-r)ds+\int_0^t\tilde{S}_s\sigma dB_S=\tilde{S}_0+\int_0^t\tilde{S}_s\sigma(\frac{\mu-r}{\sigma}ds+dB_s)=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$. 
And this is a martingale under Q.$\square$
My problems are these:


*

*I agree that $\tilde{S}_0+\int_0^t\tilde{S}\sigma d\tilde{B}_t$. is a martingale under Q. But how do we have that $\tilde{S}_0+\int_0^t\tilde{S}_s\sigma(\frac{\mu-r}{\sigma}ds+dB_s)=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$. I do see that it is logical, and where it comes from. But we can't just play around with differentials like that? I see where it comes from if we can manipulate differentials in the formula like this, but we can't just do algebra on differentials?

*If we in some way are able to deduce that $\tilde{S}_0+\int_0^t\tilde{S}_s\sigma(\frac{\mu-r}{\sigma}ds+dB_s)=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$ (from point 1), we still can't know that the integral considered as a r.v. will be the same under the measure Q, because in the definition of the stochastic integral when we defined it, it depended very much on what probability-space we started with(it realies heavily on cauchy and convergence of random variables). So how do we know that the integral $\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$ is the same considered in both $(\Omega, \mathcal{F},P)$ and $(\Omega, \mathcal{F},Q)$? Is $\int_0^t\tilde{S}_s\sigma d\tilde{B}_s[P]=\int_0^t\tilde{S}_s\sigma d\tilde{B}_s[Q]?$
Can you guys please help me? Do you see how to make the problem complete?
In summery the problem is this: How do we see that 
$\tilde{S}_0+\int_0^t\tilde{S}_s(\mu-r)ds+\int_0^t\tilde{S}_s\sigma dB_s=\tilde{S}_0+\int_0^t\tilde{S}_S\sigma d\tilde{B}_s$.
When the first stochastic integral is created using $(\Omega,\mathcal{F},P)$, and the second stochastic integral is created using $(\Omega, \mathcal{F},Q)$?
 A: I'll rewrite the proof in terms of the process $\tilde{S}_t$, rather than using stochastic integrals. The idea behind Girsanov's theorem is to 'eliminate' the drift term of $\tilde{S}_t$ by changing the probability measure (to $\mathbb{Q}$). We have,
$$d\tilde{S}_t = \tilde{S}_t(\mu-r)dt + \tilde{S}_t \sigma dB_t.$$
Since $\tilde{B}_t = B_t - \frac{r-\mu}{\sigma}t$, we obtain by Itô's lemma
$$d\tilde{B}_t = \left(\frac{\mu-r}{\sigma}\right)dt+dB_t.$$
Hence, we obtain
\begin{align}
d\tilde{S}_t & = \tilde{S}_t(\mu-r)dt + \tilde{S}_t \sigma dB_t \\
&=\tilde{S}_t(\mu-r)dt + \tilde{S}_t\sigma\left[d\tilde{B}_t - \left(\frac{\mu-r}{\sigma}\right)dt\right] \\
&= \tilde{S}_t  \sigma d\tilde{B}_t.
\end{align}
We have showed that $\tilde{S}_t$ is a martingale w.r.t $\tilde{B}_t$, hence a $\mathbb{Q}$-martingale (since $\tilde{B}_t$ is a Brownian motion under $\mathbb{Q}$ by Girsanov and a Brownian motion is a martingale). 
A: To answer your first question, stochastic integrals of this type are Ito integrals. They are defined as a specialized limit of a Riemann-like sum with respect to a partition. Under appropriate conditions on processes $X_s$ and $B_s,$ we can define the integral as a limit of left-hand (non-anticipatory) sums:
$$\int_0^t X_s dB_s = \lim_{n \to \infty} \sum_{i=1}^n X_{s_{i-1}}[B_{s_i}- B_{s_{i-1}}].$$
If $B_s = Y_s + Z_s$ then it should be obvious how we can justify decomposing the integral as
$$\int_0^t X_s dB_s = \int_0^t X_s dY_s + \int_0^t X_s dZ_s$$
