I have just started reading about stochastic calculus, and I am confused about transforming certain types of SDE using the Girsanov theorem. Here is an example from the book "Elementary stochastic calculus with finance in view" by T.Mikosch p.180-182:
Suppose that we have an underlying probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and $(W_{t})_{t\ge 0}$ be a Brownian motion wrt $\mathbb{P}$. Consider the linear SDE
$$dX_{t}=cX_{t}dt+\sigma X_{t}dW_{t}$$
where $c$ and $\sigma$ are constants. The rigorous meaning of the above is that $(X_{t})_{t}$ is a stochastic process such that
$$X_{t}=X_{0}+\int_{0}^{t} cX_{t}dt+\int_{0}^{t} \sigma X_{t}dW_{t}$$
The last term refers to the Ito integral, into which the probability measure $\mathbb{P}$ participates directly in the definition.
Consider the process $\widetilde{X}_{t}=e^{-rt}X_t$, where $r$ is a constant. By Ito lemma
$$d\widetilde{X}_{t}=\widetilde{X}_{t}[(c-r)dt+\sigma dW_{t}]$$
Now consider the substitution $\widetilde{W_{t}}=\dfrac{c-r}{\sigma}t+W_{t}$, hence $d\widetilde{W_{t}}=\dfrac{c-r}{\sigma}dt+dW_{t}$ and thus
$$d\widetilde{X}_{t}=\sigma \widetilde{X}_{t}d\widetilde{W_{t}} \ \ \ (*)$$
So far, so good. Everything is still under the measure $\mathbb{P}$. Using the Girsanov theorem, there exists a probability measure $\widetilde{\mathbb{P}}$, equivalent to $\mathbb{P}$, with respect to which $\widetilde{W}_{t}$ is a martingale.
It is then claimed that $\widetilde{X}_{t}$ is a martingale wrt $\widetilde{\mathbb{P}}$, since the Ito integral $\int_{0}^{t}\sigma \widetilde{X}_{t}d\widetilde{W_{t}}$ is a martingale wrt to $\widetilde{\mathbb{P}}$. However, this seems to mean that the equality $(*)$ is interpreted as saying that
$$ \widetilde{X}_{t}=\widetilde{X}_{0}+\int_{0}^{t}\sigma \widetilde{X}_{t}d\widetilde{W_{t}} $$ where the Ito integral $\int_{0}^{t}\sigma \widetilde{X}_{t}d\widetilde{W_{t}}$ was computed using $\widetilde{\mathbb{P}}$. To me, however, such an equality so far is only justified if we used the measure $\mathbb{P}$, as it was the only measure used to reach $(*)$.
So, my question is the following:
Do we get the same random variable when we compute an Ito integral such as $\int_{0}^{t}\sigma \widetilde{X}_{t}d\widetilde{W_{t}}$, using two different, but equivalent probability measures like $\mathbb{P}$ and $\widetilde{\mathbb{P}}$ ?
The definition of Ito integral uses the underlying probability measure from what I know, and convergence in $L^2(\widetilde{\mathbb{P}})$ and $L^2(\mathbb{P})$ is not the same as this post shows.
The post here seems to say the answer for my question is yes, but I would like for someone to confirm this.
In case I am terribly confusing something, comments would be appreciated.
$$d\widetilde{X}_{t}=\sigma \widetilde{X}_{t}d\widetilde{W_{t}} \ \ \ (*)$$
still holds under the new measure $\widetilde{\mathbb{P}}$. This measure is defined to be equivalent to $\mathbb{P}$ so that the measures agree on events that have measure zero (and one). This implies that all mechanical relationships (equalities that hold for sure) between different stochastic processes under $\mathbb{P}$ must also hold under $\widetilde{\mathbb{P}}$. Conditional on the realization for $\widetilde{W}$ say $\omega\in \Omega$ you hence get the same value for $\int_{0}^{t}\sigma \widetilde{X}_{t}d\widetilde{W_{s}}$ under both measures.