What does $\mathbb{P}(d\omega)=dw$ actually mean? I am currently reading S. Shreve's book Stochastic Calculus II, and I have a question regarding Example 1.6.4 (p.35-36) which describes a change of measure, but I am puzzled by the notation.
$\Omega=[0,1], \mathbb{P}$ is the uniform measure and 
$$\tilde{\mathbb{P}}[a,b]=\int_a^b 2\omega d\omega=b^2-a^2$$
for $0\le a\le b \le 1$.
The first step is:

We may use the fact that $\mathbb{P}(d\omega) = d\omega$ to rewrite the equation above as
  $$\tilde{\mathbb{P}}[a,b]=\int_a^b 2\omega d\mathbb{P}(\omega).$$

I have several questions regarding this statement:


*

*$\mathbb{P}$ is a probability measure from $\mathcal{F}\rightarrow\mathbb{R}$, I cannot find a clear definition in the book what $\mathbb{P}(d\omega)$ actually means. Is $d\omega$ actually an event?

*Why does this $\mathbb{P}(d\omega) = d\omega$ help in rewriting, shouldn't we need $d\mathbb{P}(\omega) = d\omega$?

*What does $\mathbb{P}(d\omega) = d\omega$ even mean, and why is this true?

 A: This kind of stuff has often bugged me and here are some thoughts that I have had over the time.


*

*Before asking what $\mathbb{P}(d\omega)$ means, one should clarify what $d\omega$ means. If you ask a physicist, they will tell you that it is a "differential" or "infinitesimal increment". However, I have never really understood what that actually is, nor have I seen a precise definition. One way to give $d\omega$ a precise meaning is to define it as a differential form using Cartan's calculus, i.e. you define it as the exterior derivative of the identity function. Thus $d\omega$ is simply a 1-form. On the other hand one should not bother to much about what $d\omega$ means by itself. Instead, understand $\int\ldots d\omega$ as one notation for the integral with respect to the Lebesgue measure using $\omega$ as a variable. (More generally, this denotes the integral with respect to the Haar measure of the underlying group.) Note in particular, that $d\omega$ determines both the measure and the variable of integration. Now we are in the case where we want to integrate with respect to some other (probability) measure $\mathbb{P}$. Then we write $\int\ldots\mathbb{P}(d\omega)$ or $\int\ldots d\mathbb{P}(\omega)$ in order to express that $\mathbb{P}$ is the measure that we want to use and $\omega$ is the variable. 

*There is no difference between writing $\mathbb{P}(d\omega)$ or $d\mathbb{P}(\omega)$. Both are used to write down the same integral and, again, one should not bother to much about the "literal" meaning of those symbols.

*If the author writes $\mathbb{P}(d\omega)=d\omega$, then this is a more or less sloppy shorthand notation to tell you that $\mathbb{P}$ is already the (restriction to $[0,1]$) of the Lebesgue measure and hence, by convention, we do not mention the measure explicitely. 
Just one final remark: Integration and Differentiation has such a long history and so many points of view have established over the time (measure theory, differential forms, integrals as linear functionals etc.) that the (historical) notation becomes blurry and this little $dx$ is much less innocent than it appears.
