I am trying to get a better picture of what is the intuition behind the Doléans measure.
Is it analogue to the measure used for random variables but it is for random process?
The only explicit way where I see the Doléans measure to appear is in definition of norms. Further, in the Lecture Notes, I see again an indirect appearance of this measure, when we define function $I$, $I: \mathcal{P} \to \mathcal{M}^2_c$. We have that $||X|| = ||X \cdot M||$, where the first norm uses Doléans measure, second norm - simple one.
Can somebody explain also how to read, understand this: $\mu_M$ on $([0,\infty) \times \Omega, \mathcal{B}([0,\infty))\times \mathcal{F})$ is defined by $\mu_M(A) = \int_\Omega \int^\infty_0 1_A(t,w) d \langle M \rangle _t(w)P(dw)$. I do not understand what the two products of sets mean. Where $\mathcal{B}$ -borel sigma algebra,$\Omega$ - sample space, $\mathcal{F}$ - sigma algebra.
I don't know what exactly they're doing with the norms, but the point is to give a norm for the process $X$ in relation to the process $M$. You might consider the analogous situation in linear algebra: if $A$ is a positive definite matrix then we often define $\| x \|_A = (x^T A x)^{1/2}$. It's just a different way to measure the size of a vector in relation to a matrix.
The last thing is just a way to make $[0,\infty) \times \Omega$ into a measure space. You have a "natural" choice of $\sigma$-algebra on each factor, namely $\mathcal{B}([0,\infty))$ and $\mathcal{F}$ respectively. So you give $[0,\infty) \times \Omega$ the product of these two $\sigma$-algebras (which, by the way, is not the Cartesian product of the two $\sigma$-algebras as sets). With that $\sigma$-algebra in hand you have a measurable space. You now give this measurable space the measure $\mu_M$ to make it into a measure space. I don't know what they're doing with $\mu_M$, but I can rewrite it as $\mu_M(A) = E \left [ \int_0^\infty 1_A d \langle M \rangle_t \right ]$, which might give you some intuition (or at least reduce the notational burden).