I'm reading Chung's Markov Processes, Brownian Motion, and Time Symmetry. Let's say we have a probability space denoted by $(\Omega, \mathcal{F}, \mathbb{P})$. We also have an initial distribution $\mu = \delta_x$ and a transition semigroup $(P_t)$. We'll denote the Markov process with $(P_t)$ as its transition semigroup and initial distribution $\mu$ and let $\mathcal{F}^o$ be a $\sigma$-field generated by $(X_t)$. In this context, what is the probability measure $\mathbb{P}^x$ defined on the $\sigma$-field generated by this process?
The reason I want to understand this accurately is because of the statement on page 51 that reads, "for all $\Lambda \in \mathcal{F}^o$, $\mathbb{P}(\Lambda) = \mathbb{P}^{\mu}(\Lambda)$". The value of $\mathbb{P}^\mu$ changes with different $\mu$, so I want to understand how this can be equal to $\mathbb{P}$. For example, if $\mu = \delta_x$, then $\mathbb{P}^\mu = \mathbb{P}^x$ by definition. Is there a misunderstanding on my part?
And in the book, they define $\mathbb{P}^x$ and for any probability measure $\mu$, they define $\mathbb{P}^\mu(\Lambda) = \int_E \mathbb{P}^x(\Lambda) \mu (dx)$ for $\Lambda \in \mathcal{F}^o$. Here, $\mathbb{P}^\mu$ is defined as the probability generated by a Markov process with initial distribution $\mu$. It's as if we can generate the probability $\mathbb{P}^\mu$ with initial distribution $\mu$ (which was asked about in the first paragraph) from $\mathbb{P}^x$. I'm having difficulty understanding which one is defined first and what follows as a result.
"We also have an initial distribution $\mu = \delta_x$." The phrase "initial distribution" and consequently, the symbol $\mu$ refers to a probability measure over all possible initial conditions $x$. Whereas the transition semigroup $(P_t)_{t\geq 0}$ is a collection of kernels which encode conditional expectation with respect to $\mu$. Namely, $$E_\mu[f(X_t)\mid X_0] = \int P_t(X_0,dy)f(y)$$ In particular, $P_0(x,\cdot) = \delta_x(\cdot)$ so that $$E_\mu[f(X_0)\mid X_0] = \int \delta_{X_0}(dy)f(y) = f(X_0)$$ One also has $$P_\mu(\Lambda) = E_\mu[P_\mu(\Lambda \mid X_0)] = \int P_x(\Lambda)\mu(dx)$$ , where $P_x$ is the measure on the chain associated with initial condition $x$. Since the initial condition $X_0$ is random (with $X_0\sim \mu$), the measure $P_x$ is a random measure.