A Markov chain $(X\_i)\_{i\in \mathbb{N}}$ on a measurable space $(E,\Sigma)$ is (see e.g. Revuz or Meyn/Tweedie) constructed on the following probabilty space.
$$ \Omega = \{ (x\_l)\_{l \in \mathbb{N}} \mid x\_l \in E \text{ for all } l \in \mathbb{N} \}$$ 
and $X\_i$ is definied as
$$X\_i ((x\_l)\_{l \in \mathbb{N}}) = x\_i$$ and $\Sigma$ is constructed suitable. Then one can show, for any [Markov kernel](https://en.wikipedia.org/wiki/Markov_kernel) p and probability measure $\mu$ exists a a probability measure $\mathbb{P}\_\mu$ on $\Omega$ with
$$ \mathbb{P}\_\mu [X\_0 \in A\_0, X\_1 \in A\_1 ,\dots X\_n \in A\_n] =$$
$$ \int\_{A\_0} \dots \int\_{A\_n} p(y\_{n-1},A\_n) \, p(y\_{n-2}, dy\_{n-1}) \dots p(y\_0, dy\_1) \, \mu(dy\_0).$$ Thus by this particular construction for all Markov chains the space $(\Omega,\Sigma)$ and the random variable $(X\_i)\_{i \in \mathbb{N}}$ are identical and two Markov chains only differ in the associated probability measure $\mathbb{P}\_\mu$.


**My question:**
Assume two measures $\mu\_1$ and $\mu\_2$ and two Markov kernels $p\_1$ and $p\_2$ are given. Then I have two seperate probability measures $\mathbb{P}\_{\mu\_1}$ and $\mathbb{P}\_{\mu\_2}$ where the projection process lives on.
Can I conclude from this that there exits a single probability space $(\hat{\Omega},\hat{\Sigma},\mathbb{P})$ together with two stochastic processes $(Z\_i)$ and $(Y\_i)$ with $$ \mathbb{P} [Z\_0 \in A\_0, Z\_1 \in A\_1 ,\dots Z\_n \in A\_n] = \mathbb{P}\_{\mu\_1} [X\_0 \in A\_0, X\_1 \in A\_1 ,\dots X\_n \in A\_n]$$


and
$$ \mathbb{P} [Y\_0 \in A\_0, Y\_1 \in A\_1 ,\dots Y\_n \in A\_n] = \mathbb{P}\_{\mu\_2} [X\_0 \in A\_0, X\_1 \in A\_1 ,\dots X\_n \in A\_n]?$$ 



General Markov Chains on same Probability Space?
The answer is yes: both processes can be constructed simultaneously on the same probability space. This is a standard fact (true in much greater generality - look up "product measure" and "Kolmogorov extension theorem" on Wikipedia and in any graduate-level textbook on probability theory).


To add a bit of detail, the space $\hat{\mathbb{P}}$ can be constructed as the product space of $(\Omega,\mathbb{P}\_{\mu\_1})$ and $(\Omega,\mathbb{P}\_{\mu\_1})$ (with the accompanying $\sigma$-algebras, which I will omit). That is, we define
$$ \hat{\Omega} = \Omega \times \Omega $$
where $\Omega = \{ (x\_l)\_{l \in \mathbb{N}}\ |\ x\_l \in E \textrm{ for all }l\in\mathbb{N}\}$
as in the question, and we equip $\hat{\Omega}$ with the obvious $\sigma$-algebra together with the unique probability measure $\hat{\mathbb{P}}$
satisfying
$$ \hat{\mathbb{P}}(A\times B) = \mathbb{P}\_{\mu\_1}(A) \mathbb{P}\_{\mu\_2}(B) $$
for all $A,B \subset \Omega$ that belong to the $\sigma$-algebra $\Omega$ comes equipped with. (The existence and uniqueness of such a measure is a standard result in measure theory.) If you now define stochastic processes $(Z\_i)\_{i\in\mathbb{N}}$ and $(Y\_i)\_{i\in\mathbb{N}}$ on $\hat{\Omega}$ by
$$ Z\_i( (a\_l)\_{l\in\mathbb{N}},(b\_l)\_{l\in\mathbb{N}} ) = a\_i, $$
$$ Y\_i( (a\_l)\_{l\in\mathbb{N}},(b\_l)\_{l\in\mathbb{N}} ) = b\_i, $$
(noting that a generic element of $\hat{\Omega}$ is of the form $((a\_l)\_{l\in\mathbb{N}},(b\_l)\_{l\in\mathbb{N}})$ ), then it is easy to check that $(Z\_i)\_{i\in\mathbb{N}}$ is a realization of the Markov chain with distribution $\mathbb{P}\_{\mu\_1}$ and $(Y\_i)\_{i\in\mathbb{N}}$ is a realization of the Markov chain with distribution $\mathbb{P}\_{\mu\_2}$, in the sense mentioned in the question.


