# How to formulate an integral on the probability space of a Markov shift?

For a problem I am currently working on, I would like to calculate the integral over the space $(\{1,2,\dots,n\}^\mathbb{N},\mathcal{F},\nu)$ where $\nu$ is the markov measure.

My problem is that that it doesn't seem possible to work with the traditional form one would use. I have:

$$\int_X f(x) \,d\nu(x) = \sum_{x_0=1}^n\sum_{x_1=1}^n\sum_{x_2=1}^n\dots f((x_0,x_1,x_2,\dots)) \nu(x_0,x_1,x_2,\dots)$$

which doesn't really allow much computational flexibility due to the infinite number sums. Are there any useful tricks or general forms which make this more computable?

PS: in my case I have $f(x) = I_{\alpha|\mathcal{F}}(x)$, where $I_{\alpha}(x) = -\sum_{A \in \alpha}1_{A}(x) \log \nu(A)$ and $\alpha$ a partition of $X$.

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Since I'm more comfortable with the probabilistic approach I would write $\mathbb{E}[f(X_0,X_1,\ldots)]$ where $(X_0,X_1,\ldots)$ is a Markov process having law $\nu$. In your case $f$ is linear combination of indicator functions and the expectation is $-\sum_{A \in \alpha}\mathbb{P}[(X_0,X_1,\ldots)\in A] \log\nu(A)=-\sum_{A \in \alpha} \nu(A)\log\nu(A)$ which is nothing but the entropy of the partition $\alpha$.