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For a recurrent Markov chain, start at $k$ and stop at its first positive return $T_k$. The expected visits to each state before that return form the excursion occupation measure. Its $k$th coordinate is one. Counting the return endpoint instead of the start gives the same measure. It is invariant by shifting the occupation sum one time step and using the Markov property. Any Markov chain invariant measure normalized by $x_k=1$ dominates it: iterate the stationarity equations on the state space with $k$ removed, then retain the nonnegative excursion-path terms.

For a Markov chain invariant measure, $xP=x$, and hence $x{P}^n=x$. The irreducible Markov chain property supplies, for each $j$, an $n$ with $(P^n{)}_{kj}>0$. Nonnegativity gives $x_j\ge x_k(P^n{)}_{kj}>0$. Every coordinate of a nonzero invariant measure is positive.

Start the Markov chain at $k$, and let $T_k=\inf \{n\ge 1:X_n=k\}$ be its first return time. Each product in the series is the probability of a path with no intermediate visit to $k$. ThusThis is the expected number of visits to $j$ in an excursion, including the return endpoint and excluding the starting point. Because the chain is recurrent Markov chain, $T_k<\infty$ almost surely, and exactly one visit to $k$ is counted: ${\gamma }_k^k=1$. Counting the starting point instead of the return endpoint gives the equivalent usual excursion occupation measure.

For the lower bound, let $J=I\setminus \{k\}$, $Q=(p_{ij}{)}_{i,j\in J}$, and $r=(p_{kj}{)}_{j\in J}$. The stationarity equations on $J$, with $x_k=1$, are $x_J=r+x_{J}Q$. Iterating and using nonnegativity yieldsThe $j$th coordinate on the right is precisely the partial excursion series for $j\ne k$. Taking its increasing limit gives $x_j\ge {\gamma }_j^k$; for $j=k$ equality follows from $x_k={\gamma }_k^k=1$. The iteration is valid also on a countable state space because all summands are nonnegative.