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Let $T_0=0$ and let $T_{r+1}$ be the first visit to $i$ strictly after $T_r$, with $T_{r+1}=\infty$ if there is none. Put $q={\mathbb{P}}_i(T_1<\infty )$. The Strong Markov property at each finite first return time gives, by induction,The number of visits, including time zero, is $N={\sum }_{r\ge 0}{1}_{\{T_r<\infty \}}={\sum }_{n\ge 0}{1}_{\{X_n=i\}}$. By Tonelli theorem for nonnegative terms,If $q<1$, this geometric series is $1/(1-q)$, and $1-q$ is the probability of never returning. If $q=1$, every term of the final series is one and both sides are infinite. Thuswith the prescribed infinite-value convention. This covers both transient states and recurrent states.

Starting at zero, the next renewal is at time $Y_1$, and there are no intervening renewals because $Y_1\ge 1$. Thus the first return time obeys $T_0=Y_1$ in distribution, indeed under the original renewal construction. Its mean recurrence time isThe chain returns to zero almost surely and has a finite expected return time, so zero is a positive recurrent state. The assumption $\mu <\infty$ is what upgrades recurrence to positive recurrence.

Conditional on $X_0=x$, the random variable $T$ is the first return time $T_x^{+}$. The Kac's lemma mean-recurrence formula for a finite irreducible Markov chain givesAveraging over $X_0\sim \pi$ therefore yields

A continuous-time Markov chain is irreducible if every state can reach every other state with positive probability. A state $i$ is recurrent if, after leaving $i$, the chain returns to $i$ with probability one; an irreducible chain is recurrent if one, and hence every, state is recurrent. A recurrent state is positive recurrent if its expected continuous first return time is finite, and an irreducible chain is positive recurrent when all its states are.

False. Again use two states, but now takeThe support condition holds. Under $P$, the first return time to state one is identically one, so ${\mu }_1=1$. Under $Q$, the chain is irreducible with stationary distribution $(1/2,1/2)$. The mean recurrence time formula givesThus enlarging transition support does not imply a smaller mean return time.