Yes, let's assume that sup (which I think should be in the denominator in the last expression) is finite, as otherwise it is always true. Suppose it is bounded by A, the each return time is also quite finite, as by chebyshev inequality $P(T > 2A) < \frac 1 2$, and using the markov property this makes each return time subgeometric, compared with the same geometric distribution. Now using the ergodic theorem calculate $\mu(C)$ by counting the number of time the process is in C. If $T_1, T_1 + T_2, $ etc are those times, then $ {\sum^n 1_C(X_i)} < na = P(T_1 + ... + T_{na} > n) $. However, by the above remarks, you can count on $T_1 + ... + T_{na}$ to be tightly concentrated around its mean, and each mean is smaller than $\sup E_x(\tau^C)$

Yes, let's assume that sup (which I think should be in the denominator in the last expression) is finite, as otherwise it is always true. Suppose it is bounded by A, the each return time is also quite finite, as by chebyshev inequality $P(T > 2A) < \frac 1 2$, and using the markov property this makes each return time subgeometric, compared with the same geometric distribution. Now using the ergodic theorem calculate $\mu(C)$ by counting the number of time the process is in C. Let $a > \mu(C)$. If $T_1, T_1 + T_2, $ etc are those times, then $ {\sum^n 1_C(X_i)} < na = P(T_1 + ... + T_{na} > n) $ and so both are about 1. Therefore you expect that $n < E(T_1 + ... + T_{na}) < na \sup(E(T) $ and therefore $a > \frac 1 {\sup(E(T)}$. I had started out arguing that they returns times were subgeometric, which I though I was going to use, but seem not to have, but I'm going to leave it there in case it occurs to me why I wanted it.