In the section on Markov chains in Billingsley's Probability and Measure (3e) we have the following inequality on page 120 in the proof of Theorem 8.3,
$$ \begin{align*} p_{ji}^{(m)} &= P_j([X_m=i] \cap [X_n = j \text{ i.o.}])\\ &\le \sum_{n>m} P_j (X_m = i, X_{m+1} \ne j,\ldots, X_{n-1} \ne j,X_n=j) \end{align*} $$
[Notation: $p_{ji}^{(m)} = P(X_m =i \mid X_0=j)$, $P_j(A) = P(A \mid X_0=j)$.]
My question is regarding the inequality sign on the second line.
$X_n=j$ infinitely often if and only if $X_n=j$ infinitely often for $n>m$. Each such path must hit $j$ for the first time after period $m$ at some finite time $n$. So it seems to me that the events on right-hand side of the second line form a partition of the event on right-hand side of the first line and therefore we must have an equality here.
Am I going wrong somewhere?
The sum on the second line computes the probability that $X_n$ hits $j$ at least once. If there is a positive probability that $X_n$ hits $j$ at least once but not infinitely often (e.g. if $j$ is transient), then the inequality is strict.