Adding to my series of renewal theory questions, I'm trying to understand a heuristic given for the stationary distribution of the forward recurrence time of a renewal process.
In what follows, the renewal process is given by $\{S_n\}$. Let the total number of renewals by time $t$ be given by $N_t$ and define the forward recurrence time by $B_t = S_{N_t} - t$, i.e. the time until the next renewal.
The passage from Asmussen's text which I'm confused about is the following:
The form of the stationary distribution $F_0$ for $\{B_t\}$ is easily guessed by a level crossing argument. Namely, in a stationary situation the average number of upcrossings of a level $x>0$ should be the same as the average number of downcrossings, which in turn leads to the rate of upcrossings being equal to the rate of downcrossings. Assume that $F_0$ exists and has a density $f_0$. An upcrossing of a stationary version of $\{B_t\}$ in $[0,h]$ occurs if $B_0 \in (0,h]$ and the jump out of time $B_0$ exceeds $x$ so that the rate is $$ f_0(0) (1-F(x)) h + o(h) $$
But how does the last expression follow from the argument he gave right above it? For example, how exactly does one argue the $o(h)$ term? It's not very obvious to me.