Doob's upcrosssing inequality states that, for martingale $X_n$, if $U_n(a,b)$ counts the number of upcrossings through $(a,b)$ up to time $n$, then $E[U_n(a,b)]\leq \frac{1}{b-a}E[(a-X_n)_+]$.
We get this result by looking at stopping times that correspond to each up-crossing: namely, $\tau_{2k-1}$ corresponds to the beginning of the $k$th up-crossing, where $X_{\tau_{2k-1}} \leq a$ and $\tau_{2k}$ corresponds to the end of that up-crossing, so that it's the first time after $\tau_{2k-1}$ where $X_{\tau_{2k}}\geq b$ (and we cap all the stopping times by time $n$).
If we sum up the corresponding jumps: $\sum_k X_{\tau_{2k}}-X_{\tau_{2k-1}}$, then of course by design, we will have $U_n(a,b)$ jumps of size at least $b-a$. We might additionally get a "partial jump" at the end, where $\tau_{2k-1}<n$ but we get to $n$ before the system makes it past $b$.
Taking expectations, we see that by Doob's optional stopping theorem, each of the full jumps has expectation zero, so we end up with a bound that's only dependent on the potential partial jump. I'm eliding some details here, since they don't pertain directly to my question, but this is the general idea.
My question is really about intuitions. There's something very weird about the fact that we designed our stopping times specifically so that the jumps $X_{\tau_{2k}}-X_{\tau_{2k-1}}$ are of size at least b-a, but in expectation they are of size zero. This feels very odd to me.
One explanation might be that really what this saying is that actually in expectation these jumps just don't occur, but if they don't occur in expectation, then shouldn't our expected number of up-crossings be zero, since that's precisely what that statement means?
Is there a better way to understand how jumps that are certainly positive end up being zero in expectation? I really want an intuitions-based answer; I understand the application of the optional stopping theorem, I'm just struggling to make sense of what is happening morally.