the upcrossing inequality
If a sequence of numbers fails to settle to a limit, it must instead oscillate — coming back down below some level a and climbing back up above some higher level b, over and over, forever. Each such down-then-up sweep across the gap [a, b] is called an upcrossing. The slick idea behind the martingale convergence theorem is to count these oscillations: if a process can only cross any gap a finite number of times, it has no room left to oscillate and therefore must converge. The upcrossing inequality is the tool that bounds, on average, how many times a martingale can cross a fixed gap.
Let U_n[a, b] be the number of complete upcrossings of the interval [a, b] that the process M_0, ..., M_n makes by time n — each time it goes from at-or-below a up to at-or-above b. Doob's upcrossing inequality bounds the expected count: for a submartingale, (b - a) * E[U_n[a, b]] <= E[(M_n - a)^+], where x^+ means the positive part max(x, 0). The proof is gorgeously concrete: imagine a gambling strategy that buys (bets one unit long) the moment M drops to a and sells (stops) the moment it reaches b. Each completed upcrossing earns you at least b - a; this strategy is a predictable bet, so the martingale transform shows your expected winnings are at most E[(M_n - a)^+]; dividing by b - a bounds the expected number of upcrossings. You literally price the oscillations as gambling profits.
Why bound upcrossings? Because if for EVERY rational pair a < b the expected (hence almost-surely finite) number of upcrossings is finite, the process cannot oscillate across any gap infinitely often, and a sequence that never oscillates infinitely must converge (to a possibly infinite limit). Combined with an L^1 bound to rule out escape to infinity, this delivers the martingale convergence theorem directly, without any compactness magic. The honest note: the inequality bounds the EXPECTED number of upcrossings; it guarantees finiteness almost surely but says nothing about WHERE the limit lands or how fast — convergence, not a rate.
Track a fair walk and the band [a, b] = [2, 5]. Each time the walk slides down to 2 and later climbs to 5, that is one upcrossing worth at least 3 to a buy-at-2, sell-at-5 strategy. The upcrossing inequality says 3 * E[number of such sweeps by time n] <= E[(S_n - 2)^+], so the band can be crossed only finitely often on average — leaving no room for endless oscillation.
Counting oscillations as gambling profits: finitely many upcrossings means the process must converge.
It bounds the EXPECTED number of upcrossings, forcing finiteness almost surely; it says nothing about where the limit is or how fast convergence happens — existence, not rate.