martingale. In particular, is a supermartingale which is bounded below, so by the martingale convergence theorem it converges pointwise almost surely to a random variable . But if then , so is almost surely zero. This means that . However, for every , since is a random walk which starts at and subsequently makes mean-zero moves (alternately, note that since is a martingale). Therefore cannot converge to in mean. Moreover, if were to converge in mean to any random variable , then some subsequence converges to almost surely. So by the above argument almost surely, which contradicts convergence in mean. Statements for the general case In the following, will be a filtered probability space where , and will be a right-continuous supermartingale with respect to the filtration ; in other words, for all , Doob's first martingale convergence theorem Doob's first martingale convergence theorem provides a sufficient condition for the random variables to have a limit as in a pointwise sense, i.e. for each in the sample space individually. For , let and suppose that Then the pointwise limit exists and is finite for -almost all . Doob's second martingale convergence theorem It is important to note that the convergence in Doob's first martingale convergence theorem is pointwise, not uniform, and is unrelated to convergence in mean square, or indeed in any Lp space. In order to obtain convergence in L1 (i.e., convergence in mean), one requires uniform integrability of the random variables . By Chebyshev's inequality, convergence in L1 implies convergence in probability and convergence in distribution. The following are equivalent: is uniformly integrable, i.e. there exists an integrable random variable such that as both -almost surely and in , i.e. Doob's upcrossing inequality The following result, called Doob's upcrossing inequality or, sometimes, Doob's upcrossing lemma, is used in proving Doob's martingale convergence theorems. A "gambling" argument shows that for uniformly bounded supermartingales, the number of upcrossings is bounded; the upcrossing lemma generalizes this argument to supermartingales with bounded expectation of their negative parts. Let be a natural number. Let be a supermartingale with respect to a filtration . Let , be two real numbers with . Define the random variables so that is the maximum number of disjoint intervals with , such that . These are called upcrossings with respect to interval . Then where is the negative part of , defined by . Applications Convergence in Lp