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Once a nonnegative martingale hits zero, it stays zero almost surely. Conditional expectation on $\{M_n=0\}$ gives $\mathbb{E}[M_{n+1}{1}_{\{M_n=0\}}]=0$; nonnegativity forces the next value to vanish there. A countable intersection makes these assertions simultaneous for all times, proving absorption at the first zero.

Let a nonnegative martingale start at one, let $T$ be its first zero, and let $T_R$ be its first hit of $[R,\infty )$. Suppose $\mathbb{E}[(M_{n+1}-M_n{)}^2\mid {\mathcal{F}}_n]\ge {\sigma }^2{1}_{\{T>n\}}$ and $M_{T_R}\le \tau R$ on finite hits, for thresholds $R\ge 1$. Then $M_{T\wedge T_R\wedge n}^2-{\sigma }^2(T\wedge T_R\wedge n)$ is a submartingale. The bounded stopped values have mean one, so their second moment is at most $\tau R$. Consequently $\mathbb{E}(T\wedge T_R)\le (\tau R-1)/{\sigma }^2$. For $\tau \ge 1$ this implies the weaker bound ${\tau }^{2}R/{\sigma }^2$. The threshold restriction prevents an impossible overshoot requirement below the starting value.

For an adapted process with right-continuous paths and a stopping time $T$, $X_t^T=X_{t\wedge T}$ is adapted. At fixed $t$, round $t\wedge T$ upward on finite partitions of $[0,t]$. Every approximating value is ${\mathcal{F}}_t$-measurable, and right continuity gives the limit. Completeness handles a null exceptional set if right continuity holds only almost surely. A martingale assumption and boundedness of $T$ are unnecessary.

A submartingale with ${\sup }_n\mathbb{E}{Y}_n^{+}<\infty$ converges almost surely to a finite integrable random variable. The positive-part bound also bounds negative parts because $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$. The result specializes to an $L^1$-bounded martingale, and hence to every nonnegative martingale. It does not by itself give convergence in L1; uniform integrability supplies that stronger conclusion.

Suppose $\epsilon =o(1)$ and $\epsilon \ge n^{-1/6}$. The breadth-first exploration of a binomial random graph has deterministic drift $F(t)=\epsilon t-t^2/(2n)+O({\epsilon }^{3}n+\epsilon )$ up to $3\epsilon n$ steps. Its weighted martingale has variance $O(\epsilon n)$. For $h=\sqrt{\epsilon }+(n{\epsilon }^3{)}^{-1/8}$, the Doob L2 maximal inequality makes its maximum smaller than $h{\epsilon }^{2}n/6$ with high probability. Then $A_t>0$ throughout $[h\epsilon n,(2-h)\epsilon n]$, giving a graph component of order $(2-o(1))\epsilon n$.

For the upper expectation bound, let $K=\lceil {\epsilon }^{-3}\rceil$. The dominating binomial branching process has branching survival probability $(2+o(1))\epsilon$ by the binomial branching survival correction. Its branching process conditioned on extinction has mean $1-\epsilon +O({\epsilon }^2+\epsilon /n)$ and total-progeny expected value $O(1/\epsilon )$. Therefore $\mathbb{E}{N}_{\ge K}\le n\rho +O(n/(\epsilon K))$. Since $L_1\le K+N_{\ge K}$ and $K=o(\epsilon n)$, this yields the matching upper bound. Controlling rare large components is necessary to conclude an expected value asymptotic from a typical-size statement.

For a zero-start continuous local martingale $M$ with $[M{]}_{\infty }\le C$ deterministically, its stochastic exponential is a uniformly integrable martingale. For each $p>1$, ${\sup }_t\mathbb{E}\mathcal{E}(M{)}_t^p\le e^{p(p-1)C/2}$. Its terminal value is strictly positive, so it defines an equivalent infinite-horizon change of measure by the Girsanov theorem.

Every bounded harmonic function on a graph is constant if the connected locally finite graph is recurrent. Along its random walk on a graph, the harmonic values form a bounded martingale and converge almost surely. Recurrence visits each vertex infinitely often, making every vertex value a subsequential limit of the same convergent sequence. Thus every value agrees.

If $u$ is a bounded harmonic function on a domain $D$, continuous on its closure, and the Brownian exit time $T_D$ is almost surely finite from $z$, then $u(z)={\mathbb{E}}_z[u(B_{T_D})]$. The Itô formula and compact localization make $u(B_{t\wedge T_D})$ a bounded martingale. bounded convergence theorem at the exit time proves the representation even if the domain is unbounded. Removing boundedness of $u$ can invalidate this conclusion through loss of uniform integrability.

A discrete-time martingale with a common deterministic bound on all increments either converges finitely or has $\mathrm{liminf}{X}_n=-\infty$ and $\mathrm{limsup}{X}_n=+\infty$, almost surely. Stopping on first crossing below $-A$ gives a lower bound $-A-M$ on the overshoot. Adding $A+M+X_0^{-}$ produces a nonnegative martingale, so the Martingale convergence theorem forces finite convergence on the event of never crossing. Apply the same argument to $-X$.

A uniformly bounded local martingale is a true uniformly integrable martingale: a localizing sequence gives uniformly dominated stopped values, so their conditional identities pass to the limit. A deterministic bound on each compact time interval gives a true martingale on each such interval, even when the bound grows with time.

A bounded stopping time is a stopping time bounded almost surely by a deterministic finite constant. In discrete time, a bound $T\le N$ makes $M_T$ integrable whenever $M_0,\dots ,M_N$ are integrable random variables, and the optional stopping theorem applies to a martingale without additional limiting hypotheses. Almost-sure finiteness alone is weaker than boundedness.

During Breadth-first search of $G(n,p)$ let $U_t$ count unseen vertices, $A_t$ active vertices, and $t$ explored vertices. When $A_{t-1}=0$ start a new unseen root, recorded by $b_t=1$; otherwise $b_t=0$. The newly discovered count is conditionally $\mathrm{Bin}(U_{t-1}-b_t,p)$. Subtract its conditional mean to obtain a martingale difference $D_t$. With $q=1-p$, $U_0=n$ and $A_0=0$, iteration givesThe last sum is a martingale. Positive $A_t$ over an interval means that the graph component under exploration does not finish there. Independently adding phantom children couples each component exploration below a binomial branching process with offspring law $\mathrm{Bin}(n,p)$.

A càdlàg martingale is a continuous-time martingale represented by a càdlàg process: its paths are right-continuous and have finite left limits at positive times, on one common probability-one event. This path regularity permits approximation of stopping times from above.

An integrable adapted process is a martingale if all its values at bounded stopping times are integrable with the same expectation as $X_0$. For the converse, compare deterministic $t$ with $U=s{1}_A+t{1}_{A^c}$, where $s<t$ and $A\in {\mathcal{F}}_s$. The resulting identity $\mathbb{E}[(X_t-X_s)1_A]=0$ is precisely the defining test for conditional expectation. In the forward direction, use the optional stopping theorem for a càdlàg martingale under the usual conditions for a filtration.

Let an integrable process $(M_n)$ be adapted to $({\mathcal{F}}_n)$. It is a martingale if and only if $\mathbb{E}[M_{n\wedge \tau }]=\mathbb{E}[M_0]$ for every $n$ and every stopping time $\tau$. For the reverse implication, fix $A\in {\mathcal{F}}_n$ and take $\tau =n$ on $A$ and $\tau =n+1$ on $A^c$. Comparing its stopped expectation at $n+1$ with the expectation obtained from the deterministic stopping time $n+1$ gives $\mathbb{E}[1_A(M_{n+1}-M_n)]=0$. Since this holds for every $A\in {\mathcal{F}}_n$, it is exactly $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$.

For an integrable random variable $Z$ and an increasing filtration $({\mathcal{F}}_n)$, the process $M_n=\mathbb{E}[Z\mid {\mathcal{F}}_n]$ is a martingale. The tower property of conditional expectation gives the martingale identity.

A continuous local martingale $M$ with $M_0=0$ and deterministically bounded quadratic variation on every finite horizon is a square-integrable martingale, with $\mathbb{E}{M}_t^2=\mathbb{E}[M{]}_t$. Localization, the Doob L2 maximal inequality, and the resulting integrable bound on the squared supremum justify removing the localizing sequence.

A continuous martingale is a continuous-time martingale whose paths are continuous outside one null event. Continuity does not remove the integrability or conditional-expectation requirements in the martingale definition.