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For a L2-bounded continuous martingale, the terminal and maximal norms satisfyThe first inequality follows from almost sure convergence. Apply the Doob L2 maximal inequality on $[0,T]$ and then the monotone convergence theorem as $T\to \infty$ for the second. These are equivalent norms on the space modulo indistinguishability of stochastic processes; neither inequality needs a zero initial value.

For fair independent and identically distributed random variables ${\xi }_j$ with the Bernoulli distribution, $M_n=2^n{1}_{\{{\xi }_1=\cdots ={\xi }_n=1\}}$ is a nonnegative martingale. It has almost sure convergence to zero but $\mathbb{E}{M}_n=1$, so it lacks convergence in L1.

A continuous martingale $M$ is L2-bounded when ${\sup }_{t\ge 0}\mathbb{E}|M_t{|}^2<\infty$. The L2 martingale convergence theorem gives a terminal value $M_{\infty }$ with convergence in $L^2$ and almost sure convergence. The resulting conditional expectation identity is $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$. Boundedness on each separate finite horizon is a weaker condition. The subspace with $M_0=0$ is commonly denoted ${\mathcal{M}}_0^2$.

A martingale satisfying ${\sup }_n\mathbb{E}|M_n{|}^p<\infty$, for $1<p<\infty$, has almost sure convergence and convergence in Lp to a limit $M_{\infty }\in L^p$. Also $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. The Doob Lp maximal inequality makes ${\sup }_n|M_n|$ integrable to the power $p$; the Martingale convergence theorem gives the almost-sure limit, and the dominated convergence theorem gives convergence in the Lp norm. Boundedness in $L^1$ alone does not imply convergence in L1.

Define the stopped martingaleIt is nonnegative: before the first visit to $1$, the nearest-neighbour random walk is at most $0$, and afterwards the stopped position is $1$. Each $M_n$ is integrable, with $|M_n|\le n+1$. MoreoverThe indicator is ${\mathcal{F}}_n$-measurable and the increment has conditional mean zero, so $M_n$ is a martingale with $\mathbb{E}{M}_n=M_0=1$.

Part (a) shows $T_1<\infty$ almost surely, so every such path eventually has $M_n=0$. Thus $M_n\to 0$ almost surely. But $\mathbb{E}|M_n-0|=1$ for every $n$, so there is no convergence in L1. Any possible L1 convergence limit would agree with the almost sure limit. This is a nonnegative martingale with almost sure convergence but no convergence in L1, and hence it is not uniformly integrable.

The Martingale convergence theorem says that a discrete-time martingale $(M_n,{\mathcal{F}}_n)$ satisfyinghas an almost sure convergence limit $M_{\infty }$, finite almost surely and in $L^1$. The theorem asserts that the limit is integrable; it does not assert convergence in L1. More generally, the almost sure submartingale convergence theorem applies to a submartingale with ${\sup }_n\mathbb{E}{M}_n^{+}<\infty$. Uniform integrability is the additional condition that upgrades a martingale's convergence to convergence in L1.

For the requested distinction, let $({\xi }_k)$ be independent fair Bernoulli variables and use their natural filtration. The coin-doubling martingaleis a nonnegative martingale: conditionally on ${\mathcal{F}}_n$, the next factor is $2{\xi }_{n+1}$ with mean one, so $\mathbb{E}(M_{n+1}\mid {\mathcal{F}}_n)=M_n$. Also $\mathbb{E}|M_n|=\mathbb{E}{M}_n=1$ for every $n$, giving the required uniform $L^1$ bound. The probability that all the Bernoulli variables equal one is ${\lim }_n{2}^{-n}=0$. Therefore a zero is eventually encountered almost surely, after which $M_n$ stays zero. ThusThere can be no other $L^1$ limit, since convergence in L1 implies convergence in probability, whose limit must agree with the almost sure limit. This martingale satisfies the almost sure theorem but does not converge in $L^1$.

The almost sure submartingale convergence theorem says that if a submartingale $(Y_n)$ satisfiesthen there is an integrable random variable $Y_{\infty }$ such thatIndeed the positive-part bound also bounds the negative parts, since $\mathbb{E}{Y}_n\ge \mathbb{E}{Y}_0$; the theorem gives a finite limit. In particular, the Martingale convergence theorem applies to any martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$.

For a nonnegative martingale, $\mathbb{E}{M}_n=\mathbb{E}{M}_0$ supplies this bound automatically. Its limit obeys $\mathbb{E}{M}_{\infty }\le \mathbb{E}{M}_0$ by the Fatou lemma. Almost sure convergence alone need not preserve the expectation or give convergence in L1. Uniform integrability is the additional condition that supplies convergence in L1 and the identity $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.

For every $w\in C^{\alpha }[0,1]$, its finite Hölder seminorm gives the deterministic boundWhen $\alpha >1/2$, the right-hand side tends to zero. Thus under any probability measure on the stated coordinate sigma-algebra, these measurable finite-coordinate sums converge to zero on every path, and hence almost surely.

If the coordinate process were a Brownian motion, part (b) would make the same sums converge to one in $L^2$, and hence in probability. Almost sure convergence to zero also implies convergence in probability to zero. Uniqueness of a limit in probability is a contradiction. Therefore no such Brownian measure exists:This is the quadratic variation obstruction to Hölder regularity. It does not address the critical exponent $\alpha =1/2$, where the displayed deterministic bound does not tend to zero.

Work on a filtered probability space with the usual conditions for a filtration. A continuous martingale belongs to the class of L2-bounded continuous martingales whenThis means a uniform bound over the entire time interval, not merely being square-integrable at each individual time. The L2 martingale convergence theorem gives a terminal variable $M_{\infty }$ and almost sure convergence and convergence in the Lebesgue space $L^2$; moreover $M_t=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_t]$ by conditional expectation.

The predictable sigma-algebra $\mathcal{P}$ on $\Omega \times [0,\infty )$ is the smallest sigma-algebra making every left-continuous adapted process measurable. Equivalently, it is generated bywhere the sample coordinate comes first. A previsible process is precisely a process measurable for this predictable sigma-algebra. Values at time zero matter for its definition, although they make no contribution to integration against a continuous quadratic variation starting at zero.

Let ${\xi }_1,{\xi }_2,\dots$ be independent and identically distributed random variables with the fair Bernoulli distribution, and putRelative to ${\mathcal{F}}_n=\sigma ({\xi }_1,\dots ,{\xi }_n)$, this fair-coin doubling martingale satisfies $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$: while alive it doubles with probability $1/2$ and becomes zero otherwise. Its expected value is $\mathbb{E}{M}_n=2^n{2}^{-n}=1$.

The probability of an infinite run of ones is ${\lim }_n{2}^{-n}=0$, so $M_n$ is eventually zero almost surely. However,Thus the martingale lacks convergence in L1 to its limit from almost sure convergence. Nor can it have convergence in L1 to any other limit: convergence in L1 implies convergence in probability, whose limit is unique up to almost sure equality.

A sequence of integrable random variables $(X_n)$ has uniform integrability exactly whenThe uniformly integrable martingale convergence theorem states that a uniformly integrable discrete-time martingale $(M_n)$ has an integrable random variable $M_{\infty }$ withIndeed, uniform integrability implies ${\sup }_n\mathbb{E}|M_n|<\infty$, so the Martingale convergence theorem gives almost sure convergence. The combination of uniform integrability and almost sure convergence gives convergence in L1. For $m\ge n$, the martingale identity $\mathbb{E}[M_m\mid {\mathcal{F}}_n]=M_n$ passes to the limit by the L1 contraction of conditional expectation.

Conversely, uniform integrability of conditional expectations shows that a martingale of the form $M_n=\mathbb{E}[Y\mid {\mathcal{F}}_n]$ for one integrable random variable $Y$ is uniformly integrable.

Suppose $X_n\to X$ with convergence in probability. For real $t$ and $\delta >0$, the elementary inequality $|e^{iu}-e^{iv}|\le \min (2,|u-v|)$ givesFirst let $n\to \infty$, then let $\delta \downarrow 0$. The characteristic functions converge pointwise to that of $X$, which is continuous at zero. The Lévy continuity theorem gives convergence in probability implies convergence in distribution.

For the converse, let $Z$ have the fair Bernoulli distribution, put $X=Z$, and set $X_n=1-Z$ for every $n$. All $X_n$ have the same probability distribution as $X$, so $X_n\stackrel{d}{\to }X$. But $|X_n-X|=1$ always, henceThere is no convergence in probability to $X$. In this example $X_n$ does have convergence in probability to $1-Z$: the counterexample concerns the converse with the specified limit $X$, as usual. If one interprets the question as denying convergence in probability to any limit, take $X_n=Z$ for even $n$ and $X_n=1-Z$ for odd $n$. Its probability distribution is still constant, while the two subsequences have distinct limits from almost sure convergence, ruling out any common limit in probability by uniqueness of a limit in probability.