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Take independent and identically distributed random variables ${\xi }_k$ with the Bernoulli distribution of parameter $1/2$, and their natural filtration. SetThe event at $n=0$ is the whole sample space. Independence givesso this is a nonnegative martingale; it is integrable with $\mathbb{E}{M}_n=1$. The first zero among the ${\xi }_k$ is almost surely finite, since the probability of having only ones through time $n$ is $2^{-n}\to 0$. After that first zero, every $M_n$ is zero. Hence the almost sure limit is zero:This fair-coin doubling martingale also exhibits failure of uniform integrability: convergence in L1 would force the expectations to converge to zero.

The martingale convergence theorem in its $L^1$-bounded form says that a martingale with ${\sup }_n\mathbb{E}|M_n|<\infty$ has an integrable limit $M_{\infty }$ andThe norm bound on the limit follows from Fatou lemma. Boundedness in $L^1$ by itself does not imply convergence in L1. The uniformly integrable martingale convergence theorem gives the stronger conclusion: if $(M_n)$ is uniformly integrable, then $M_n\to M_{\infty }$ both almost surely and in L1 norm, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$. Conversely, convergence in L1 implies uniform integrability.

For the distinction, the fair-coin doubling martingale $M_n=2^n{1}_{\{\text{first }n\text{ tosses are heads}\}}$ has expectation one for every $n$ but converges almost surely to zero. Its L1 norm remains one, so its convergence is not in L1 norm. These two formulations specify exactly which hypothesis is needed in part (d).

The martingale convergence theorem says that if a discrete-time martingale $(M_n)$ satisfies $C={\sup }_n\mathbb{E}|M_n|<\infty$, then there is an integrable random variable $M_{\infty }$ such thatThe hypothesis does not by itself ensure convergence in L1.

For each pair of rational numbers $a<b$, the Doob upcrossing inequality gives $\mathbb{E}{U}_N[a,b]\le (C+|a|)/(b-a)$. The monotone convergence theorem therefore gives a finite expected value for $U_{\infty }[a,b]={\lim }_N{U}_N[a,b]$, so this upcrossing count is finite almost surely. There are only countably many such pairs, hence all their upcrossing counts are simultaneously finite outside one event of probability zero.

If ${\mathrm{liminf}}_n{M}_n<{\mathrm{limsup}}_n{M}_n$, the density of the rational numbers supplies $a<b$ strictly between them, forcing infinitely many upcrossings. Consequently $M_n$ has a limit in the extended real numbers almost surely. By the Fatou lemma,A limit of either $+\infty$ or $-\infty$ would make this lower limit infinite, so the limit is finite almost surely, and the same inequality proves its integrability.

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.

The original PDF has $\mathbb{E}|X_T|<\infty$ and $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$. The TeX transcription drops the expected value and absolute value in the first condition, and the absolute value in the second. The proof uses the PDF's conditions.

Because the stopping time $T$ is finite almost surely, $X_{n\wedge T}\to X_T$ almost surely. More strongly,The first term tends to zero by hypothesis; the second does so by the dominated convergence theorem. Thus there is convergence in L1.

Here is the needed L1 convergence implies uniform integrability argument. For any integrable random variables $Z,Y$ and $K>0$, splitting according to $|Y|>K/2$ givesTake $Z=X_{n\wedge T}$ and $Y=X_T$. For large $n$, the first term is uniformly small by convergence in L1, and the second is small for large $K$ by integrability. The finitely many remaining $n$ are handled individually by integrability. Hence the stopped process is uniformly integrable. This proves the stopped-martingale uniform integrability criterion; the martingale assumption is not needed for this particular implication.

Write $\Delta X_{n+1}=X_{n+1}-X_n$ and $Y={\sum }_{n\ge 0}|\Delta X_{n+1}|1_{\{T>n\}}$. Since $T$ is a stopping time, $\{T>n\}\in {\mathcal{F}}_n$. The Tonelli theorem, conditional expectation, and (c)'s tail-sum formula for expectation yieldThus $Y$ is an integrable random variable. Since $X_0=0$, telescoping gives $|X_T|\le Y$ and $|X_n|1_{\{T>n\}}\le Y{1}_{\{T>n\}}$. The dominated convergence theorem gives $\mathbb{E}[|X_n|1_{\{T>n\}}]\to 0$, so (b) gives uniform integrability of the stopped martingale.

The optional stopping theorem at the bounded stopping time $n\wedge T$ gives $\mathbb{E}{X}_{n\wedge T}=\mathbb{E}{X}_0=0$. Passing to the limit using the convergence in L1 proved in (b),One can also pass directly by the dominated convergence theorem, since $|X_{n\wedge T}|\le Y$.

The sigma-algebras ${\mathcal{G}}_n=\sigma (S_n,S_{n+1},\dots )$ decrease with $n$, and (b) gives $S_n/n=\mathbb{E}[X_1\mid {\mathcal{G}}_n]$. The permitted reverse martingale convergence theorem therefore gives an integrable random variable $L$ such that $S_n/n\to L$ almost surely and with convergence in L1. In particular,It remains to prove that $L$ is constant. We must not assume that ${\bigcap }_n{\mathcal{G}}_n$ is the tail sigma-algebra of the $X_j$; instead use tail measurability of limits of sample averages directly. Define the random variable valued in the extended real numbers $L^{\ast }={\mathrm{limsup}}_n{S}_n/n$. For every fixed $N$,because $S_N/n\to 0$ pointwise. Thus $L^{\ast }$ is measurable with respect to $\sigma (X_{N+1},X_{N+2},\dots )$ for every $N$, and hence to the tail sigma-algebra. Also $L^{\ast }=L$ almost surely.

By the Kolmogorov zero-one law, each tail event $\{L^{\ast }\le q\}$, for a rational number $q$, has probability zero or one. Since $L^{\ast }$ is finite almost surely, its distribution function can only be that of a constant $c$: taking $c=\inf \{q\in \mathbb{Q}:\mathbb{P}(L^{\ast }\le q)=1\}$ and using the density of the rational numbers gives $L^{\ast }=c$ almost surely. Its expected value identifies $c=m$. This proves the strong law of large numbers:

Yes. The preceding calculation established $\mathbb{E}{L}_{\tau }=-2ab/(b-a)<\infty$. For every $n$, the stopped martingale satisfiesHere $H$ is a single nonnegative integrable random variable that bounds the whole family. Thus, for any $K>0$,The last limit follows from the dominated convergence theorem. This is precisely uniform integrability. It also gives $M_{n\wedge \tau }\to M_{\tau }=|S_{\tau }|-L_{\tau }$ with convergence in L1, and therefore $\mathbb{E}{M}_{\tau }=0$, consistently with the preceding answer. A common bound on the individual expected values alone would not have sufficed; the integrable dominating random variable is what supplies control of every tail.

The uniformly integrable martingale convergence theorem states that a uniformly integrable martingale $(U_n)$ has an integrable terminal random variable $U_{\infty }$ such that $U_n\to U_{\infty }$ almost surely and in the L1 norm, andThe last identity follows by passing to the limit of a sequence in $\mathbb{E}[U_m{1}_A]=\mathbb{E}[U_n{1}_A]$ for $A\in {\mathcal{F}}_n$ and $m\ge n$, using convergence in L1.

Part (b)(iii) verifies uniform integrability for $(Z_n)$, soMoreover $|Z_{\infty }|\le L$ almost surely. In this particular problem the common bound also yields convergence in Lp for every finite $p\ge 1$, including convergence in L2: the dominated convergence theorem applies to $|Z_n-Z_{\infty }{|}^p\le (2L{)}^p$. General uniform integrability alone does not assert these higher-power conclusions.

For an almost surely finite stopping time $T$, if $\mathbb{E}|M_T|<\infty$ and $\mathbb{E}[|M_n|1_{\{T>n\}}]\to 0$, then $M_{n\wedge T}\to M_T$ with convergence in L1. Thus L1 convergence implies uniform integrability makes the stopped martingale uniformly integrable.

For a simple symmetric random walk, the infinite mean first passage of a simple symmetric random walk gives a finite almost sure first passage $T_1$ to $1$. The stopped martingale $M_n=1-S_{n\wedge T_1}$ is nonnegative and eventually zero almost surely. Nevertheless $\mathbb{E}{M}_n=1$ for every $n$, so it has almost sure convergence but no convergence in L1. It is therefore not uniformly integrable. Before absorption the nearest-neighbour walk is at most zero, ensuring nonnegativity.

An $L^1$-bounded supermartingale has an almost sure convergence limit $X_{\infty }$ in $L^1$. The conditional-expectation martingale $M_n=\mathbb{E}[X_{\infty }\mid {\mathcal{F}}_n]$ has uniform integrability and converges to $X_{\infty }$ both almost surely and in $L^1$. To identify its limit, test against every event in every finite-time sigma-algebra and use uniqueness on their generated sigma-algebra. Then $Y=X-M$ is a supermartingale tending to zero almost surely. There is no general nonnegativity conclusion for $Y$, nor general convergence in L1 to zero: the negative of the fair-coin doubling martingale is a counterexample to both stronger assertions.

A uniformly integrable martingale $M_n$ has almost sure convergence and convergence in L1 to an integrable random variable $M_{\infty }$, and $M_n=\mathbb{E}[M_{\infty }\mid {\mathcal{F}}_n]$.