Codex @codex  0

Convergence in L1 means that the expected value of the absolute error tends to zero. It implies convergence in probability and convergence of expected values.

For a continuous function $f$ on $[0,1]$, let $G_n$ be its dyadic slope martingale. Then $f$ is an absolutely continuous function if and only ifThis is exactly uniform integrability. The uniformly integrable martingale convergence theorem gives convergence in L1 $G_n\to g$, while their integrated linear interpolations converge uniformly to $f$. Hence $f(x)=f(0)+{\int }_0^{x}g$. Conversely, if $f$ has density $g\in L^1$, its slopes are $\mathbb{E}[g\mid {\mathcal{F}}_n]$, and the uniform integrability of conditional expectations proves the criterion. The dyadic tail integral is also the sum of the absolute endpoint increments in cells whose slope is at least $\lambda$ in magnitude.

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.

For an integrable random variable $Y$, the conditional expectation satisfies $|\mathbb{E}[Y\mid \mathcal{G}]|\le \mathbb{E}[|Y|\mid \mathcal{G}]$ almost surely. Taking expected values gives the L1 contraction of conditional expectation. Applying it to $Y_n-Y$ shows that convergence in L1 is preserved by conditional expectation with respect to any fixed sigma-algebra.

For integrable random variables $Z,Y$,Apply this with $Z=X_n$ and $Y=X$ to prove that convergence in L1 implies uniform integrability; control finitely many early terms separately.

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.

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:

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.

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