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For a real sequence of exchangeable random variables, every event $A$ in its invariant sigma-algebra of an exchangeable sequence agrees almost surely with an event of its tail sigma-algebra. Put $F=1_A$. Approximate $F$ in $L^1$ by its conditional expectation on the first $n+1$ coordinates. Swap that block with the next $n+1$ coordinates. Invariance of $A$ and the law shows that the conditional expectation on the second block is an equally good approximation. It is measurable with respect to the tail after index $n$. The L1 contraction of conditional expectation then bounds $\Vert F-\mathbb{E}[F\mid {\mathcal{G}}_n]{\Vert }_1$ by twice the approximation error. The reverse martingale convergence theorem gives $F=\mathbb{E}[F\mid {\bigcap }_n{\mathcal{G}}_n]$ almost surely. Thresholding the latter at $1/2$ supplies the required tail event. Together with the reverse inclusion, the two sigma-algebras have identical probability completions.

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.

Let $Y$ be an integrable random variable, let $H$ be a finite $\mathcal{G}$-measurable random variable, and suppose $HY$ is an integrable random variable. Then a known factor can be taken outside the conditional expectation:For a bounded $H$, first verify the defining integral identity when $H$ is the indicator function of a measurable set in $\mathcal{G}$, extend by linearity to simple functions, and then use bounded approximation and the dominated convergence theorem. For nonnegative $H,Y$, truncating $H$ and using the conditional monotone convergence theorem proves the identity with extended nonnegative conditional expectations. In particular,For general $H,Y$ under the stated integrability assumptions, the conditional Jensen inequality bounds $|H\mathbb{E}[Y\mid \mathcal{G}]|$ by that integrable random variable. Apply the bounded-factor result to truncations of $H$, then use the dominated convergence theorem on the products and the L1 contraction of conditional expectation. This proves the identity without assuming $H$ itself is an integrable random variable.

The product-integrability condition is essential to the ordinary finite conditional expectation statement. An unbounded factor cannot be moved through an expected value simply because its partner is an integrable random variable.^[ref]

We first construct a terminal integrable function without using the martingale convergence theorem. Put $G={\sup }_n|g_n|$ and $d\eta =Gdx$ on $(0,1{]}^d$. This is a finite positive measure. If $G=0$ almost everywhere the conclusion is immediate. Otherwise, for a bounded dyadic step function $h$ measurable with respect to ${\mathcal{F}}_n$ in the dyadic filtration, defineThe martingale property makes the definition independent of the choice of a sufficiently large $n$. Moreover $|\Lambda (h)|\le \int |h|d\eta$. Dyadic step functions are dense in $L^1(\eta )$: continuous functions on the closed unit cube are dense for this finite Borel measure (extend the measure by zero onto the omitted faces), and uniform continuity approximates each continuous function by step functions on fine dyadic partitions. Thus $\Lambda$ extends to a bounded linear functional on $L^1(\eta )$ of norm at most one. The duality of Lp spaces representation of this functional gives $H\in L^{\infty }(\eta )$ with $|H|\le 1$ and $\Lambda (h)=\int hHd\eta$. Set $g=HG$, taking $g=0$ where $G=0$. Then $g\in L^1(dx)$ and testing indicators of all dyadic atoms provesThis argument also works for complex-valued martingales, with a complex linear functional.

To prove convergence, choose a dyadic step function $h$ with $\Vert g-h{\Vert }_1$ arbitrarily small. For all $n$ beyond its dyadic level, $\mathbb{E}(h\mid {\mathcal{F}}_n)=h$. The L1 contraction of conditional expectation yieldsThis proves convergence in $L^1$. For convergence almost everywhere, Doob maximal inequality applied to the conditional expectations of $g-h$ givesLet the approximation error tend to zero, and then take positive rational $a$. Hence $g_n\to g$ both in $L^1$ and almost everywhere.

For the second assertion, write $M={\sup }_n\mathbb{E}|f_n|<\infty$ and take the given first-exit stopping time $\tau$. Each $g_n=f_{\tau \wedge n}$ is integrable, being a finite sum of restrictions of integrable variables. Its increments satisfyThe indicator is ${\mathcal{F}}_n$-measurable, so taking its conditional expectation proves that $(g_n)$ is a martingale.

The overshoot at $\tau$ need not be bounded by $N$. Instead, for a fixed $m$, on $\{\tau =j\}$ with $j\le m$ we have $|f_j|\le \mathbb{E}(|f_m|\mid {\mathcal{F}}_j)$. Integrating over this ${\mathcal{F}}_j$-measurable event, summing the disjoint events, and including $\{\tau >m\}$ givesThe variables $Z_m=|f_{\tau }|1_{\{\tau \le m\}}$ increase to $Z$, and $Z_m\le |g_m|$. The monotone convergence theorem gives $\mathbb{E}Z\le M$. Before stopping, the process has magnitude less than $N$; after stopping it remains at $f_{\tau }$. ConsequentlyThis is the integrable overshoot of a stopped L1-bounded martingale. The first part now proves that every such stopped process converges almost everywhere.

Finally, Doob maximal inequality gives $|\{{\sup }_n|f_n|\ge N\}|\le M/N$, by applying the finite-horizon estimate and then increasing the horizon. Thus ${\sup }_n|f_n|$ is finite almost everywhere. Outside the union of the null sets for all positive integer stopping thresholds, choose an integer $N$ larger than this supremum. No stopping occurs at that point, so $f_n=g_n$ for all $n$, and $f_n$ converges to a finite limit almost everywhere. We have not assumed, or concluded, $L^1$ convergence for this general $L^1$-bounded martingale.

Here is an alternative route for the ergodicity step using the Birkhoff ergodic theorem. After invariance has been proved, the theorem identifies the almost-everywhere limit of $A_n\phi$ with ${\mathbb{E}}_{\mu }[\phi \mid \mathcal{I}]$. The hypothesis therefore givesfor every continuous function $\phi$. For an invariant Borel set $B$, approximate $h=1_B$ in $L^1$ by such a $\phi$. Since $h$ is measurable with respect to the invariant sigma-algebra, its conditional expectation is $h$ itself. The L1 contraction of conditional expectation givesArbitrarily accurate continuous approximation forces $h=\mu (B)$ almost everywhere. Thus this use of the theorem reaches the same zero-or-one criterion as the direct averaging proof above, without assuming in advance that $\mu$ is ergodic.

Let $T$ swap coordinate blocks $0,\dots ,n$ and $n+1,\dots ,2n+1$, fixing all other coordinates. The symmetric law makes $T$ measure preserving, and invariance of the event $A$ gives $F\circ T=F$ pointwise. Put ${\mathcal{B}}_n=\sigma (X_{n+1},\dots ,X_{2n+1})$.

Conditional expectation transforms covariantly under this block swap:To check it, the left side is ${\mathcal{B}}_n$-measurable, and for $C\in {\mathcal{B}}_n$ the change of variables by $T$ sends its defining integral to the conditional-expectation identity on $T(C)\in {\mathcal{E}}_n$, with $F\circ T=F$. Thus the two versions agree. Measure preservation now givesThis is the first requested estimate.

The L1 contraction of conditional expectation follows from the pointwise bound $|\mathbb{E}[Z\mid \mathcal{H}]|\le \mathbb{E}[|Z|\mid \mathcal{H}]$: take expectations to get $\Vert \mathbb{E}[Z\mid \mathcal{H}]{\Vert }_1\le \Vert Z{\Vert }_1$. Applied to a difference, it makes conditional expectation a contraction between any two integrable inputs.

Let $U_n=\mathbb{E}[F\mid {\mathcal{B}}_n]$. Since ${\mathcal{B}}_n\subseteq {\mathcal{G}}_n$, $U_n$ is ${\mathcal{G}}_n$-measurable. HenceCombining this with the other given approximation bound yieldsThe left side does not depend on $n$, and $\epsilon >0$ was arbitrary. It is therefore zero. For a $\mathcal{G}$-measurable version $Y=\mathbb{E}[F\mid \mathcal{G}]$, the event $C=\{Y>1/2\}$ belongs to $\mathcal{G}$ and satisfiesThis proves that exchangeable invariant events are tail events modulo null sets. It does not assert literal pointwise equality of the two events or sigma-algebras before completing them with null sets.

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.