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For a discrete group, one standard definition of amenability is the fixed-point property: every continuous affine action on a nonempty compact convex set in a locally convex space has a fixed point. The equivalent state formulation, which may be stated without proof here, is that there is a left-invariant invariant mean on ${\ell }^{\infty }(\Gamma )$. Equivalently, every unital left-translation-invariant star-subalgebra of ${\ell }^{\infty }(\Gamma )$ admits a left-invariant state, with continuity for the supremum norm; one may equally pass to its norm closure. A state means a positive linear functional of norm one with value one on the constant function one. Restriction proves the subalgebra formulation from the full-algebra version, and taking the full algebra proves the converse. Such a mean is finitely additive on indicator functions, not necessarily countably additive.

Write $(\lambda (g)\mu )(h)=\mu (g^{-1}h)$ on ${\ell }^1(\Gamma )$ and $(L_{g}f)(h)=f(g^{-1}h)$ on bounded functions. We prove equivalence of this mean condition, the Reiter condition, and existence of a Følner sequence.

Assume first that an invariant mean $m$ exists. Fix a finite set $K\subseteq \Gamma$. In the real Banach space ${⨁}_{g\in K}{\ell }^1(\Gamma )$ with sum norm, consider the convex setIf zero were outside its norm closure, the Hahn-Banach separation theorem would give bounded real functions $f_g$ and $\delta >0$ such thatfor every such $\mu$. Testing point masses at $h$ showsApplying the positive invariant mean makes the left side zero and the right side at least $\delta$, a contradiction. Therefore zero is in the closure: for any $\epsilon >0$ a finitely supported probability measure makes the sum of the translation defects over $K$ less than $\epsilon$. Enumerate the countable group, take the first $n$ elements for $K$ and $\epsilon =1/n$, and obtain a sequence satisfying the Reiter condition. This proves (1) implies (2).

Next assume (2). An ${\ell }^1$ probability measure can be approximated in norm by finitely supported probability measures, by restricting to a large finite set and renormalizing. Replacing $\mu$ by $\nu$ changes each defect by at most $2\Vert \mu -\nu {\Vert }_1$. Thus simultaneous small defects for finitely many translations can be achieved with finite support.

For such $\nu$, set $F_t=\{h:\nu (h)>t\}$. The elementary identity $|r-s|={\int }_0^{\infty }|1_{r>t}-1_{s>t}|dt$ for $r,s\ge 0$ givesIf the sum of defects over $K$ is less than $\epsilon$, some $t$ with $F_t\ne \varnothing$ must satisfy ${\sum }_{g\in K}|F_t△g{F}_t|<\epsilon |F_t|$; otherwise integration would contradict that strict inequality. Applying this layer-cake extraction of Følner sets to successive finite sets of translations gives nonempty finite $F_n$ with each relative boundary tending to zero. Hence (2) implies (3).

Conversely, normalized counting measures ${\mu }_n=1_{F_n}/|F_n|$ satisfyso (3) implies (2). Finally, from (2) define states $m_n(f)={\sum }_h{\mu }_n(h)f(h)$. The dual unit ball of ${\ell }^{\infty }(\Gamma )$ is weak-star compact by the Banach-Alaoglu theorem; take a convergent subnet, since this compactness need not be metrizable. Its limit $m$ is positive with $m(1)=1$, andThus $m$ is invariant, proving (2) implies (1). Together these implications establish all three equivalences. Countability is used to obtain sequences from finite tests; it is not a justification for replacing the compactness subnet by a subsequence.

An equivalent martingale measure is a probability measure $Q$ equivalent to the objective measure $P$, meaning the two have exactly the same null events, under which the discounted stock ${\tilde{S}}_t=S_t/B_t$ is a martingale. In this one-period model the condition isThe discounted bank account is identically one. Integrability of the discounted stock is part of the martingale condition. In an augmented market the discounted price of every additional asset must satisfy the same condition.

By part (a), $f$ is a bounded continuous function, hence is Borel measurable and integrable for the given probability measure. By part (e), $l\in E^{\prime }$, and its restriction to the unit ball is also bounded and integrable. For every $u\in U$, the supporting inequality says $f(u)\ge f(x)+l(u)-l(x)$. Taking expectations and using the barycentre hypothesis for this particular $l$ givesThus $\boxed{\mathbb{E}f\ge f(x)}$, the Jensen inequality. This proof only uses the stated weak barycentre identity; a vector-valued integral for the identity map is not needed.

Identify the vertices of the Boolean hypercube $Q_n$ with subsets of $[n]$, adjacent when their symmetric difference has size one. Write $N(\mathcal{A})$ for the closed graph neighbourhood, including $\mathcal{A}$ itself. In simplicial order on the discrete cube, smaller sets come first, and equal-sized sets are ordered by lexicographic order: the least element of the symmetric difference belongs to the earlier member. Let $I_m$ denote the first $m$ vertices. The vertex-isoperimetric inequality in the discrete cube statesSubtracting $|\mathcal{A}|$ gives the equivalent assertion for the external vertex boundary.

We first check that $N(I_m)$ is also an initial segment in simplicial order on the discrete cube. Apart from the empty and full cases, $I_m$ consists of complete lower layers and a lexicographic initial segment $\mathcal{F}$ in its top rank $r$. Its closed graph neighbourhood contains all ranks at most $r$ and the upper shadow $\nabla \mathcal{F}$ in rank $r+1$. This upper shadow is lexicographic initial: an $(r+1)$-subset $C$ belongs to it precisely when its first $r$-subset in lexicographic order, obtained by deleting its largest element, belongs to $\mathcal{F}$. Taking the first $r$ elements preserves weak lexicographic order, proving the assertion. This also handles $r=0$, where $N(\{\varnothing \})$ consists of ranks zero and one.

We prove the vertex-isoperimetric inequality in the discrete cube by mathematical induction on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write ${\mathcal{A}}_0,{\mathcal{A}}_1\subseteq Q_{n-1}$ for the sections without and with $i$. The two sections of its closed graph neighbourhood areA simplicial section compression replaces ${\mathcal{A}}_j$ by the initial segment $I_{a_j}$ of size $a_j=|{\mathcal{A}}_j|$ in the other coordinates. By the preceding observation, both $N(I_{a_j})$ and $I_{a_{1-j}}$ are initial segments, so their union has size $\max \{|N(I_{a_j})|,a_{1-j}\}$. The mathematical induction hypothesis givesAdding the two section bounds proves that simplicial section compression cannot increase the closed graph neighbourhood.

Apply changing simplicial section compressions until none remains. The sum of the global simplicial order on the discrete cube positions strictly decreases at each step: within a section the order is exactly the restriction of the global order. Thus the procedure terminates at a set family $\mathcal{B}$ every coordinate section of which is initial. We now classify the possible terminal families for simplicial section compression, rather than assuming this terminal set family must be globally initial.

If $X<Y$, $X\notin \mathcal{B}$ and $Y\in \mathcal{B}$, then $X,Y$ cannot agree in any coordinate, since that coordinate section would have an omitted earlier member and an included later one. Therefore $Y=X^c$. Let $X$ be the first omitted vertex and $Y$ the last included vertex. If $\mathcal{B}$ is noninitial, $X<Y$ and $Y=X^c$. Any vertex $Z$ strictly between them would have to be included, since an omitted $Z$ paired with $Y$ would have to equal $Y^c=X$. But it would also have to be omitted, since an included $Z$ paired with $X$ would have to equal $X^c=Y$. Thus there is no such $Z$: $X,Y$ are consecutive complementary vertices. Everything before $X$ is included and everything after $Y$ omitted. Hence $\mathcal{B}$ is obtained from the initial segment of its size by replacing $X$ with its immediate successor $X^c$.

There are only two forms of this exception. If $n=2r+1$, consecutive complementary vertices must straddle ranks $r,r+1$; otherwise an intermediate rank or a further vertex in one of their ranks separates them. They are consequentlywith $X$ the last $r$-subset and $Y$ the first $(r+1)$-subset. The corresponding initial segment is all ranks at most $r$. For $r\ge 1$, the omitted $X$ remains in $N(\mathcal{B})$ through a lower neighbour, and every $(r+1)$-subset has at least two $r$-subsets, only one of which was omitted. Thus $N(\mathcal{B})$ contains every rank at most $r+1$, namely $N(I_{|\mathcal{B}|})$. For $r=0$ both singleton closed graph neighbourhoods are the whole $Q_1$.

If $n=2r$, the two complementary vertices must both have rank $r$. The complement map reverses the lexicographic order in that rank, so a consecutive complementary pair must be its central pair. Exactly half the $r$-subsets contain $1$, and all of those precede those missing $1$. Therefore the pair isHere $I_{|\mathcal{B}|}$ comprises all ranks below $r$ and all $r$-subsets containing $1$. For $r\ge 2$, $N(\mathcal{B})$ contains every rank at most $r$, because all lower ranks were retained. Every $(r+1)$-subset containing $1$ has $r\ge 2$ different $r$-subsets containing $1$, so deletion of the one member $X$ removes no vertex of the upper shadow. Consequently $N(\mathcal{B})\supseteq N(I_{|\mathcal{B}|})$. For $r=1$, the two closed graph neighbourhoods both have size three in $Q_2$. These cases exhaust the exceptions. Since the successive simplicial section compressions never increased the closed graph neighbourhood, this proves the vertex-isoperimetric inequality in the discrete cube.

The same argument gives all integer radii. Write $N^t$ for $t$ repetitions of the closed graph neighbourhood. Its application to an initial segment is again initial; the sizes of these initial segments are monotone in the original size. Inducting on $t$ therefore gives

For the concentration question, use uniform probability measure on a finite connected graph and normalize its graph distance by its graph diameter. A sequence is a Lévy family of graphs if, for every fixed $\epsilon >0$ and $\delta >0$, the radius-$\epsilon$ neighbourhood of every set of measure at least $\delta$ has measure tending to one, uniformly over those sets. One may equivalently require this only for sets of measure at least one half. To see the reverse implication, first the radius-$\epsilon /2$ neighbourhood of every positive-density set must eventually have measure greater than one half. Otherwise its complement has measure at least one half, while the radius-$\epsilon /4$ neighbourhood of that complement misses the original positive-density set, contradicting the half-measure condition. Apply the half-measure condition once more to the radius-$\epsilon /2$ neighbourhood, enlarging it by another $\epsilon /2$, to obtain the full claim. The metric normalization is essential here.

For $Q_n$, the graph diameter is $n$. A uniformly chosen vertex has size $Z_n$ with binomial distribution $\mathrm{Bin}(n,1/2)$, expectation $n/2$ and variance $n/4$. Fix $\delta >0$ and choose $C>0$ so that $1/(4C^2)<\delta$. By Chebyshev's inequality,For sufficiently large $n$, put $k_n=\lfloor n/2-C\sqrt{n}\rfloor \ge 0$. Every initial segment of measure at least $\delta$ contains the complete ranks through $k_n$. Its radius-$t$ graph neighbourhood therefore contains every rank at most $k_n+t$. Taking $t=\lfloor \epsilon n\rfloor$, the all-radius vertex-isoperimetric inequality in the discrete cube gives, uniformly for $|\mathcal{A}|\ge \delta 2^n$,The last Chebyshev's inequality is used only when its denominator has positive square root. These are precisely the radius-$\epsilon$ neighbourhoods in the normalized graph distance. The discrete cubes form a Lévy family. In fact, taking $t=\lfloor n^{3/4}\rfloor$ in the same bound makes the omitted measure $O(n^{-1/2})$ for each fixed $\delta$: a radius that is $o(n)$ already suffices.

For fixed $d\ge 1$, the nearest-neighbour grid $[n{]}^d$ has graph distance ${\sum }_i|x_i-y_i|$ and graph diameter $d(n-1)$. Take the slab ${\mathcal{A}}_n=\{x:x_1\le \lceil n/2\rceil \}$, whose uniform measure is at least one half. Its integer radius-$t$ graph neighbourhood is exactly $\{x:x_1\le \min (n,\lceil n/2\rceil +t)\}$: changing the first coordinate alone attains the minimum distance to the slab. For $0<\epsilon <1/(2d)$ and $t=\lfloor \epsilon d(n-1)\rfloor$, the omitted proportion tends toThus there is failure of concentration in fixed-dimensional grids, and the fixed-dimensional grids do not form a Lévy family with normalized graph distance.

On the real line, Prohorov's theorem states that a sequence of probability measures is uniformly tight if and only if every subsequence has a further subsequence converging weakly to a probability measure. Here uniformly tight means that, for every $\epsilon >0$, some compact interval contains mass at least $1-\epsilon$ for every measure in the sequence. This is relative sequential compactness for weak convergence of probability measures, not merely the existence of one convergent subsequence.

We first prove the characteristic-function tail bound. Put $c_0=1-\sin 1>0$. For $y\ne 0$,At $y=0$ the expression is zero by continuity. If $|y|\ge \lambda$, the supplied elementary bound, with $t=|y|/\lambda$, makes it at least $c_0$. The integrand is nonnegative, so Tonelli's theorem yieldsThus one valid universal constant isThe estimate controls spatial tails using only the characteristic function near zero.

Apply it to each ${\mu }_n$. For fixed $\lambda$, pointwise convergence of the characteristic functions and the bound $|{\varphi }_n|\le 1$ allow dominated convergence on the finite integration interval, givingContinuity of a characteristic function and $\varphi (0)=1$ make the last expression tend to zero as $\lambda \to \infty$. Given $\epsilon >0$, choose $\lambda$ making it less than $\epsilon /2$; then for all sufficiently large $n$, the preceding tail is less than $\epsilon$. Enlarge the compact interval to handle the finitely many remaining probability measures. This proves uniform tightness of the whole sequence.

By Prohorov's theorem, every subsequence has a further weakly convergent subsequence, say with limit $\nu$. The real and imaginary parts of $e^{iuy}$ are bounded continuous functions, so along that subsequenceThe uniqueness theorem for characteristic functions implies $\nu =\mu$. All subsequential weak limits are therefore $\mu$. If the original sequence failed to converge weakly to $\mu$, a bounded continuous test function and a subsequence would keep its integrals a fixed distance from the limiting integral; a further weakly convergent subsequence would contradict that. HenceThis supplies the required proof of the Lévy continuity theorem in the stated case.

For the final request, the printed $Q$ is not defined; the intended function is $\varphi$, the characteristic function of a summand. We prove that ${\varphi }^{\prime }(0)=ia$ without assuming integrability. Write $Y_n=S_n/n$. Independence gives $\mathbb{E}{e}^{iu{Y}_n}=\varphi (u/n{)}^n$. Moreover convergence in probability to $a$ implies uniform convergence of these characteristic functions on each bounded frequency interval: for $|u|\le T$ and $\delta >0$,First let $n\to \infty$, then $\delta \downarrow 0$.

The potential logarithm-branch ambiguity needs to be resolved, not discarded. By continuity of $\varphi$ near zero, its principal complex logarithm is defined at $u/n$ for all $|u|\le 1$ and all large $n$, with value zero at $u=0$. Alsouniformly on $[-1,1]$, so $\mathrm{Log}{\Psi }_n(u)\to 0$ uniformly there. The two continuous logarithms differ by an integer multiple of $2\pi i$:That integer is continuous in $u$ and zero at zero, hence identically zero. ConsequentlyFor arbitrary nonzero $h\to 0$, take $n=\lfloor 1/|h|\rfloor$ and $u=nh$. Eventually $1/2\le |u|\le 1$, so dividing the preceding error by $u$ shows $\mathrm{Log}\varphi (h)/h\to ia$. Exponentiating at zero proves the derivative forced by the weak law:This is the weak law forces a characteristic-function derivative argument. Its estimates use bounded exponentials, not $\mathbb{E}|X_1|$. If $X_1$ is integrable, dominated convergence permits differentiation inside the expectation and gives ${\varphi }^{\prime }(0)=i\mathbb{E}{X}_1$, so then $a=\mathbb{E}{X}_1$.

Absolute integrability is genuinely unnecessary. For an explicit weak law without an absolute first moment, take a symmetric variable withwith mass $1-1/e$ at zero and no mass with $0<|X|<e$. This is a valid law: the specified tail decreases from $1/e$ to zero, and its derivative defines the continuous part beyond $e$. Its absolute first moment is infinite because ${\int }_e^{\infty }dx/(x\log x)=\infty$. For independent copies, set $Y_{n,j}=X_j{1}_{\{|X_j|\le n\}}$. Symmetry gives $\mathbb{E}{Y}_{n,j}=0$, and integration of the tail givesFor the last estimate, split the integral at $\sqrt{n}$; the initial part is $O(\sqrt{n})$ and the remainder is at most $2n/\log n$, apart from a fixed initial constant. Chebyshev's inequality now gives $n^{-1}{\sum }_{j\le n}{Y}_{n,j}\to 0$ in probability. The probability that any truncation changes a summand is at most $n\mathbb{P}(|X|>n)=1/\log n\to 0$, so $S_n/n\to 0$ in probability although $\mathbb{E}|X|=\infty$. The derivative is zero, but the usual expectation of $X$ does not exist.

If either the positive or negative part is integrable, the finite weak-law limit does force full integrability. For example, if $\mathbb{E}{X}^{-}<\infty$, then $X\wedge A$ is integrable and its sample averages have the weak law of large numbers limit $\mathbb{E}(X\wedge A)$. Since they are bounded above by the original averages, $\mathbb{E}(X\wedge A)\le a$ for every $A$. Letting $A\to \infty$ shows $\mathbb{E}{X}^{+}<\infty$. The symmetric argument handles the other one-sided hypothesis. The nonintegrable example necessarily has both one-sided expectations infinite.

Use the uniform probability measure on the symmetric group $S_n$ and the unnormalized Hamming distanceLet $f:S_n\to \mathbb{R}$ have Lipschitz constant $K>0$ for this distance. We will prove the one-sided concentration inequalitiesConsequently $\mathbb{P}(|f-\mathbb{E}f|\ge t)\le 2e^{-t^2/(2n{K}^2)}$ for $t>0$. A zero Lipschitz constant makes $f$ constant, so that case is immediate.

Let $\sigma$ be a uniform permutation, reveal $\sigma (1),\dots ,\sigma (k)$ successively, and define the Doob exposure martingaleFix the first $k-1$ images. For an available value $a$, let $h(a)$ be the conditional mean when $\sigma (k)=a$. Given two available values $a,b$, left-composition with the transposition exchanging $a,b$ is a bijection from completions with next image $a$ to completions with next image $b$. It fixes the previously exposed images and changes exactly two coordinates of each complete permutation. Since conditional completions are uniform, averaging the Lipschitz condition over this pairing gives $|h(a)-h(b)|\le 2K$.

Thus, conditionally on the first $k-1$ images, $M_k-M_{k-1}$ has mean zero and takes values in an interval of length at most $2K$. We need the range-length version of the Hoeffding lemma: if a real random variable $Z$ has mean zero and lies in an interval of length $\ell$, thenHere is a proof. For $\psi (\lambda )=\log \mathbb{E}{e}^{\lambda Z}$, differentiation gives ${\psi }^{\prime \prime }(\lambda )={\mathrm{Var}}_{\lambda }(Z)$ under the exponentially tilted probability distribution. Any distribution on an interval of length $\ell$ has variance at most ${\ell }^2/4$, because its squared distance from the interval midpoint is at most ${\ell }^2/4$. Since $\psi (0)={\psi }^{\prime }(0)=0$, integrating this second-derivative bound gives the lemma for both signs of $\lambda$. The same argument holds for every conditional distribution.

Apply it conditionally to each martingale difference and use the tower property of conditional expectation repeatedly:The exponential Markov bound now gives, for $\lambda >0$,Choose $\lambda =t/(n{K}^2)$ to obtain the upper tail in the box, and apply the argument to $-f$ for the lower tail. This proves concentration of Lipschitz functions on the symmetric group. Using only the bound $|M_k-M_{k-1}|\le 2K$ in the absolute-increment Azuma-Hoeffding inequality would lose a factor four in the exponent; the conditional range length is what yields the constant needed in part (ii).

The Doob maximal inequality for a nonnegative submartingale states that, for a nonnegative submartingale $(M_j{)}_{0\le j\le N}$ and $u>0$,For completeness, partition the crossing event according to its first crossing time $j$. On that event $M_j\ge u$, while $\mathbb{E}[M_N\mid {\mathcal{F}}_j]\ge M_j$. Summing the corresponding expectations over $j$ proves $u\mathbb{P}(\max M_j\ge u)\le \mathbb{E}[M_N{1}_{\{\max M_j\ge u\}}]\le \mathbb{E}{M}_N$.

If $q>p$, the exponential $r^{S_j}$, with $r=q/p>1$, is increasing as a function of $S_j$. Apply the inequality to the martingale from part (a), with $u=r^k$:Letting $N\to \infty$ and using continuity of probability measures on increasing events givesThe same event is obtained if the supremum excludes time zero, since $k\ge 1$. If $p\ge q$, the right side is at least one and the bound is automatic; the nontrivial Doob maximal inequality argument is for $q>p$.

Let $G$ be the cumulative distribution function of the arbitrary Borel probability measure $\mu$. It may have atoms or flat intervals. Use its quantile functionThe limits of $G$ at the two infinities make $Q(u)$ finite. Its monotonicity and right continuity imply the exact equivalence $Q(u)\le x⟺u\le G(x)$. For the nontrivial direction, if $Q(u)\le x$, points with $G\ge u$ lie arbitrarily close above $Q(u)$; if the infimum itself is $x$, right continuity gives $G(x)\ge u$. The other direction follows directly from the defining infimum.

Take independent uniform variables $U_k$ and set $Y_k^{\prime }=Q(U_k)$. Thenso $(Y_k^{\prime })$ has the same entire sequence law as the original iid sample $(Y_k)$. On this construction, the empirical distribution function satisfiesThus part (ii) givesThe supremum is measurable: since both distribution functions are right-continuous, it equals the supremum over rational $x$. The convergence event consequently depends measurably on the coordinate sequence, and equality of the sequence laws transfers its probability one to $(Y_k)$. HenceThis is the Quantile proof of the Glivenko-Cantelli theorem, including arbitrary distributions with atoms. The original PDF supplies the full definition and absolute-value signs in this part, which the TeX conversion incompletely records.

A real Gaussian process $(Z_t{)}_{t\in T}$ is a family of random variables such that every finite vector $(Z_{t_1},\dots ,Z_{t_d})$ has a multivariate normal distribution, allowing degenerate distributions. Equivalently every finite real linear combination is a Gaussian random variable. Its mean and covariance kernel are $m(t)=\mathbb{E}{Z}_t$ and $C(s,t)=\mathbb{E}[(Z_s-m(s))(Z_t-m(t))]$.

For any finite set of indices, these functions give the whole mean vector and covariance matrix. The characteristic function of the vector isThe standard uniqueness theorem for characteristic functions therefore determines every finite-dimensional distribution, including the singular cases. Cylinder events form a generating class for the product sigma-algebra on ${\mathbb{R}}^T$ and are closed under finite intersections, so agreement on all cylinder events gives agreement of the process laws by the uniqueness theorem for probability measures. Thus the mean and covariance determine the law on this coordinate sigma-algebra even when $T$ is uncountable. This statement does not claim uniqueness on an arbitrary finer sigma-algebra of path properties.

Here $F=1_A$ is bounded and hence lies in $L^2$ as well as $L^1$. Conditional expectation on a sub-sigma-algebra is the orthogonal projection in $L^2$ onto its measurable subspace. We can use these projections to justify both limits without any symmetry assumption on the probability measure.

The increasing coordinate sigma-algebras ${\mathcal{E}}_n$ generate the whole product sigma-algebra $\mathcal{F}$. Functions depending on finitely many coordinates are dense in $L^2(\mathcal{F})$: their cylinder indicators generate $\mathcal{F}$, and the usual monotone-class approximation proves density of their linear span. If a cylinder function $V$ is ${\mathcal{E}}_N$-measurable, then for $n\ge N$ the $L^2$ projection contraction givesApproximate $F$ arbitrarily well by such $V$ to prove the first $L^2$ limit.

For the decreasing tail sigma-algebras ${\mathcal{G}}_n$, put $Y_n=\mathbb{E}[F\mid {\mathcal{G}}_n]$. If $m\ge n$, the tower property gives $\mathbb{E}[Y_n\mid {\mathcal{G}}_m]=Y_m$. Projection orthogonality therefore yieldsThe norms on the right decrease and are nonnegative, so $(Y_n)$ is Cauchy in $L^2$ and has a limit $Y$. Choose an almost-surely convergent subsequence and take its pointwise upper limit. For every fixed $n$ its tail consists of ${\mathcal{G}}_n$-measurable functions, so this supplies a version of $Y$ measurable with respect to every ${\mathcal{G}}_n$, hence to $\mathcal{G}={\bigcap }_n{\mathcal{G}}_n$. For any $C\in \mathcal{G}$, $\mathbb{E}[Y_n{1}_C]=\mathbb{E}[F{1}_C]$; passing to the limit identifies $Y=\mathbb{E}[F\mid \mathcal{G}]$.

Finally $\Vert W{\Vert }_1\le \Vert W{\Vert }_2$ on a probability space, by the Cauchy-Schwarz inequality. ThusThese are the increasing-filtration and reverse martingale convergence theorem limits; the argument above establishes the requested norm convergence directly.

Use the following form of the Girsanov theorem. If $W$ is Brownian motion, $\vartheta$ is predictable, andis a uniformly integrable martingale with $Z_0=1$, then the probability measure with terminal density $Z_{\infty }$ makes $W_t-{\int }_0^t{\vartheta }_{s}ds$ a Brownian motion. The analogous finite-horizon statement uses the terminal density at that horizon.

Here take $W_t=X_t-x$ and ${\vartheta }_s=1_{\{s<\tau \}}/X_s$. Before $\tau$, $X_s\in [\epsilon ,M]$, so the stochastic integral is well-defined. The assumed uniform integrability of the stopped stochastic exponential givesThus the proposed weighting really defines a probability measure, and the Girsanov theorem gives the Brownian motionunder that measure. In particularAll calculations are first made on this compact stopped interval. The integral involving $1/X$ is naturally defined on $[0,T_0)$; at $T_0$ its quadratic variation may diverge. The density will have a zero limiting extension there, as established below, rather than a finite stochastic logarithm at the hitting time.

A lambda-system contains the whole space and is closed under complements and countable disjoint unions. Let $\mathcal{D}$ be the smallest lambda-system containing a pi-system $\mathcal{P}$. For $A\in \mathcal{P}$, the family of $B\in \mathcal{D}$ with $A\cap B\in \mathcal{D}$ is a lambda-system containing $\mathcal{P}$, hence is all of $\mathcal{D}$. Fixing $B\in \mathcal{D}$ and repeating the argument shows $\mathcal{D}$ is closed under intersections. A lambda-system closed under intersections is a sigma-algebra, so $\mathcal{D}=\sigma (\mathcal{P})$. This proves the pi-lambda theorem. If two probability measures agree on $\mathcal{P}$, the sets on which they agree form a lambda-system; thus they agree on $\sigma (\mathcal{P})$.

The nonnegative form of Fubini's theorem, also called Tonelli theorem, states that for a nonnegative product-measurable function on two sigma-finite measure spaces, the product integral equals either iterated integral, allowing the value infinity.

For Lebesgue measure on the plane, a bounded interval rectangle $I\times J$ has measure $|I||J|$. Its preimage under $f$ is $(\lambda I)\times ({\lambda }^{-1}J)$, which has the same measure. Under $g$, its vertical slice at $x\in I$ is $J-sx$, whose one-dimensional Lebesgue measure is $|J|$ by translation invariance. Tonelli therefore gives $\mu (g^{-1}(I\times J))=|I||J|$. Horizontal slices likewise prove the equality for $h$. The interval rectangles form a generating pi-system. Sigma-finite measure uniqueness, obtained by the finite-measure pi-lambda argument on an increasing rectangular exhaustion, extends these equalities to all Borel sets. Completion extends them to all Lebesgue-measurable sets, since Borel null sets and their subsets remain null.

Writing ${\lambda }^2=c$, matrix multiplication givesThus rotations through angles in $[0,\pi /2)$ preserve measure. Their inverses and compositions do too; every plane rotation is a finite composition of these, including a right angle as two rotations through $\pi /4$. Lebesgue measure is invariant under every rotation.

Define $U(x)=U_0({\delta }_x)$, where ${\delta }_x$ is the Dirac measure at $x$. For the full sigma-algebra of subsets of a finite $E$, affineness directly gives $U_0(\lambda )={\sum }_{x\in E}\lambda (\{x\})U(x)$.

The statement also holds for an arbitrary sigma-algebra on a finite set. List each atom of a sigma-algebra as $A_1,\dots ,A_r$ and choose $x_j\in A_j$. Points in one atom have identical Dirac measures on this sigma-algebra, so $U$ is constant on each atom and is measurable. Every probability measure decomposes as $\lambda ={\sum }_j\lambda (A_j){\delta }_{x_j}$, whenceThis is the expected utility representation on a finite measurable space. It does not silently assume that each singleton is measurable.

Use the uniform probability measure on the Boolean hypercube. The influence of a variable is the flip probabilityThe polynomial defining $f$ is interpreted over ${\mathbb{F}}_2$. Flipping the first coordinate adds $x_2+\cdots +x_n$ modulo two. This nonempty parity is balanced for $n\ge 2$, so

An explicit Tribes function works for every $n\ge 2$. Choose the largest integer $w\ge 1$ with $w{2}^w\le n$, let $m=\lfloor 2^w\log 2\rfloor$, and take the OR of $m$ disjoint ANDs of $w$ coordinates. Ignore the unused coordinates. Its zero probability is $(1-2^{-w}{)}^m$, between $1/4$ and $3/4$, and a used coordinate has influenceHere $n<2^{w+1}(w+1)$ by maximality. For the balance claim, $w=1$ gives probability $1/2$; for $w\ge 2$, $m\ge 2^w\log 2-1$ gives zero probability at most $e^{-\log 2+2^{-w}}<3/4$, while $(1-1/q{)}^q\ge 1/4$ for $q=2^w\ge 2$ gives a lower bound $(1/4{)}^{\log 2}>1/4$. Only the example was requested, but these checks fix its parameters unambiguously.

For the lower bound we prove the required consequence of Beckner's inequality with constants. Write the Fourier-Walsh transform as $f={\sum }_S\hat{f}(S){\chi }_S$, where ${\chi }_S(x)=(-1{)}^{{\sum }_{i\in S}{x}_i}$ and $\hat{f}(S)=\mathbb{E}(f{\chi }_S)$. The discrete derivative of a Boolean function is $D_{i}f=(f-f(\cdot +e_i))/2$. Since $f$ takes values zero and one, $D_{i}f$ takes values $0,\pm 1/2$. Consequently Parseval's identity givesThe factor four is required for the printed $0/1$ convention. Also $V:=\mathrm{Var}(f)={\sum }_{S\ne \varnothing }|\hat{f}(S){|}^2\ge 3/16$ under the balance hypothesis.

The true Beckner's inequality states $\Vert T_{\rho }h{\Vert }_q\le \Vert h{\Vert }_p$ for $1<p\le q<\infty$, $0\le \rho \le \sqrt{(p-1)/(q-1)}$, with $T_{\rho }h={\sum }_S{\rho }^{|S|}\hat{h}(S){\chi }_S$. Apply it with $p=3/2$, $q=2$, $\rho =1/\sqrt{2}$ to $h_i={\chi }_{\{i\}}{D}_{i}f$, a function independent of coordinate $i$. Its nonzero values still have magnitude $1/2$. We obtainLet $a={\max }_i{I}_i(f)>0$. For an integer $\ell \ge 1$, the low-degree Fourier weight satisfiesFor the high-degree part the total influence identity givesTogether these imply $4V\le na(2^{\ell -1}{a}^{1/3}+1/(\ell +1))$.

Suppose, seeking a contradiction, that $a<\log n/(100n)$, with natural logarithms. Then $a<1/(100e)$ and $\ell =\lfloor \log (1/a)/(6\log 2)\rfloor \ge 1$. Hence $2^{\ell -1}{a}^{1/3}\le a^{1/6}/2$. Put $y=\log n$. The first term on the right is bounded bythe middle expression is maximized at $y=7$. Moreover $\log (1/a)\ge y/2$, since $y\le 100e^{y/2}$. ThusWe would have $4V<1/10$, contradicting $4V\ge 3/4$. ThereforeThis derives the stated Kahn-Kalai-Linial theorem bound from the actual noise operator on the Boolean hypercube inequality, rather than using the desired conclusion as an input.

Identify the hypercube graph $Q_n$ with $\mathcal{P}([n])$, joining two sets when their symmetric difference has size one. Write $N[A]$ for the closed graph neighbourhood of $A$, and $N_t[A]$ for its radius-$t$ neighbourhood in Hamming distance. In the simplicial order on the discrete cube, smaller sets come first, with lexicographic order within each rank: the smallest differing coordinate belongs to the earlier set. Let $I_m$ be the first $m$ vertices in this order. Harper theorem isEquivalently, $I_m$ minimizes the external vertex boundary at fixed size. We will also obtain $|N_t[A]|\ge |N_t[I_{|A|}]|$ for every integer $t\ge 0$.

Here is a complete proof by simplicial section compression. First note two properties of the simplicial order on the discrete cube. Its restriction to either face obtained by fixing one coordinate is the corresponding order on the smaller hypercube graph. Also, the closed graph neighbourhood of an initial segment is an initial segment. To verify the latter, a nonempty initial segment contains every level below its last level $r$, and a lexicographic prefix $\mathcal{L}$ at level $r$. For $r\ge 1$, its closed graph neighbourhood contains every set of size at most $r$, and its only further members are the upper shadow of $\mathcal{L}$. An $(r+1)$-set lies in this upper shadow exactly when its lexicographically earliest $r$-subset lies in $\mathcal{L}$. That earliest subset is obtained by deleting the largest element. These earliest subsets vary monotonically in lexicographic order, so the upper shadow is another lexicographic prefix. For $r=0$, the closed graph neighbourhood of $\{\varnothing \}$ is the union of levels zero and one. The empty initial segment causes no exception.

Induct on $n$, with $n=0,1$ immediate. Fix a coordinate $i$ and write the sections of $A$ as $A_0,A_1\subseteq Q_{n-1}$, deleting $i$ from the latter section. Replace them by the initial segments $I_a,I_b$ with $a=|A_0|$, $b=|A_1|$. The sections of the old closed graph neighbourhood areBy induction their sizes are at least $\max \{|N[I_a]|,b\}$ and $\max \{|N[I_b]|,a\}$, respectively. These are exactly the sizes of the new sections, because both sets in each new union are initial segments and hence nested. Thus simplicial section compression preserves size and never enlarges the closed graph neighbourhood.

Repeatedly compress any section that is not already an initial segment. Every change strictly reduces the sum of the global simplicial positions of the included vertices. This nonnegative integer decreases only finitely often, so the process ends at a family $B$ whose sections in every coordinate are initial segments. If $B$ itself is an initial segment, the induction is complete. Otherwise choose the earliest absent vertex $X$ and the latest present vertex $Y$, with $X\prec Y$. They cannot agree in any coordinate: agreement would put them in one compressed section, where an earlier absent vertex cannot precede a later present one. Hence $Y=X^c$. Nor can a vertex $Z$ lie between them: if $Z$ is present, pairing it with $X$ would force $Z=X^c=Y$; if absent, pairing it with $Y$ would force $Z=Y^c=X$. Thus $X,Y$ are consecutive, and $B$ is obtained from $I_{|B|}$ by replacing its last vertex $X$ with the next vertex $Y$.