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For independent random variables, $\{\min (X,Y)>u\}=\{X>u,Y>u\}$ and $\{\max (X,Y)\le v\}=\{X\le v,Y\le v\}$. Factor their probabilities to obtain the distribution functions of independent minima and maxima:These formulas also allow atoms; no probability density function assumption is needed.

For independent unit-rate exponential distributions, $P(U>u)=e^{-2u}$ for $u\ge 0$, so $U\sim \mathrm{Exp}(2)$. To establish the independent minimum and gap of two exponential variables, set $D=V-U$. On either ordering branch, the change of variables is $(X,Y)=(u,u+d)$ or $(u+d,u)$ with unit absolute Jacobian determinant. Both contribute $e^{-2u-d}$; ties have probability zero. Thus the joint probability density isIt factors into normalized exponential probability density functions, proving $U\sim \mathrm{Exp}(2)$, $D\sim \mathrm{Exp}(1)$ and their independence.

The maximum is $V=U+D$. A half-scaled unit-rate exponential has rate two, so the independent pair $(D,U)$ has the same joint law as $(X,Y/2)$. ConsequentlyUsing the expected values, variances and independence of the summands,The spacing factorization explains why an order statistic can be represented as a sum of independent waiting times.

For a nonnegative integer-valued random variable, the probability generating function iswith $0\le s\le 1$ always allowed. Its derivative at $1$ from below gives the expectation when finite.

The waiting time $N$ has the positive-integer geometric distribution $P(N=j)=p{q}^{j-1}$. For the capped geometric waiting time, $X=j<k$ exactly when $N=j$, while $X=k$ occurs whenever $N\ge k$. Thus the terminal mass is $q^{k-1}$, givingThe finite sum is a polynomial; any apparent singularity of the rational expression is removable. This includes $k=1$, where $G_X(s)=s$.

Differentiating the rational form at $s=1$ yieldsAs a second interpretation, the expectation is the sum of the tail probabilities $P(X\ge j)=q^{j-1}$ for $1\le j\le k$. Their finite geometric sum gives the same answer.

The moment-generating function is $M_X(t)=\mathbb{E}[e^{tX}]$ at real $t$ for which that expectation is finite; $M_X(0)=1$. For independent random variables, the joint law factorizes, soThis is the transform form of independence. Its ordinary finite-valued interpretation applies where both factors are finite, with the transform of a zero multiple equal to $1$.

The Burton-Keane theorem says that for independent bond percolation on the hypercubic lattice ${\mathbb{Z}}^d$, $d\ge 2$, at every fixed $p\in [0,1]$, there is almost surely at most one infinite percolation cluster. It does not assert that such a cluster exists, nor does it assert absence of percolation at the critical parameter. The cases $p=0$ and $p=1$ are immediate; assume $0<p<1$ for the proof.

Let $N_{\infty }$ be the number of infinite percolation clusters. This random variable is translation invariant. By translation ergodicity of Bernoulli percolation, it is almost surely a constant $k\in \{0,1,2,\dots ,\infty \}$. One way to see ergodicity is to approximate an invariant event by events on finite sets of edges, translate one approximation far enough to make the two approximations independent, and then let the approximation error vanish. Its probability equals its square, hence is zero or one.

The other probabilistic input is the finite-energy property of Bernoulli percolation. Conditionally on all edges outside a finite set, every prescribed inside configuration has positive probability, namely a product of powers of $p$ and $1-p$. Thus an event of positive probability can be followed by a specified finite modification and still gives a positive-probability image event. Outside infinite arms are unaffected.

First exclude a finite $k\ge 2$. Some finite box meets at least two infinite percolation clusters with positive probability. Open all edges in and incident to that box. These clusters merge. Opening finitely many edges cannot create an infinite percolation cluster solely out of finite clusters, since it joins only finitely many such clusters. Hence the number of infinite percolation clusters decreases with positive probability, contradicting its almost-sure constant value. This rules out every finite multiplicity greater than one.

Now suppose $k=\infty$. With positive probability a finite box meets at least three distinct infinite percolation clusters. Retain an exterior infinite ray from each of three of them. Replace the finitely many interior and boundary edges by a minimal tree joining their three attachment graph vertices, and close all unused edges there. The exterior rays belong to distinct original clusters, so have no outside connections to each other. The resulting joined cluster has a branch graph vertex $v$ with exactly three incident open edges, and deleting $v$ leaves three infinite components. If a terminal lies in the middle of the interior connecting path, its exterior ray supplies the third branch there. Such a graph vertex is a trifurcation vertex in percolation. A finite box has only finitely many candidate locations and modifications; therefore some fixed graph vertex has positive trifurcation probability. By translation invariance, that probability is a common number $\tau >0$ at every graph vertex.

The decisive deterministic input is the trifurcation boundary-counting lemma. Let $T$ be all these degree-three trifurcation vertices in percolation in a finite box $B_n$. In each infinite percolation cluster meeting $T$, delete the graph vertices of $T$, contract each remaining connected component of a graph adjacent to $T$ to a node, and retain the incidences with the deleted graph vertices, including direct edges between graph vertices of $T$. This is a finite tree: a cycle would connect two neighbours of an original trifurcation vertex in percolation without passing through it, contradicting its three-way separation. Every deleted graph vertex has degree three. Every leaf component is infinite, since a finite leaf would give a finite branch after removing its adjacent trifurcation vertex in percolation. The elementary tree degree identity therefore supplies at least $|T\cap C|+2$ infinite leaf components for a cluster containing $T$.

Each such infinite component reaches the outer graph vertex boundary ${\partial }^{+}{B}_n$, and different components reach different boundary graph vertices. Summing over the clusters yields, in particular,For a nearest-neighbour lattice box,Taking expectations and using the common trifurcation probability givesThis contradicts $\tau >0$ and excludes infinitely many infinite percolation clusters. Thus only $k=0$ or $k=1$ remains:The proof explains the role of the amenable boundary growth of lattice boxes: finite-energy property of Bernoulli percolation would make branching points have positive volume density, while their disjoint infinite branches can only be supplied through a boundary of lower order. It uses no planar duality and works in every indicated dimension.

Relative to a filtration $({\mathcal{F}}_t)$, progressive measurability means that, for every $T<\infty$, the mapis measurable for the product sigma-algebra $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$ and the Borel sigma-algebra on $\mathbb{R}$.

Fix $T>0$ and divide $[0,T]$ into $2^m$ equal subintervals with mesh $h_m=T/2^m$. DefineSince $X$ is adapted, every random variable $X_{k{h}_m}$ is ${\mathcal{F}}_{k{h}_m}$-measurable, hence ${\mathcal{F}}_T$-measurable. Each approximation is consequently $\mathcal{B}([0,T])\otimes {\mathcal{F}}_T$-measurable. For $t<T$, its sampling time lies strictly to the right of $t$, tends to $t$, and never exceeds $T$. Right continuity implies $X_t^{(m)}\to X_t$; at $T$ equality is exact. Thus $X$ is the pointwise limit of measurable functions on this product space. As $T$ was arbitrary, $X$ is progressively measurable. This is the theorem that right-continuous adapted processes are progressively measurable.

The right-endpoint approximations need not themselves be adapted at their intermediate times. What the proof requires is their joint measurability with respect to the single terminal sigma-algebra ${\mathcal{F}}_T$. The proof uses the pathwise càdlàg convention. If path regularity is assumed only almost surely, under a completed filtration setting the stochastic process to zero on its common exceptional null event gives an indistinguishable progressively measurable version. Arbitrary values on that null event need not make the original stochastic process progressively measurable: even with a complete filtration, a null sample point may be assigned a non-Borel time function. This is why almost sure path regularity does not ensure progressive measurability.

Fix $t$ and put $\sigma =t\wedge \tau$. The stopping time property implies that $\sigma$ is an ${\mathcal{F}}_t$-measurable random variable with values in $[0,t]$: for $s<t$,and for $s\ge t$ the event is the whole space. By part (c), the restriction of $X$ to $[0,t]\times \Omega$ is $\mathcal{B}([0,t])\otimes {\mathcal{F}}_t$-measurable.

The evaluation map $\omega \mapsto (\sigma (\omega ),\omega )$ is measurable from $(\Omega ,{\mathcal{F}}_t)$ to this product space: the inverse image of a measurable rectangle $A\times C$ is $\{\sigma \in A\}\cap C$. Composing it with the jointly measurable stochastic process gives an ${\mathcal{F}}_t$-measurable random variableSince this holds for every fixed $t$, the stopped process is adapted. If path regularity holds only almost surely, first apply the proof to its pathwise regular representative; with a completed filtration, the original stopped variable differs only on a null event and is also ${\mathcal{F}}_t$-measurable. The adaptedness of a stopped right-continuous process requires neither boundedness of $\tau$ nor a martingale assumption; $\tau =\infty$ is harmless because $t\wedge \tau \le t$.

Let $W_n$ be the interpolation in part (a). The operationis continuous, since $|I(f)-I(g)|\le \Vert f-g{\Vert }_{\infty }$. By the Donsker invariance principle and the continuous mapping theorem,The exact trapezoidal integral of the linear interpolation isTherefore the statistic in question differs from $I(W_n)$ by $S_n/(2n^{3/2})$. Using independence, zero means, and unit variances givesThis error tends to zero in $L^2$, hence in probability. The Slutsky theorem now proves the integrated random-walk limit:The limiting law can also be made explicit. The time integral is a Gaussian random variable, as a mean-square limit of linear combinations of a Gaussian process. It is centered, and the covariance identity $\mathbb{E}(B_s{B}_t)=\min (s,t)$ givesThus the terminal value of integrated Brownian motion here has law $N(0,1/3)$.

Use the natural filtration ${\mathcal{F}}_n=\sigma (X_1,\dots ,X_n)$. The first exit time is a stopping time. If $r\le x$, it is zero, so its expectation is already finite. Now assume $0<x<r$.

The positive-increment hypothesis gives a number $h>0$ with $\mathbb{P}(X_1>h)>0$. Choose an integer $m$ so large that $r/m<h$, and put $p=\mathbb{P}(X_1>r/m)>0$. A block of $m$ increments all exceeding $r/m$ has probability $p^m$. From any point still in $(0,r)$, that block forces an upper exit before the block ends.

By independence, conditional on ${\mathcal{F}}_{km}$ and survival to time $km$, the next block has this same probability. Thus the geometric tail bound from a uniform escape probability givesUsing the tail-sum formula for the expected value of a nonnegative integer-valued random variable,This is the random-walk exit bound from a positive-increment block. In particular, the exit occurs with probability one. The mean-zero assumption is not needed for this first bound.