Skorokhod Representation and Tightness on Path Space
Work on a filtered probability space satisfying the usual conditions. \(B\) denotes Brownian motion, and stochastic integrands are predictable with the stated integrability. A local martingale need not be a true martingale: the moment conditions in each claim are essential. See Brownian motion and martingale inequalities.
Weak convergence concerns distributions and does not specify how random variables are coupled. Skorokhod’s representation theorem supplies a new coupling, on a common probability space, under which convergence is almost sure. The original joint distribution need not be preserved.
Let \((S,d)\) be a complete separable metric space, and let \(\mu_n\Rightarrow\mu\) be Borel probability measures on \(S\). We construct measurable \(\widetilde X_n,\widetilde X:[0,1]\to S\), with their respective laws \(\mu_n,\mu\) under Lebesgue measure, such that \(\widetilde X_n\to\widetilde X\) almost surely. This is Skorokhod’s representation theorem.
Choose refining countable Borel partitions \(\mathcal A_k\) whose cells have diameter at most \(2^{-k}\) and whose boundaries have \(\mu\)-measure zero. Here is the construction. Around a countable dense set of centers choose radii in \((2^{-k-3},2^{-k-2})\). For each center, at most countably many radii have spheres of positive \(\mu\)-measure, since those spheres are disjoint. We can therefore choose zero-boundary radii in this interval, and the resulting balls cover \(S\). Disjointise the balls in their fixed enumeration, subtracting only earlier balls, and intersect with the previous partition. Boundaries of each resulting cell lie in a finite union of zero-measure boundaries from its ancestors. Discard empty cells.
First take any Borel probability measure \(\nu\) on \(S\); the partitions need not be \(\nu\)-continuity sets for this step. Give each first-level cell a consecutive half-open interval in \([0,1)\) of length \(\nu(A)\). Subdivide its interval among its children, in their fixed enumeration, with lengths \(\nu(A')\), and repeat. This forms nested intervals \(I_A^\nu\), one for each cell \(A\), with \(|I_A^\nu|=\nu(A)\). Ignore the countable set of all endpoints and choose one representative \(x_A\in A\) for each nonempty cell.
For a point \(u\) outside the endpoints, let \(X_\nu^k(u)=x_A\) when \(u\in I_A^\nu\) at level \(k\). Nested cells imply \(d(X_\nu^j(u),X_\nu^k(u))\leq2^{-k}\) for \(j\geq k\). Completeness gives a limit \(X_\nu(u)\). Give all exceptional points a fixed value in \(S\). The approximants and their limit are measurable. For a bounded continuous function \(f\),
The last convergence follows because \(d(x_{A_k(x)},x)\leq2^{-k}\). Bounded convergence on the unit interval also gives convergence to \(\int_0^1 f(X_\nu(u))du\). Therefore \(X_\nu\) has law \(\nu\). This identifies the law directly; it does not require the incorrect assertion that the probability of an arbitrary closed set tends to zero.
Apply the construction to \(\mu\) and every \(\mu_n\), using exactly the same cell tree and ordering. For each cell \(A\), the continuity-set part of the Portmanteau theorem gives \(\mu_n(A)\to\mu(A)\). Each interval endpoint also converges: at the first level it is a finite sum of earlier cell masses; at later levels it is the parent’s left endpoint plus a finite sum of earlier sibling masses. Induction on the level proves the claim. Equivalently, at each fixed level the countable mass vectors satisfy
by domination by the summable masses \(\mu(A)\). Both the absolute values and the parentheses in this identity matter.
Outside the countable union of interval endpoints for all measures, fix \(k\) and let \(u\) lie in the interior of \(I_A^\mu\). Convergence of the two endpoints puts \(u\) in \(I_A^{\mu_n}\) for every sufficiently large \(n\). Both limits \(X_\mu(u)\) and \(X_{\mu_n}(u)\) lie in the closure of \(A\), so their distance is at most \(2^{-k}\). First let \(n\to\infty\), then \(k\to\infty\). This proves the almost sure convergence of the full sequence on the newly constructed probability space.
1. Tightness
For continuous processes, the random variable is the whole path. Equip \(C[0,1]\) with the uniform metric, and write \(\omega_f(\delta)=\sup_{|t-s|\leq\delta}|f(t)-f(s)|\). A family of laws is tight if, for every \(\eta>0\), one compact set carries probability at least \(1-\eta\) under every law. Arzelà–Ascoli suggests controlling the starting values and a sequence of moduli of continuity.
Suppose that, for every \(\varepsilon,\eta>0\), some \(A,\delta>0\) and \(n_0\) satisfy
These bounds can be made uniform over all \(n\). There are only finitely many omitted processes; each has finite starting value almost surely and continuous paths, so increasing \(A\) and decreasing \(\delta\) handle them. Fix \(\eta>0\), choose \(A\) with uniform starting-tail bound \(\eta/2\), and choose \(\delta_m\downarrow0\) with
The set
is closed and equicontinuous. It is uniformly bounded as well: use finitely many intervals of length at most \(\delta_1\) to bound every value from \(f(0)\). Arzelà–Ascoli makes \(K\) compact, and the union bound gives \(\sup_n\mathbb P(X_n\notin K)\leq\eta\). This proves the criterion.
Consider now i.i.d. Rademacher variables \(Y_i\), with \(\mathbb P(Y_i=1)=\mathbb P(Y_i=-1)=1/2\), and \(S_k=\sum_{i=1}^kY_i\). Define \(Z_n(k/n)=S_k/\sqrt n\) and interpolate linearly between adjacent mesh points. We will prove that the laws of \(Z_n\) on \(C[0,1]\) converge to Wiener measure.
The fourth moment is \(\mathbb{E} S_m^4=m+6\binom m2=3m^2-2m\). The same calculation handles all interpolated increments at once. If \(s<t\), then
where \(a_i\) is \(n\) times the length of the portion of the \(i\)-th mesh interval lying in \([s,t]\). Independence gives
To see explicitly why this gives tightness, choose \(0<\alpha<1/4\). On the dyadic grid of mesh \(2^{-m}\), Markov’s inequality and a union bound give, uniformly in \(n\),
The sum over \(m\geq M\) tends to zero. If none of these events occurs, dyadic approximation of each endpoint, followed by continuity, gives \(\omega_{Z_n}(2^{-M})\leq C_\alpha2^{-M\alpha}\): the successive approximation errors are bounded by the geometric sum \(\sum_{m\geq M}2^{-m\alpha}\), and the two level-\(M\) approximations are at most one mesh interval apart. Since \(Z_n(0)=0\), the preceding criterion proves tightness.
It remains to identify every joint distribution, not only the one-time marginals. Fix \(0=t_0<t_1<\cdots<t_r\leq1\). The disjoint random-walk blocks
are independent, and their joint characteristic function is
The limit follows from \(\log\cos x=-x^2/2+O(x^4)\). It identifies independent centered Gaussian increments of the required variances. Each interpolated value differs from \(S_{\lfloor nt\rfloor}/\sqrt n\) by at most \(n^{-1/2}\), so the same joint convergence holds for the increments of \(Z_n\), and hence for its finite-dimensional vectors.
By Prokhorov’s theorem, tightness gives a weakly convergent subsubsequence of every subsequence. Evaluations at finitely many times are continuous maps on \(C[0,1]\), so the limiting law has the Brownian finite-dimensional distributions just computed. These distributions determine the law on \(C[0,1]\): rational-time evaluations generate its Borel sigma-field, because uniform distance between continuous functions is the supremum of their rational-time distances and the space is separable. Thus every subsequential limit is Wiener measure. If the full sequence failed to converge, a subsequence separated from that law would have a further convergent subsequence, a contradiction. Therefore the entire sequence converges weakly to Brownian motion.
The compactness step uses the tightness implication in Bass (2011, Theorem 30.4). The representation construction above proves the stronger coupling assertion separately; convergence in law by itself makes no assertion about the original sample paths.
Reading: Le Gall (2016), Brownian Motion, Martingales, and Stochastic Calculus.