On the completion of Skorokhod space

We consider the classical Skorokhod space $D[0,1]$ and the space of continuous functions $C[0,1]$ equipped with the standard Skorokhod distance $\rho$. It is well known that neither $(D[0,1],\rho)$ nor $(C[0,1],\rho)$ is complete. We provide an explicit description of the corresponding completions. The elements of these completions can be regarded as usual functions on $[0,1]$ except for a countable number of instants where their values vary"instantly".


Introduction
We consider the classical Skorokhod

|s − γ(s)|
where the infimum is taken over all continuous strictly increasing maps γ acting from [0, 1] onto itself. C is a closed linear subspace in (D, ρ).
The space D equipped with the Skorokhod topology generated by ρ is a de facto standard framework in the theory of stochastic processes with jumps, such as Lévy processes, random walks, etc. (see Skorokhod [4,5]).
It is well known that neither (C, ρ) nor (D, ρ) is complete (see Example 1 below). Usually this issue is solved by introducing the other, more delicate, distance generating the same topology (see e.g. Billingsley [1,Section 14]). The other option is to enlarge the spaces C and D completing them with respect to ρ. The goal of this paper is to describe the completions explicitly.
We mention a related study by Whitt [6,Section 15], aimed to enlarge the space D in order to let more sequences converge (in the Skorokhod topologies M 1 and M 2 ). Whitt's motivation was in queueing theory.
The completions of the spaces (C, ρ) and (D, ρ) appeared, respectively, in the important functional large deviations principles of Mogulskii for random walks [2] and Lévy processes [3]. Both papers used the completed spaces without caring about their nature. However, it is not quite clear how to work in such abstract setting, e.g. how to calculate the large deviations rate function on the entire completed space.
We finish this introduction with a simple but representative and instructive example.
Notice that every function g θ takes all values between zero and one going first up, then down. Informally, one may say that the limiting "function" should take all these values in the same order (up and down) at the single time instant s * = 1 2 . We call this paradoxical behaviour an "instanton" and formalize it in the following sections.

Turbofunctions
Let us denote I := [0, 1] and I (t) := [0, 1], stressing the following subtle but important difference. We consider I as a usual time interval equipped with the standard distance while I (t) is considered as a topological space, without any distance on it. Use the standard notation C[I (t) ] and D[I (t) ] for the spaces of continuous and càdlàg functions on I (t) , respectively.
We denote by Γ the class of all increasing homeomorphisms of I, that is strictly increasing continuous functions acting from I onto I. Similarly, Γ (t) denotes the class of all increasing homeomorphisms of I (t) . Denote by Σ the class of all continuous non-decreasing mappings acting from I (t) onto I.
The extended Skorokhod space is defined as D + := D[I (t) ] ⊕ Σ. Its elements, the pairs will be called turbofunctions. Similarly, put C + := C[I (t) ] ⊕ Σ. If σ ∈ Σ is strictly monotone, hence invertible, then every turbofunction F σ may be visualized as a usual function F σ ∈ D[I] defined by If σ is not strictly monotone, such simple visualization is not possible. In this case, there exists a finite or countable number of s ∈ I such that σ −1 (s) is a non-degenerate interval in I (t) . This corresponds to an instanton at time s, meaning that F σ takes all values {F (t) : t ∈ σ −1 (s)} "instantly" at time s.
In the opposite direction, every function f ∈ D[I] can be interpreted as a turbofunction where ς ∈ Σ is defined by ς(t) := t. Here and elsewhere the symbol • stands for superposition of mappings. In the following we will equip D + with a relevant Skorokhod-type semidistance and show that, after the natural factorization, D + can be interpreted as the completion of (D, ρ).
For continuous functions the situation is similar. If f ∈ C, then f + ∈ C + and the natural factorization of C + can be interpreted as the completion of (C, ρ).

Distance and factorizations
Let us define the Skorokhod semi-distance 2 ρ + on turbofunctions in D + by For the restriction of ρ + on C + , we can replace the suprema in the above definition by the maxima; however, this will play no role in the following. It is easy to show that all properties of semi-distance are verified. Indeed, the symmetry follows from is a non-negative symmetric function satisfying triangle inequality and ρ(x, x) = 0 but allowing ρ(x, y) = 0 for x = y.
where we used the variable change t = γ −1 (τ ). To prove the triangle inequality note that for all t ∈ I (t) and all γ 12 , γ 23 ∈ Γ (t) we have Taking supremum over t ∈ I (t) and using the variable change τ = γ 23 (t) in the first supremum on the right-hand side, we get sup Adding the two inequalities and subsequently optimizing over γ 12 , γ 23 yields the triangle inequality (3).
Remark 2 It is obvious that C + is a closed subset of (D + , ρ + ).
The natural equivalence generated by the semi-distance ρ + is as follows: It is worthwhile to compare this equivalence with the other one, which is more natural and very close to ≡ but still different from it. Namely, let Proof: The first claim follows from the definitions of the semi-distance ρ + and the equivalences. For the second claim, let us choose a sequence {γ n } n∈N in Γ (t) such that Then for an arbitrary The latter assumption may cease to hold only for finitely or countably many t ′ ∈ I (t) \ {0, 1} since F 1 is a càdlàg function and σ −1 1 is a continuous one. On the other hand, we know that ( , and it follows that 2 , as required.
This claim immediately follows from Proposition 3 and triangle inequality.
Somewhat surprisingly, the full converse in Proposition 3 is not true, i.e.
Here is a counter-example.
Then for every strictly increasing σ ∈ Σ it is true that (F 1 ) σ (t) ≡ (F 1 ) ς . However, let us consider some σ that is not strictly increasing and approximate it uniformly with strictly increasing σ δ ∈ Σ, 0 < δ < 1, so that sup This can be done by letting We have ρ + ((F 1 ) σ δ , (F 1 ) σ ) ≤ δ, which can be seen if we let γ ∈ Γ (t) be the identical homeomorphism of I (t) in the definition of ρ + . Furthermore, On the other hand, (F 1 ) ς (t) ≡ (F 1 ) σ does not hold. Indeed, assuming otherwise would imply that there is a representation σ = ς • γ with some γ ∈ Γ (t) . The right-hand side of this equality is strictly increasing, while the left-hand side is not, which is a contradiction.
The next proposition shows that both equivalences ≡ and (t) ≡ respect the visualization mapping.
Assume that σ 1 , σ 2 are strictly increasing and F σ 1 Proof: By Proposition 3 we have F σ 1
Proof: Let f 1 , f 2 ∈ D. Then which proves the property of isometry. Now we prove that the image {f + : f ∈ D} is dense in D + . To see this, let us consider some arbitrary F σ ∈ D + with strictly increasing σ. Then we have We see that F σ is (t) ≡ -equivalent to an element of the image. Finally, let F σ be an arbitrary element of D + . Consider its approximations F σ δ with strictly increasing σ δ introduced in (5). Then we have as δ → 0.
By Corollary 4 we have which proves that {f + : f ∈ D} is dense in (D + , ρ + ).
In Example 1, we constructed a family of functions {g θ } θ>2 in D with the Cauchy property but not having a limit in (D, ρ), as θ → ∞. We show now that the isometrically embedded family {g + θ } θ>2 has a limit in (D + , ρ + ). This is a good hint at completeness of (D + , ρ + ).
We use the notation from the mentioned example. Let Introduce the time changes σ θ ∈ Σ as piecewise linear functions with nodes Then we have g θ (s) = F (σ −1 θ )(s) = F σ θ (s), s ∈ I, and the sequence g + θ = [ F σ θ ] + (t) ≡ F σ θ (where we used (6) at the last step) has a limit F σ in C + (as θ → ∞) with σ ∈ Σ being the piecewise linear function with nodes Since the limiting variable change σ is not strictly increasing, the limiting turbofunction F σ has an instanton and can not be interpreted as a usual function.
4 Completeness of (D + , ρ + ) In the following proposition we essentially reach our goal by showing that (D + , ρ + ) is complete.
Proposition 9 Let {F σn n } n∈N be a Cauchy sequence in (D + , ρ + ). Then there exists a limit F σ ∈ D + such that lim n→∞ ρ + F σn n , F σ = 0. Moreover, if F σn n ∈ C + for all n ∈ N, then F σ ∈ C + . Proof: It suffices to show that {F σn n } n∈N contains a converging subsequence. Choose a subsequence {n k } k∈N in N such that ρ + (F σn k n k , F σn k+1 n k+1 ) < 1/2 k . By definition of the semi-metric ρ + , for every k ∈ N there exists a γ k ∈ Γ (t) such that Define λ k := γ k−1 • . . . • γ 1 for integer k ≥ 2 and λ 1 be the identical homeomorphism of I (t) . Since λ k ∈ Γ (t) , the change of variables t = λ k (τ ) yields Hence the sequences {F n k • λ k } k∈N and {σ n k • λ k } k∈N are Cauchy sequences in D[I (t) ] (resp. C[I (t) ]) equipped with the uniform distance. These metric spaces are complete. Therefore, we can find F ∈ D[I (t) ] and σ ∈ C[I (t) ] such that We also have σ ∈ Σ, since σ is non-decreasing as a pointwise limit of non-decreasing functions σ n k • λ k . Finally, the above equality implies that ρ + (F σn k n k , F σ ) → 0, as k → ∞, as required. Under the additional assumption {F σn
Remark 11 An equivalence class, i.e. an element of D + ≡ , may be viewed as an element of the completion, while the turbofunctions from this class may be viewed as its different parametrizations. This is quite similar to a curve on a manifold having a multitude of parametrizations.
By doing so, we extend definition (1) of the visualization mapping to the whole D. Then F σ ∈ D for any F σ ∈ D + .
We first give two corollaries.
Corollary 13 Any Cauchy sequence in (D, ρ) converges pointwise to an element of D except, possibly, for at most countable set of points in [0, 1). This follows from Theorem 12 combined with Proposition 7 and the fact that the functions σ −1 and F have at most countably many discontinuities.

Remark 14
We stress that the limiting element in D does not characterize completely the limiting turbofunction F σ because it skips the instantons. In Example 1 the limiting function is zero but F σ is not degenerate.
The next corollary asserts that Proposition 6 still holds for the extended notion of visualization.
This follows from Theorem 12 once we take F σ 2 2 = F σ 3 3 = . . . and F σ = F σ 1 1 and use the fact that a càdlàg function on [0, 1] is identified by its values on any dense set that includes 1.