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1 Introduction

For a stochastic process X and a subset B of the real numbers, the random variable T = inf{t ≥ 0 | X t  ∈ B} is called the first hitting time of B by X. A classical result in the general theory of processes is the début theorem, which has as a corollary that under the usual conditions, the first hitting time of a Borel set for a progressively measurable process is a stopping time, see [3], Sect. III.44 for a proof of this theorem, or [1] and [2] for a recent simpler proof. For many purposes, however, the general début theorem is not needed, and weaker results may suffice, where elementary methods may be used to obtain the results. For example, it is elementary to show that the first hitting time of an open set by a càdlàg adapted process is a stopping time, see [4], Theorem I.3. Using somewhat more advanced, yet relatively elementary methods, Lemma II.75.1 of [5] shows that the first hitting time of a compact set by a càdlàg adapted process is a stopping time.

These elementary proofs show stopping time properties for the first hitting times of a càdlàg adapted process X. However, the jump process \(\Delta X\) in general has paths with neither left limits nor right limits, and so the previous elementary results do not apply. In this note, we give a short and elementary proof that the first hitting time of an open set by \(\Delta X\) is a stopping time when the filtration is right-continuous and X is càdlàg adapted. This result may be used to give an elementary proof that the jumps of a càdlàg adapted process are covered by the graphs of a countable sequence of stopping times.

2 A Stopping Time Result

Assume given a filtered probability space \((\Omega,\mathcal{F},(\mathcal{F}_{t}),P)\) such that the filtration \((\mathcal{F}_{t})_{t\geq 0}\) is right-continuous in the sense that \(\mathcal{F}_{t} = \cap _{s>t}\mathcal{F}_{s}\) for all t ≥ 0. We use the convention that X 0 −  = X 0, so that there is no jump at the timepoint zero.

Theorem 1.

Let X be a càdlàg adapted process, and let U be an open set in \(\mathbb{R}\) . Define \(T =\inf \{ t \geq 0\vert \Delta X_{t} \in U\}\) . Then T is a stopping time.

As X has càdlàg, \(\Delta X\) is zero everywhere except for on a countable set, and so T is identically zero if U contains zero. In this case, T is trivially a stopping time. Thus, it suffices to prove the result in the case where U does not contain zero. Therefore, assume that U is an open set not containing zero. As the filtration is right-continuous, an elementary argument yields that to show the stopping time property of T, it suffices to show \((T < t) \in \mathcal{F}_{t}\) for t > 0, see Theorem I.1 of [4].

To this end, fix t > 0 and note that

$$\displaystyle\begin{array}{rcl} (T < t) = (\exists \;s \in (0,\infty ): s < t\ \text{ and }\ X_{s} - X_{s-}\in U)\;.& &{}\end{array}$$
(1)

Let \(F_{m} =\{ x \in \mathbb{R}\vert \;\forall \;y \in {U}^{c}: \vert x - y\vert \geq 1/m\}\), F m is an intersection of closed sets and therefore itself closed. Clearly, (F m ) m ≥ 1 is increasing, and since U is open, \(U = \cup _{m=1}^{\infty }F_{m}\). Also, \(F_{m} \subseteq F_{m+1}^{\circ }\), where F m + 1  ∘  denotes the interior of F m + 1. Let Θ k be the subset of \({\mathbb{Q}}^{2}\) defined by \(\Theta _{k} =\{ (p,q) \in {\mathbb{Q}}^{2}\vert 0 < p < q < t,\vert p - q\vert \leq \frac{1} {k}\}\). We will prove the result by showing that

$$\displaystyle\begin{array}{rcl} & & (\exists \;s \in (0,\infty ): s < t\ \text{ and }\ X_{s} - X_{s-}\in U) \\ & & = \cup _{m=1}^{\infty }\cup _{ n=1}^{\infty }\cap _{ k=n}^{\infty }\cup _{ (p,q)\in \Theta _{k}}(X_{q} - X_{p} \in F_{m})\;.{}\end{array}$$
(2)

To obtain this, first consider the inclusion towards the right. Assume that there is 0 < s < t such that X s  − X s −  ∈ U. Take m such that \(X_{s} - X_{s-}\in F_{m}\). As \(F_{m} \subseteq F_{m+1}^{\circ }\), we then have \(X_{s} - X_{s-}\in F_{m+1}^{\circ }\) as well. As F m + 1  ∘  is open and as X is càdlàg, it holds that there is \(\varepsilon > 0\) such that whenever p, q ≥ 0 with \(p \in (s-\varepsilon,s)\) and \(q \in (s,s+\varepsilon )\), \(X_{q} - X_{p} \in F_{m+1}^{\circ }\). Take \(n \in \mathbb{N}\) such that \(1/2n <\varepsilon\). We now claim that for k ≥ n, there is (p, q) ∈ Θ k such that \(X_{q} - X_{p} \in F_{m+1}\). To prove this, let k ≥ n be given. By the density properties of \(\mathbb{Q}_{+}\) in \(\mathbb{R}_{+}\), there are elements \(p,q \in \mathbb{Q}\) with p, q ∈ (0, t) such that p ∈ (s − 1 ∕ 2k, s) and q ∈ (s, s + 1 ∕ 2k). In particular, then 0 < p < q < t and | p − q | ≤ | p − s |  +  | s − q | ≤ 1 ∕ k, so (p, q) ∈ Θ k . As \(1/2k \leq 1/2n <\varepsilon\), we have \(p \in (s-\varepsilon,s)\) and \(q \in (s,s+\varepsilon )\), and so \(X_{q} - X_{p} \in F_{m+1}^{\circ }\subseteq F_{m+1}\). This proves the inclusion towards the right.

Now consider the inclusion towards the left. Assume that there is m ≥ 1 and n ≥ 1 such that for all k ≥ n, there exists (p, q) ∈ Θ k with \(X_{q} - X_{p} \in F_{m}\). We may use this to obtain sequences (p k ) k ≥ n and (q k ) k ≥ n with the properties that \(p_{k},q_{k} \in \mathbb{Q}\), 0 < p k  < q k  < t, \(\vert p_{k} - q_{k}\vert \leq \frac{1} {k}\) and \(X_{q_{k}} - X_{p_{k}} \in F_{m}\). Putting p k  = p n and q k  = q n for k < n, we then find that the sequences (p k ) k ≥ 1 and (q k ) k ≥ 1 satisfy \(p_{k},q_{k} \in \mathbb{Q}\), 0 < p k  < q k  < t, \(\lim _{k}\vert p_{k} - q_{k}\vert = 0\) and \(X_{q_{k}} - X_{p_{k}} \in F_{m}\). As all sequences of real numbers contain a monotone subsequence, we may by taking two consecutive subsequences and renaming our sequences obtain the existence of two monotone sequences (p k ) and (q k ) in \(\mathbb{Q}\) with 0 < p k  < q k  < t, \(\lim _{k}\vert p_{k} - q_{k}\vert = 0\) and \(X_{q_{k}} - X_{p_{k}} \in F_{m}\). As bounded monotone sequences are convergent, both (p k ) are (q k ) are then convergent, and as \(\lim _{k}\vert p_{k} - q_{k}\vert = 0\), the limit s ≥ 0 is the same for both sequences.

We wish to argue that s > 0, that \(X_{s-} =\lim _{k}X_{p_{k}}\) and that \(X_{s} =\lim _{k}X_{q_{k}}\). To this end, recall that U does not contain zero, and so as F m  ⊆ U, F m does not contain zero either. Also note that as both (p k ) and (q k ) are monotone, the limits \(\lim _{k}X_{p_{k}}\) and \(\lim _{k}X_{q_{k}}\) exist and are either equal to X s or X s − . As \(X_{q_{k}} - X_{p_{k}} \in F_{m}\) and F m is closed and does not contain zero, \(\lim _{k}X_{q_{k}} -\lim _{k}X_{p_{k}} =\lim _{k}X_{q_{k}} - X_{p_{k}}\neq 0\). From this, we can immediately conclude that s > 0, as if s = 0, we would obtain that both \(\lim _{k}X_{q_{k}}\) and \(\lim _{k}X_{p_{k}}\) were equal to X s , yielding \(\lim _{k}X_{q_{k}} -\lim _{k}X_{p_{k}} = 0\), a contradiction. Also, we cannot have that both limits are X s or that both limits are X s − , and so only two cases are possible, namely that \(X_{s} =\lim _{k}X_{q_{k}}\) and \(X_{s-} =\lim _{k}X_{p_{k}}\) or that \(X_{s} =\lim _{k}X_{p_{k}}\) and \(X_{s-} =\lim _{k}X_{q_{k}}\). We wish to argue that the former holds. If \(X_{s} = X_{s-}\), this is trivially the case. Assume that X s ≠X s −  and that \(X_{s} =\lim _{k}X_{p_{k}}\) and \(X_{s-} =\lim _{k}X_{q_{k}}\). If q k  ≥ s from a point onwards or p k  < s from a point onwards, we obtain X s  = X s − , a contradiction. Therefore, q k  < s infinitely often and p k  ≥ s infinitely often. By monotonicity, q k  < s and p k  ≥ s from a point onwards, a contradiction with p k  < q k . We conclude \(X_{s} =\lim _{k}X_{q_{k}}\) and \(X_{s-} =\lim _{k}X_{p_{k}}\), as desired.

In particular, \(X_{s} - X_{s-} =\lim _{k}X_{q_{k}} - X_{p_{k}}\). As \(X_{q_{k}} - X_{p_{k}} \in F_{m}\) and F m is closed, we obtain \(X_{s} - X_{s-}\in F_{m} \subseteq U\). Next, note that if s = t, we have p k , q k  < s for all k, yielding that both sequences must be increasing and \(X_{s} =\lim X_{q_{k}} = X_{s-}\), a contradiction with the fact that \(X_{s} - X_{s-}\neq 0\) as X s  − X s −  ∈ U. Thus, 0 < s < t. This proves the existence of s ∈ (0, ∞) with s < t such that \(X_{s} - X_{s-}\in U\), and so proves the inclusion towards the right.

We have now shown (2). Now, as X s is \(\mathcal{F}_{t}\) measurable for all 0 ≤ s ≤ t, it holds that the set \(\cup _{m=1}^{\infty }\cup _{n=1}^{\infty }\cap _{k=n}^{\infty }\cup _{(p,q)\in \Theta _{k}}(X_{q} - X_{p} \in F_{m})\) is \(\mathcal{F}_{t}\) measurable as well. We conclude that \((T < t) \in \mathcal{F}_{t}\) and so T is a stopping time.