• $(\Omega,\mathcal{A},\operatorname{P})$ be a probability space
  • $I\subseteq\mathbb{R}$
  • $E$ be a metric space and $\mathcal{E}:=\mathcal{B}(E)$ be the Borel-$\sigma$-algebra on $E$
  • $X:=(X_t)_{t\in I}$ and $Y:=(Y_t)_{t\in I}$ be stochastic processes on $(\Omega,\mathcal{E}$ with values in$(E,\mathcal{E})$
  • $N_t:=\left\{X_t\ne Y_t\right\}\color{blue}{:=\left\{\omega\in\Omega:X_t(\omega)\ne Y_t(\omega)\right\}}$ for $t\in I$

Now, suppose

  • $I$ is an intervall
  • $\operatorname{P}[N_t]=0$ for all $t\in I$ $\color{blue}{\text{(We say that }X\text{ and }Y\text{ are modifications of each other})}$
  • For $\operatorname{P}$-almost every $\omega\in\Omega$ $$I\to E\;,\;\;\;t\mapsto X_t(\omega)$$ is continuous from the right $\color{blue}{\text{(We say that the paths of }X\text{ are almost surely continuous from the right)}}$ and the same holds for $Y$

Let $$R':=\left\{X,Y\text{ are continuous from the right}\right\}$$ By our assumption, there exists a $R\in\mathcal{A}$ such that $R\subseteq R'$ and $\operatorname{P}[R]=1$. Now, let $$\tilde{I}:=\begin{cases}I\cap\mathbb{Q}&\text{, if }I\text{ is right-open}\\I\cap\mathbb{Q}\cup\max I&\text{, otherwise}\end{cases}$$ and $\tilde{N}:=\bigcup_{q\in\tilde{I}}N_q$. Since $X$ and $Y$ are modifications of each other, it's easy to prove $\operatorname{P}[\tilde{N}]=0$.

My question: Clearly, the inclusion $$N_t\cap R\subseteq\bigcup_{q\in\tilde{I}:q\ge t}\left(N_q\cap R\right)\;\;\;\text{for all }t\in I\tag{1}$$ is somehow implied by the continuity from the right. But how exactly can we derive $(1)$?


1 Answer 1


Lemma: Let $I=[a,b)$ be an interval and $f,g: [a,b) \to E$ functions which are continuous from the right. Then $$f=g \iff \forall x \in \mathbb{Q} \cap [a,b): f(x) = g(x).$$

Proof: "$\Rightarrow$" is obvious. For "$\Leftarrow$" we fix $x \in [a,b)$. Then there exists a sequence $(x_n)_{n \in \mathbb{N}} \subseteq I \cap \mathbb{Q}$ such that $x_n \downarrow x$. By assumption, $g(x_n) = f(x_n)$. It follows from the right-continuity at $x$ that $$f(x) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} g(x_n) = g(x).$$

Applying this lemma to the paths $t \mapsto X_t(\omega)$ and $t \mapsto Y_t(\omega)$ for fixed $\omega$ yields

$$\{X_{\cdot} = Y_{\cdot}\} \cap R = R \cap \{X_{\max I} = Y_{\max I}\} \cap \bigcap_{t \in \mathbb{Q} \cap I} \{X_t = Y_t\} = R \cap \bigcap_{t \in \tilde{I}} \{X_t = Y_t\}.$$

  • $\begingroup$ So, we can conclude $$R\cap N_t\subseteq R\cap\left\{X_s=Y_s\;\text{for all }s\in I\;\text{with }s\ge t\right\}^C=R\cap\left(\bigcap_{g\in\tilde{I}:q\ge t}\left\{X_q=Y_q\right\}\right)^C$$ right? $\endgroup$
    – 0xbadf00d
    Apr 17, 2015 at 14:48
  • $\begingroup$ @0xbadf00d yes, that's correct. $\endgroup$
    – saz
    Apr 17, 2015 at 14:49
  • $\begingroup$ Maybe it's obvious, but why does it follow $$R^C\cup\bigcup_{t\in I}N_t\subseteq R^c\cup\tilde{N}\;?$$ $\endgroup$
    – 0xbadf00d
    Apr 17, 2015 at 14:58
  • $\begingroup$ @0xbadf00d Well, obviously $$\bigcup_{t \in I} N_t = \left(R \cap \bigcup_{t \in I} N_t \right) \cup R^c.$$ Now your previous comment shows $$R \cap \bigcup_{t \in I} N_t = \bigcup_{t \in I} (R \cap N_t) \subseteq R \cap \left( \bigcap_{q} \{X_q = Y_q\} \right)^c \subseteq R \cup \tilde{N}.$$ $\endgroup$
    – saz
    Apr 17, 2015 at 15:06
  • 1
    $\begingroup$ @0xbadf00d Ah, sorry, instead of "$=$" it should read "$\subseteq$" in the first equation, i.e. $$\bigcup_{t \in I} N_t \subseteq \left( R \cap \bigcup_{t \in I} N_t \right) \cup R^c.$$ (More abstractly: $A = (A \cap B) \cup (A \cap B^c) \subseteq (A \cap B) \cup B^c$ for any two sets $A,B$.) $\endgroup$
    – saz
    Apr 18, 2015 at 5:09

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