# $f(X_{T})=f(X_{T^{-}})$ a.s. for all $f \in C_0$ implies $X_{T}=X_{T^{-}}$ a.s.

During a proof of quasi-left continuity of Lévy processes, there is the next step:

$$E(f(X_{T})g(X_{T}))=E(f(X_{T^{-}})g(X_{T}))$$ for all $f,g\in C_{0},$ where $C_{0}$ is the set of all continuous functions vanishing at infinity and $T$ is stopping time.

Because of the above we conclude that $X_{T}=X_{T^{-}}$ a.s.

I was thinking in the following:

For the equality between expectation above, we have $$f(X_{T})g(X_{T})=f(X_{T^{-}})g(X_{T}) \quad \text{a.s.}$$ Then $$f(X_{T})=f(X_{T^{-}}) \quad \text{a.s.}$$ The above holds for every $f\in C_{0},$ so $X_{T}=X_{T^{-}}$ a.s. but I have doubt in this argument. I don't think this is immediatly true, isn't it?

How could be conclude this correctly?

Any kind of help is thanked in advanced.

Note that $$\mathbb{E}Y = \mathbb{E}Z$$ does not imply $Y=Z$ almost surely, and therefore the very first step of your reasoning doesn't work.

The idea is, essentially, to show that

$$\mathbb{E}(f(X_{T}) g(X_T)) = \mathbb{E}(f(X_{T-}) g(X_T)) \tag{1}$$

implies that the random vectors $U:=(X_T,X_T)$ and $V:= (X_{T-},X_T)$ have the same distribution. Once we know this, we can conclude that $X_T-X_T$ and $X_T-X_{T-}$ have the same distribution, and so $X_T-X_{T-}=0$ in distribution. This, however, already implies $X_T-X_{T-}=0$ almost surely.

There are several ways to prove that $U$ and $V$ have the same distribution; one possibility is to use a monotone class (or approximation) argument to show that $(1)$ implies

$$\mathbb{E}(h(X_T,X_T)) = \mathbb{E}(h(X_{T-},X_T))$$

for a suitable large class of functions $h$. Alternatively, we can observe that $(1)$ holds for bounded continuous functions $f,g$ (this follows from a simple truncation argument), and so, in particular,

$$\mathbb{E}\exp \big( i \xi X_T+ i \eta X_T \big)=\mathbb{E}\exp \big( i \xi X_{T-}+ i \eta X_T \big)$$

for any $\xi,\eta \in \mathbb{R}$. This means that $U$ and $V$ have the same characteristic function, and so $U=V$ in distribution.

• Many thanks @saz. You're answer has been very useful. I have a doubt. I remember for measure theory, that $E(X)=0$ implies $X=0 a.s.$ So, why this kind of argument doesn't work in my idea? It's because they have different probability measure or distribution? – Squird37 Apr 4 '18 at 16:58
• Another doubt. Why $X_{T}-X_{T^{-}}=0$ in distribution must imply that $X_{T}-X_{T^{-}}=0$ a.s? – Squird37 Apr 4 '18 at 17:01
• @Squird37 If $X$ is a non-negative random-variable, then $\mathbb{E}(X)=0$ implies $X=0$ almost surely; without the assumption of non-negativity of $X$ the assertion fails, in general, to hold (just consider for instance a random variable $X$ such that $\mathbb{P}(X=1) = \mathbb{P}(X=-1) = 1/2$). Regarding your 2nd question: Since $X_T-X_{T-}=0$ in distribution, we know that the cdf $$F(x) := \mathbb{P}(X_T-X_{T-} \leq x)$$ satisfies $$F(x) = \begin{cases} 0, & x <0, \\ 1, & x \leq 0 \end{cases}$$ and so $$\mathbb{P}(X_{T}-X_{T-}=0) = F(0)-F(0-)=1.$$ – saz Apr 4 '18 at 17:10
• You're totally right @saz. Many thanks for the help! – Squird37 Apr 4 '18 at 19:34
• @Squird37 You are welcome. If you find the answer helpful, you can upvote it by clicking on the up arrow next to it. – saz Apr 4 '18 at 19:38