# Càdlàg Feller process is quasi-left-continuous

I've been working in Chung's "Lectures from Markov Processes to Brownian Motion", and I got stuck at Exercise 1 from 2.4. The objective of the problem is to give a short proof of the quasi-left-continuity of a Feller process $$(X_t)_{t \geq 0}$$ (Markov property, càdlàg, Feller property): let $$(T_n)_n$$ a sequence of stopping times, with $$T_n \nearrow T$$. Therefore, $$\lim_{n \to \infty} X_{T_n} = X_T,$$ over the set $$\{T < +\infty \}$$.

There is a hint for the exercise: for every $$\alpha > 0$$, $$f$$ positive, bounded and continuous, $$(e^{-\alpha t}U^\alpha f(X_t))_{t \geq 0}$$ is a right continuous supermartingale (with $$U^\alpha f(x) \doteq \int_0^\infty e^{- \alpha s} P_sf(x)ds$$, where $$(P_t)_t$$ denotes the semigroup of $$(X_t)_{t \geq 0}$$). Using this, the hint states that $$\lim_{n \to \infty} U^\alpha f(X_{T_n}) = \mathbb{E}\left( U^\alpha f(X_T) \big| \bigvee_{m=1}^\infty \mathcal{F}_{T_m} \right).$$ I wonder why the last limit holds.

I don't see how to give a short proof of the stated identity (... perhaps I'm missing something). I will use the following general statement on conditional expectations which is a consequence of Lévy's convergence theorem

Lemma Let $$(Y_n)_n$$ be a sequence of random variables converging almost surely to some random variable $$Y$$ and satisfying $$|Y_n| \leq K$$ for some constant $$K$$ (not depending on $$n$$). Let $$(\mathcal{G}_n)_{n \in \mathbb{N}}$$ be a filtration and set $$\mathcal{G}_{\infty} := \bigvee_{n=1}^{\infty} \mathcal{G}_n$$. Then $$\mathbb{E}(Y \mid \mathcal{G}_{\infty}) = \lim_{n \to \infty}\mathbb{E}(Y_n \mid \mathcal{G}_n) \quad \text{a.s.}$$

Fix $$t>0$$. By the continuity of $$U_{\alpha}f$$ and the càdlàg property of the sample paths, we have $$U_{\alpha} f(X_{(T+t)-}) = \lim_{n \to \infty} U_{\alpha} f(X_{T_n+t})$$ for any sequence of stopping times $$T_n$$ with $$T_n \uparrow T$$. Applying the above lemma with $$Y := U_{\alpha} f(X_{(T+t)-}) \qquad Y_n := U_{\alpha} f(X_{T_n+t}) \qquad \mathcal{G}_n := \mathcal{F}_{T_n}$$ we find that

\begin{align*} \mathbb{E}(U_{\alpha} f(X_{(T+t)-}) \mid \bigvee_{n=1}^{\infty} \mathcal{F}_{T_n}) &= \lim_{n \to \infty} \mathbb{E}(U_{\alpha} f(X_{T_n+t}) \mid \mathcal{F}_{T_n}). \end{align*}

It now follows from the strong Markov property that

\begin{align*} \mathbb{E}(U_{\alpha} f(X_{(T+t)-}) \mid \bigvee_{n=1}^{\infty} \mathcal{F}_{T_n}) &= \lim_{n \to \infty} P_t(U_{\alpha}f)(X_{T_n}) \tag{1} \end{align*}

If we let $$t \downarrow 0$$, then it follows from the fact that $$U_{\alpha}f$$ is bounded and continuous and that $$X$$ has càdlàg sample paths that the left-hand side of $$(1)$$ converges to

$$\mathbb{E}(U_{\alpha}f(X_T) \mid \bigvee_{n=1}^{\infty} \mathcal{F}_{T_n}).$$

It remains to show that the right-hand side of $$(1)$$ converges to $$\lim_{n \to \infty} U_{\alpha} f(X_{T_n})$$ as $$t \to 0$$. To this end, we note that, by the Feller property, $$\|P_t (U_{\alpha}f)- (U_{\alpha} f)\|_{\infty} \xrightarrow[]{t \to 0} 0,$$

and so

\begin{align*} \limsup_{t \to 0} \left| \lim_{n \to \infty} P_t(U_{\alpha} f)(X_{T_n}) - \lim_{n \to \infty} U_{\alpha} f(X_{T_n}) \right| &= \limsup_{t \to 0} \lim_{n \to \infty} |P_t (U_{\alpha} f)(X_{T_n})-(U_{\alpha}f)(X_{T_n})| \\ &\leq \limsup_{t \to 0} \|P_t(U_\alpha f) - U_{\alpha} f\|_{\infty} =0. \end{align*}

• Thank you a lot! – Pablo López Rivera Apr 14 at 23:23
• @PabloLópezRivera You are welcome. – saz Apr 15 at 6:05

Another approach, also using the Lemma stated by @saz, can be based on $$e^{-\alpha T_n}U_\alpha f(X_{T_n}) =\Bbb E\left[\int_{T_n}^\infty e^{-\alpha t} f(X_t)\,dt\,|\,\mathcal F_{T_n}\right],$$ which follows from the strong Markov property. The limit on the left exists because $$X$$ is cadlag and $$U_\alpha f$$ is continuous. On the right the time integral converges boundedly to $$\int_{T}^\infty e^{-\alpha t} f(X_t)\,dt$$. Writing $$\mathcal G$$ for $$\vee_n\mathcal F_{T_n}$$, the Lemma shows that the limit of the right side is $$\Bbb E\left[\int_{T}^\infty e^{-\alpha t} f(X_t)\,dt\,|\,\mathcal G\right].$$ But $$e^{-\alpha T}U_\alpha f(X_T)=\Bbb E\left[\int_{T}^\infty e^{-\alpha t} f(X_t)\,dt\,|\,\mathcal F_T\right]$$ by the strong Markov property again. Because $$\mathcal G\subset\mathcal F_T$$, the Tower Property of conditional expctations finishes the job: \eqalign{ \Bbb E\left[\int_{T}^\infty e^{-\alpha t} f(X_t)\,dt\,|\,\mathcal G\right] &= \cr &=\Bbb E\left[\Bbb E\left[\int_{T}^\infty e^{-\alpha t} f(X_t)\,dt\,|\,\mathcal F_T\right]\,|\,\mathcal G\right]\cr &=e^{-\alpha T}\Bbb E\left[U_\alpha f(X_T)\,|\,\mathcal G\right],\cr } because $$T$$ is $$\mathcal G$$-measurable.

N.B. $$\lim_n U_\alpha f(X_{T_n+t})=U_\alpha f(X_{(T+t)-})$$ requires that the sequence $$(T_n)$$ increases strictly to $$T$$, which is not part of the q.l.c. definition.