Modulus of continuity of Levy process as jump size tends to zero

173 Views Asked by At

When reading Kallenberg's "Foundations of Modern Probability Theory", 2nd edition, I have a question regarding an argument in the proof of Lemma 15.19.

Let $X_n(t)$ be a sequence of Levy processes. Let $\Delta X_n(t)=X_n(t)-X_n(t-)$ be the jump component. The proof claims that: $$ if \sup_{ t \in [0,1]}|\Delta X_n(t)|\overset{P}{\rightarrow} 0 ~as~ n\rightarrow\infty,~~then~ w(X_n,h_n)\overset{P}{\rightarrow} 0 $$ for some sequence $h_n\downarrow 0$, where $$w(f,\delta)=\sup_{s_1,s_2\in[0,1], |s_1-s_2|<\delta} |f(s_1)-f(s_2)|. $$ is the modulus of continuity on $[0,1]$.

I don't understand how such $h_n$ is found.

I understand for each cadlag sample path $f$ with jump size bounded by $\epsilon$, one can find small $\delta>0$ so that $w(f,\delta)<2\epsilon$ through a compactness argument. But I don't know how to get the required $h_n$ for sample paths constituting a large probability.