Let
- $E$ be a normed $\mathbb R$-vector space;
- $(\Omega,\mathcal A,\operatorname P)$ be a probability space;
- $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$;
- $(X_t)_{t\ge0}$ be an $E$-valued càdlàg $(\mathcal F_t)_{t\ge0}$-Lévy process on $(\Omega,\mathcal A,\operatorname P)$;
- $$\pi_t(B):=\sum_{\substack{s\in[0,\:t]\\\Delta X_s\ne0}}1_B\left(\Delta X_s(\omega)\right)=\left|\left\{s\in(0,t]:0\ne\Delta X_s\in B\right\}\right|$$ for $B\in\mathcal B(E)$ and $t\ge0$.
Let $B\in\mathcal B(E)$ with $0\not\in\overline B$ and $$f(x):=x1_B(x)\;\;\;\text{for }x\in E.$$ Note that $$\pi_tf:=\int f\:{\rm d}\pi_t=\sum_{\substack{s\in[0,\:t]\\\Delta X_s\:\in\:B}}\Delta X_s\;\;\;\text{for all }t\ge0.$$
Let $t\ge0$ and $\varepsilon>0$. In the proof of Theorem 9.7 b) of this paper it claimed that $\operatorname P\left[\|\pi_t\|_E>\varepsilon\right]$ is bounded by the probability that $X$ has at lease one jump of size $B$ in $[0,t]$. Why is this true?
What if $E=\mathbb R$ and $B=\{-1,1\}$. Couldn't $X$ have exactly two jumps in $[0,1]$, one of size $-1$ and the other of size $1$, so that $\pi_tf=0$?
I can not explain what it is meant by "$X$ has at least one jump of size $B$" but the link that you have referred to ultimately claims that $$ \operatorname P\left[|\pi_t(f)|>\varepsilon\right] \leq \operatorname P\left[\pi_t(B)>0\right], $$ which seems correct to me. To see this you need to observe that $\{|\pi_t(f)|>\varepsilon\}\subseteq \{\pi_t(B)>0\}$. This is equivalent to $\{\pi_t(B)=0\}\subseteq \{|\pi_t(f)|=0\}$. Now if $\pi_t(B)=0$, then by the definition you wrote $\pi_t(f)=0$.
The example that you gave does not violate anything. In that case $\operatorname P\left[|\pi_t(f)|>\varepsilon\right] =0$, which is trivially upper bounded by $\operatorname P\left[\pi_t(B)>0\right]$.
Let me know if this helps you, otherwise, I will delete the answer.