The following proposition will be proved:

**Proposition** Suppose that $X,Y$ are càdlàg adapted processes and $X$ have no fixed time of discontinuity, that is
$$ \mathsf{P}(\Delta X_t\ne 0)=0, \qquad \forall t\ge 0.$$
If $X,Y$ are independent, then
$$ \mathsf{P}\Bigl(\sum_{s}|\Delta X_s| |\Delta Y_s|>0\Bigr)
=\mathsf{P}(\text{$X,Y$ have simultaneous jumps})=0.\tag{1}$$
Remark. If $X,Y$ are two independent Lévy processes, then it satisfy the conditions of above proposition.

*Proof.* **1** Since $X$ is càdlàg adapted process, then the random set
\begin{align}\{\Delta X\ne 0\}&=\bigcup_{m=1}^\infty[\![S_m]\!],\\
\{(\omega,t)\in\Omega\times\mathbb{R}_+: \Delta X_t(\omega)\ne 0\}
&=\bigcup_{m=1}^\infty\{(\omega,t)\in\Omega\times\mathbb{R}_+:S_m(\omega)=t\}.
\end{align}
where $S_m$ is a sequence stopping time
(cf. Jacod, J. and A. N. Shiryayev, *Limit Theory for Stochastic Processes*, 2ed. Springer, 2003. I.1.32, p.8).
This means
\begin{gather}
\{\omega:\Delta X_t(\omega)\ne0\}=\bigcup_{m=1}^\infty\{\omega:S_m(\omega)=t\}, \qquad \forall t\in\mathbb{R}_+.\\
\{\omega: \text{the path of $X_s(\omega)$ has jumps}\}=\bigcup_{m=1}^\infty\{\omega:S_m(\omega)<\infty\}
\end{gather}
The $\{S_m, m\ge 1\}$ is also a sequence stopping time with respect to natural filteration of $X$, that is
$$ (S_m\le t)\in \mathcal{F}^X_t\stackrel{\text{def}}{=}\bigcap_{u\ge t} \sigma(X_s,s\le u)\vee \mathcal{N},\qquad \forall t>0.$$
**2** Similarly, for $Y$,
$$\{\Delta Y\ne 0\}=\bigcup_{n=1}^\infty[\![T_n]\!], \qquad (Y_n\le t)\in \mathcal{F}^Y_t. $$

For $X$ and $Y$,
\begin{align}
&\biggl\{\omega: \sum_{s}|\Delta X_s| |\Delta Y_s|>0\biggr\}=\{\omega: \text{$X_s(\omega),Y_t(\omega)$ have simultaneous jumps}\}\\
&\quad =\bigcup_{m=1}^\infty\bigcup_{n=1}^\infty\{\omega:S_m(\omega)=T_n(\omega)<\infty\}. \tag{2}
\end{align}

**3** Since $X$ has no fixed time of discontinuity, that is
$$ \mathsf{P}(\{S_m=s\})\le \mathsf{P}\Bigl(\bigcup_{m\ge 1}\{S_m=s\}\Bigr) =\mathsf{P}(\Delta X_s\ne 0)=0, \quad\forall s>0, m\ge 1.$$
Hence $F_{S_m}(s)=\mathsf{P}(S_m\le s)$, the distribution function of $S_m$, is continuous and
$$\Delta F_{S_m}(s)=0,\qquad \forall s\ge 0, m\ge 1.\tag{3}$$

**4.** Using the independence of $X, Y$ and the measurability of $\{S_m, m\ge 1\}(\{T_n,n\ge 1\})$ with respect to
$\mathcal{F}^X_{\infty}(\mathcal{F}^Y_{\infty}$, respectively),
we find that $\{S_m, m\ge 1\},\{T_n,n\ge 1\}$ are independent and

\begin{align}
&\mathsf{P}(S_m=T_n<\infty)=\int_0^\infty\int_0^\infty1_{\{s=t\}}(s,t)\,dF_{S_m}(s)dF_{T_n}(t)\\
&\quad=\int_0^\infty \Delta F_{S_m}(t)dF_{T_n}(t)=0 \quad \forall m,n\ge 1.\qquad \text{by (3)}. \tag{4}
\end{align}
Now (1) follows from (2),(4).