Jumps of independent Lévy processes Suppose I have two independent Lévy processes $(X_t)_t$ and $(Y_t)_t$, both not continuous.

Is anyone familiar and can refer me to a result (or a counterexample) which states that

${\displaystyle \sum_{0\leq s\leq t}}|\Delta X_{s}\cdot \Delta Y_{s}|=0
 $ for all $t\in \mathbb{R}$ a.s?

A different yet equivalent formulation of this is

$\Delta X_{t}=0$  or  $\Delta Y_{t}=0$ a.s. for all $t\in \mathbb{R}$ 

In words, every two independent Lévy processes have no simultaneous jumps a.s. I know it holds for independent Poisson processes and I'm wondering if it generalizes.
Let $(X_t)_{t \geq 0}$, $(Y_t)_{t \geq0}$ be independent Lévy processes with Lévy triplet $(\ell_1,q_1,\nu_1)$ and $(\ell_2,q_2,\nu_2)$, respectively. By the independence of the processes, we know that $(X_t,Y_t)_{t \geq 0}$ is a Lévy process and using Lévy-Khinchine's formula we find that its Lévy measure equals

$$\nu(dx,dy) := \nu_1(dx) \otimes \delta_{\{0\}}(dy)+ \delta_{\{0\}}(dx)\otimes \nu_2(dy)$$

where $\delta_{\{0\}}$ denotes the Dirac measure centered at $0$ (for a proof see e.g. this question). Roughly speaking, the jumps of the process $(X,Y)$ are concentrated on the coordinate axis. Indeed: Let

$$A(\varepsilon) := \mathbb{R}^2 \backslash \bigg(B[0,\varepsilon] \cup \{(x,y) \in \mathbb{R}^2; x = 0 \vee y=0\} \bigg)$$

for $\varepsilon&gt;0$. Obviously, $A(\varepsilon)$ is open and satisfies $\nu(A(\varepsilon)) = 0$. Consequently, there exists $\Omega(\varepsilon)$, $\mathbb{P}(\Omega(\varepsilon))=1$, such that

$$\forall t \geq 0: \Delta (X_t,Y_t)(\omega) = (\Delta X_t(\omega),\Delta Y_t(\omega)) \notin A(\varepsilon)$$

for all $\omega \in \Omega(\varepsilon)$. Set $\Omega' := \bigcap_{k \in \mathbb{N}} \Omega(1/k)$. Then $\mathbb{P}(\Omega')=1$ and

$$\forall t \geq 0: (\Delta X_t(\omega),\Delta Y_t(\omega))  \notin \mathbb{R}^2 \backslash \{(x,y); x = 0 \vee y=0\}$$

for any $\omega \in \Omega'$. This shows that $X$ and $Y$ do not jump simultaneously a.s.
by Fubini's theorem we know that $\sum_{0\leq s\leq \infty}|\triangle X_{s}\triangle Y_{s}|$ is measurable. We can write it this way $\sum_{i=1}^{\infty}|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}$. Take expectation, and by independence,the MCT and the continuity in probability of the Levy processes we have

$E\Bigl(\sum_{i=1}^{\infty}|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}\Bigr)=\sum_{i=1}^{\infty}E\Bigl(|\triangle X_{s}\triangle Y_{s}|1_{\{i-1\leq s\leq i\}}\Bigr)=\sum_{i=1}^{\infty}E\Bigl(|\triangle X_{s}|\Bigr)E\Bigl(|\triangle Y_{s}|\Bigr)1_{\{i-1\leq s\leq i\}}=0$

so $\sum_{0\leq s\leq \infty}|\triangle X_{s}\triangle Y_{s}|=0$ a.s. Since the trajectories of the summand are increasing it implies for all $t&gt;0$.