simultaneous jumps of independent Levy processes Suppose I have two independent Levy processes $X_t$ and $Y_t$, 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}}|\bigtriangleup X_{s}(\omega)\bigtriangleup Y_{s}(\omega)|=0
 $ for all $t\in \mathbb{R}$ a.s?
A different yet equivalent formulation of this is
$\bigtriangleup X_{t}=0$  or  $\bigtriangleup Y_{t}=0$ a.s. for all $t\in \mathbb{R}$ 
In words, every two independent Levy processes have no simultaneous jumps a.s. I know it holds for independent Poisson processes and I'm wondering if it generalizes. 
 A: Consider a finite time interval, let $J_i$ denote the set of (times of) jumps of the $i$'th process, and let $J_{i,n}$ denote the set of jumps of the $i$'th process of size $(\frac 1 n,\frac 1{n-1}]$.  Note that $J_{i,n}$ is finite a.s. and therefore $J_{i,n}\cap J_{j,m}=\emptyset$ a.s. for all $n,m$ and $i\neq j$.  Since there are countably many such intersections, it follows that $J_i\cap J_j = \emptyset$ a.s.
A: 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). 
A: Besides the detailed answer you got from the double posting on math.stackexchange, one can even say a bit more:
When you consider the 2-dim Lévy process $(X(t), Y (t))$ without Gaussian part, the components are independent if and only if the support of the corresponding Lévy measure ν is in the set $\{(x, y) : xy = 0\}$. 
Intuitively, the measure is concentrated on the axis and the two components of the process never jump
simultaneously. (see Proposition 5.3. in [1])
As a textbook reference I recommend to have a look at Section 5.4 in [1].

References
[1] Cont, R., Tankov, P. Financial modelling with jump processes, Chapman Hall/CRC, Boca Raton, FL, 2004, xvi+535
