The interaction of Brownian motion and quadratic pure jump Levy process Let $(\Omega, \mathcal{F}, \mathcal{F}_{t},P)$ be a complete  filtered probability space. 
$B_{t}$ is a Brownian motion, $N_{t}$ is a quadratic pure jump Levy process, that is $[N_{t}, N_{t}]^{c}=0$. Then it is well-known that $B_{t}$ and $N_{t}$ are independent. 
Consider 
$$  X_{t} = B_{t} + N_{t} $$
My questions is about how to describe the jump. For example, 
For fixed $t$, 
$$ P(\omega, X_{t}(\omega)\neq X_{t-}(\omega)) = 0\quad ?$$
For a stopping time $T$, what's the probability 
$$ P(\omega, X_{T}(\omega)\neq X_{T-}(\omega)) \quad ?$$
 A: The first question is true and comes from the property of Lévy processes sometimes taken as a definition as it characterize those processes (see for example Protter's book) :
They are continuous in probability 
From this your first question point is trivially true. 
For the second the question the answer is obviously no take a $T$ as the 1st jump time of your process $X$ it's a stopping time and then $P(\omega, X_{T}(\omega)\neq X_{T-}(\omega))=1$.
Best regards
A: First of all note that $$\{X_t \neq X_{t-}\} = \{N_t \neq N_{t-}\}$$ as the Brownian motion has continuous sample paths. Moreover, since $(N_t)_{t \geq 0}$ is a Lévy process the jumps cannot occur at fixed times, i.e.
$$\mathbb{P}(X_{t} \neq X_{t-}) = \mathbb{P}(N_t \neq N_{t-}) = 0.$$
Indeed: Write $\Delta N_t := N_t-N_{t-}$. Then, by Fatou's lemma,
$$\begin{align*} \mathbb{P}(|\Delta N_t|>\varepsilon) &\leq \mathbb{P} \left( \bigcup_{j \geq 1} \bigcap_{k \geq j} \{|N_t-N_{t-1/k}|\geq \varepsilon\} \right) \\ &\leq \liminf_{k \to \infty} \mathbb{P}(|N_t-N_{t-1/k}| \geq \varepsilon) =0.\end{align*}$$
If we consider a stopping time, this is no longer true. For example, if $(N_t)_t$ is a Poisson process and
$$T:=\inf\{t \geq 0; N_t=1\},$$
then $$\mathbb{P}(N_T \neq N_{T-})=1.$$
