The jump probability for a continuous time markov chain in infinitesimal time interval is 0 I got the following problem my stochastic processes in life insurance class, which I am not suppose to show, but I want to do it anyway.
show that 
$$\sum_j (1_{\{Z(t) =j \}}-1_{\{Z(t-) =j \}})dB^j(t)=0 \,\textit{ almost surely} $$
Where Z(t) is a continuous time Markov chain with bounded transition intensities and B(t) is some kind of payment process.
One should note, that the notation above is kind of informal, and formally this translates to showing 
$$\int_A \sum_j (1_{\{Z(t) =j \}}-1_{\{Z(t-) =j \}})dB^j(t)=0  $$
where $ A \subset (0,\infty)$ is a borel set. 
My attempt was to show, that $P(\{Z(t) \neq Z(t-)\})=0$ and recall, that the amount of jumps in a cont. markov chain is countable (this is to my knowledge assumed in the definition), given us a us a countable sum of zeros. My problem is, that I am not sure how to show the first step, and if it is in fact true. Could anyone help me?
 A: Your idea is the right one. Your first step follows from the absolute continuity of the jump times, which is a consequence of the existence of transition intensities.
If the jump process has deterministic jumps (implying that transition intensities do not exist), and they occur at the same point in time as deterministic payments (discontinuities of $B$), the result does not hold.
A rigorous proof building on the same ideas is the following:
Let me start by adding a few probabilistic details. We have a Markov jump process $(X_t)_{t\geq 0}$ on a finite state space $S$ with transition intensities $\mu_{jk}$ which we assume to be measurable, locally integrable from the right, and bounded on bounded intervals (the last condition is essential: it ensures that the jump process is non-explosive: i.e. that there is at-most a finite number of jumps in any time interval $[0,t]$ for any $t$); examples include continuous transition intensities.
$B^j$ is some deterministic càdlàg function of finite variation.
We want to proof that
\begin{align}
\int_{(0,t]} \left( \mathbb{1}_{(X(s-)=j)} - \mathbb{1}_{(X(s)=j)}\right) \mathrm{d}B^j(s)
\overset{a.s.}{=}
0.
\end{align}
We know that $X$ is non-explosive: there is at-most a finite number of jumps in any time interval $[0,t]$. (A Poisson process with rate $\lambda = \max_{j,k \in S, k \neq j} \sup_{s\in[0,t]} \mu_{jk}(s)$ is more likely to jump on $[0,t]$ than $X$ but still almost surely only has a finite number of jumps on $[0,t]$.)
Denote with $(T^j_n)_{n \in \mathbb{n}}$ the jumps of $X$ from or into state $j\in S$. Then
\begin{align*}
\int_{(0,t]} \left( \mathbb{1}_{(X(s-)=j)} - \mathbb{1}_{(X(s)=j)}\right) \mathrm{d}B^j(s)
=
\int_{(0,t]\cap (T^j_1,T^j_2,\ldots)} \mathrm{d}B^j(s).
\end{align*}
Almost surely, $(0,t]\cap (T^j_1,T^j_2,\ldots)$ is a finite subset of $(0,t]$. The measure $\mathrm{d}B^j$ can be decomposed into a part which is absolutely continuous w.r.t. the Lesbesgue measure, $\nu^a$, and a part which is singular to the Lebesgue measure $\nu^s$. Furthermore, $\nu^s$ can be decomposed into a non-atomic part $\nu^c$,
\begin{align*}
\nu^c(\{s\}) = 0, \hspace{5mm} s \in [0,\infty),
\end{align*}
and a jump part $\nu^j$, such that for an at-most countable subset $J$ of $[0,\infty)$ and weights $w_k \in \mathbb{R}$, $k \in J$, it holds that
\begin{align*}
\nu^j(A)
=
\sum_{k \in A \cap J} w_k.
\end{align*}
Because almost surely, $(0,t]\cap (T^j_1,T^j_2,\ldots)$ is a finite subset of $(0,t]$, it follows that
\begin{align*}
\int_{(0,t]\cap (T^j_1,T^j_2,\ldots)} \mathrm{d}B^j(s)
&\overset{a.s.}{=}
\nu^j((0,t] \cap (T^j_1,T^j_2,\ldots)) \\
&=
\sum_{k \in (0,t] \cap (T^j_1,T^j_2,\ldots) \cap J} w_k.
\end{align*}
Thus the desired statement follows if we can show that the set $\{\omega \in \Omega : (0,t] \cap (T^j_1(\omega),T^j_2(\omega),\ldots) \cap J \neq \emptyset\}$ has probability zero. To do so, we must show that $\mathbb{P}(T^j_n\in J)=0$, but this follows from the assumption of existence of transition intensities. (Conditionally on $X_t$ being in state $j$, the next jump time has the same distribution as the first realization of a time inhomogeneous Poisson process with rate $\sum_{k\in S, k \neq j} \mu_{jk}$. Thus the distributions of the jump times are absolutely continuous w.r.t. the Lebesgue measure.). This completes the proof.
Some details concerning the last part of the argument
Let $T(s)$ be the next jump time after time $s\geq 0$,
\begin{align*}
T(s)
=
\inf\{t>s: X(t) \neq X(s)\}
\end{align*}
Now using that $X$ has transition intensities,
\begin{align*}
\mathbb{P}(t - 1/n < T(s) \leq t \, | \, X(s))
=
e^{-\int_{(s,t-1/n]} \sum_{k\in S, k \neq X(s)} \mu_{X(s)k}(u) \, \mathrm{d}u}
-
e^{-\int_{(s,t]} \sum_{k\in S, k \neq X(s)} \mu_{X(s)k}(u) \, \mathrm{d}u}
\end{align*}
for $t>s$, $n\in\mathbb{N}$ sufficiently large. By monotone convergence
\begin{align*}
\mathbb{P}(t - 1/n < T(s) \leq t \, | \, X(s))
\overset{n\to\infty}{\to}
0
\end{align*}
Furthermore, it follows from downwards continuity of measures, as
\begin{align*}
\bigcap_{n\in\mathbb{N}} (t-1/n,t]
=
\{t\}
\end{align*}
that we have the convergence
\begin{align*}
P(t-1/n < T(s) \leq t \, | \, X(s))
\overset{n \to \infty}{\to}
P(T(s)=t \, | \, X(s)),
\end{align*}
By uniqueness of limits, it holds that
\begin{align*}
P(T(s)=t \, | \, X(s))
=
0.
\end{align*}
A: Fix $j$. First consider the continuous payment rate part of $B^j(t)$. Since $Z(t)$ is CADLAG, so is $1_{\{Z(t)=j\}}$. By an example we had, which was pretty basic, we have:
$$1_{\{Z(t)=j\}} b^j(t) dt = 1_{\{Z(t-)=j\}} b^j(t) dt \, \, a.s.$$
Next consider the lump sum part of $B^j(t)$. Let $\left( t^{(j)}_m \right)_{1\leq m \leq n_j}$ be the deterministic lump sum payout times, finite in number. Note that
$$\Delta N(t)=0 \implies 1_{\{Z(t)=j\}} - 1_{\{Z(t-)=j\}} = 0.$$
Note that the set $\left(\Delta N(t^{(j)}_m)=0, 1\leq m \leq n_j \right)$ has full probability since for fixed $m$
$$P(\Delta N(t_m^{(j)})=1 )= \lim_{t \uparrow s} \int_t^s f(x) dx=0, $$
for some bounded intensity function $f$ and since $(B^j)^{(d)}(t)$ is a step function, we have
$$ \left( 1_{\{Z(t)=j\}} - 1_{\{Z(t-)=j\}} \right)d(B^j)^{(d)}(t) = 0. $$
Thus
$$ \sum_j \left( 1_{Z(t) = j} - 1_{Z(t-)=j} \right) b^j(t) = 0 \,\, a.s.$$
and the conclusion follows
