Is the stochastic integral of the jumps process equal to zero for a continuous integrator? Let $X$ be a continuous semimartingale and $H$ a progressively measurable process in $L(X)$. Assume $H$ has left limits almost surely. I claim that the jumps process of $H$, denoted by $\Delta H = H - H_- $ is also progressively measurable (this is easy to check) and further:
i) $\Delta H \in L(X)$
ii)$\int \Delta H dX = 0$  
I believe the claim holds in general but I only have a proof for the case in which $H$ has countably many jumps. This would be the case for example if $H$ has both lateral limits almost surely, but I don't think this is the case for any progressively measurable $H\in L(X)$ with left limits, or is it?. If not, then the question would be whether my statement holds in general or not and how to prove it/disprove it.
In the case of countable jumps, the proof I have in my mind is the following: 
Write $X = M + V$ for a continous local martingale $M$ and a continuous FV process $V$. Then $\Delta H$ is in $L(X)$ iff it's in $L(M)\cap L(V)$. 
Now $\Delta H\in L(V)$ because $\int_0^t |\Delta H|d|V| =\sum_i \Delta H_i |\Delta V_i|=0  $ for all $t$ since $V$ is continuous and by the same argument $\int_0^t \Delta H dV = 0$
On the other hand, $\Delta H\in L(M) $ since $\int_0^t \Delta H^2 d\langle M\rangle = 0$, again because $H$ has countable jumps and this is a Lebesgue-Stieljes integral with respect to a continous process. And the fact that $\int_0^t \Delta H^2 dM = 0$ can be concluded by a localization argument. Given a localizing sequnce of stopping times $\tau_n$ such that $\Delta H^{\tau_n}\in L^2([M])$, the integral of each $H^\tau_n$ will be zero due to Iso Isommetry and the previous argument. Hence the integral $\int \Delta H dM $ must be zero. 
Finally, this yields $\int \Delta H dX =0$.
Any ideas on the general case?
 A: @webbster: I'm going to flip this around and deal instead with right-limited functions. So let $h:[0,1]\to\Bbb R$ have right limits ate each point of $[0,1)$. Extend $h$ to all of $[0,\infty)$ by setting $h(t)=h(1)$ for $t\ge 1$. It suffices to show that for a fixed $\epsilon>0$, the set $B:=\{t\in[0,1): |h(t)-h(t+)|\ge\epsilon\}$ is countable.
For $t\in[0,1)$ define $\tau(t):=\inf\{s>t:s\in B\}$. (Convention: $\tau(t):=1$ if $B\cap(t,1)=\emptyset$.) I claim that $\tau(t)>t$ for $t\in[0,1)$. Indeed, because $h$ has a right limit, call it $b$, at $t$, there exists $\delta>0$ such that if $s\in(t,t+\delta)$ then $|h(s)-b|<\epsilon/3$. Thus for $s,s'\in(t,t+\delta)$,
$$
|h(s)-h(s')|\le|h(s)-b|+|b-h(s')|<2\epsilon/3,
$$
and in particular, $|h(s)-h(s+)|\le2\epsilon/3<\epsilon$. Thus $B\cap(t,t+\delta)=\emptyset$, so $\tau(t)\ge t+\delta>t$, as claimed.
Now define $T_0:=0$, $T_1:=\tau(0)$, $T_2:=\tau(T_1)$, proceeding recursively.
In more detail, if $\beta$ is a (countable) ordinal, define $T_\beta:=\tau(T_\alpha)$ if $\beta$ is the successor of $\alpha$, and $T_\beta:=\sup_{\alpha<\beta}T_\alpha$ if $\beta$ is a limit ordinal. Let $\Omega$ denote the first uncountable ordinal. The map $\alpha\mapsto T_\alpha$ of countable ordinals into $[0,1]$ is non-decreasing.  As such, there is an ordinal $\beta<\Omega$ such that $T_\alpha=1$ for $\alpha\ge\beta$. Thus $B\subset\{T_\alpha: \alpha<\beta\}$ is countable.
