For a signed measure $\mu$ is $Y(t):=\int_0^tX(s)\mu(ds)$ continuous? 
As in the title I wonder if  $Y(t):=\int_0^tX(s)\mu(ds)$ is continuous for a signed measure $\mu$ is? [With some condition on $X $ such that $Y(t)<\infty$

I would like to claim so, writing the integral as $\int_0^tX(s)\mu^+(ds)-\int_0^tX(s)\mu^-(ds)$ (where $\mu=\mu^+-\mu^-$) and then using that it holds for a measure $\lambda$ the map $Y(t):=\int_0^tX(s)\lambda(ds)$ that  $Y $ is continuous. [For $X $ integrable on every $[0,t]$ as this follows from the dominated convergence theorem by considering $1_{\{t+h\}}X$]
But I have the following confusing result from a book by Medvegyev.


Where The part that confuses me is that $\Delta Y= X \Delta V $ where $\Delta Y(t):= Y(t+)-Y(t-) $ since of course if $Y $ is continuous then $\Delta Y=0 $.

(Here I assume the conditions on $V $ means for every $\omega $ there corresponds a signed measure to $s \mapsto V(s,\omega) $)
Thanks in advance!
 A: It might be best to first think about the Riemann-Stieltjes integral in the deterministic setting. Checking baby Rudin Theorem 6.15, you will find similar to the following:
Let $a < s < b$, let $h$ be the unit (or Heaviside) step function, and let $\alpha(x) = h(x-s)$. If $f$ is bounded on $[a,b]$ and continuous at $s$ then
$$
\int_a^b f \,\mathrm{d}\alpha = f(s).
$$
That the integral makes sense is implicit here and justified in Rudin's proof.
The relevant part of the result you quote seems to be a generalization of this fact.
A: I will add the following: As per Nate Eldrage comment if a measure $\mu $ has an atom at a point $t$ - which happens if and only if the distribution function corresponding to $\mu $ is discontinuous at $t $- then $Y $ isn't continuous at $t$.
Consider that for a simple function $X=\sum \alpha_i 1_{A_i } $, then as $(t-h,t]$ is contained in one of the sets $A_i $ for $h $ sufficiently small we get that  $\lim_{h\to 0+ }(Y(t)-Y(t-h))=X(t)\mu(\cap_{h>0 } (t-h,t])=X(t)\mu(\{t\})$ and this may then be generalized to $X \in L^1(\mu)$ considering a sequence of simple functions converging to $X $.
