Evaluate $f(t -) := v \uparrow t \int _0^v x(s)F(ds)$ where $F$ is non-decreasing, non-negative and right-continuous? Consider the following: we have a non-decreasing, non-negative and right-continuous function $F$ from which we get a unique measure $\mu$ from $\mu((a,b]):=F(b)-F(a)$. Further assume that we have a function $x$ that is integrable w.r.t. $\mu$ on finite intervals. I would like to get the left-sided limits of the function
$$f(t) :=\int _0^t x(s)F(ds)$$
that is evaluate
$$f(t -) := v \uparrow t \int _0^v x(s)F(ds)$$
To do so we follow the usual route by considering $x$ to first be an indicator function, then a simple function, then a non-negative function and lastly writing it as a difference of its positive and negative parts.
Thus let $x(s)=1_B(s)$ and assume $B=(t_0 -v, t_0+v)$ for some $v>0$ - this is to get at the heart of the problem but other cases of course exists. Take any sequence $h_n \uparrow 0$ and consider that
$$\lim_{h_n \uparrow 0 }\int 1_{[0,t_0+h_n]} 1_B F(ds)= \lim_{h_n \uparrow 0 } \int 1_{(t_0-v, t_0+h_n ]}(s) F(ds)=\lim_{h_n \uparrow 0 } F(t_0 + h_n)-F(t_0-v)$$
Which equals $F(t_0 -)-F(t_0-v)$
For a simple function $x(s)=\sum_{j=1 } ^k \alpha_k 1_{A_k } (s)$, with the same reasoning we get that the limit of the integral equals
$$\sum_{j=1 } ^{k-1 } \alpha_j \mu(A_j) + \alpha_k(F(t_0 - ) - F(t_0 - v))$$
assuming $x$ equls $\alpha_k$ for some interval $(t_0-v,t_0+v)$ about $t_0$.

My question then is: what happens when we assume that $x$ is a non-negative function and we take a sequence of simple function converging pointwise to it?
$$\lim_{h_n \uparrow 0 } \lim_k \int 1_{[0,t_0+h_n ]} \sum_{j=1 } ^k \alpha_j 1_{A_j } (s)F(ds) $$
May we change the two limits? If this cannot generally be written in closed form consider the jump that $x$ makes at $t_0$ instead - that is consider $x(t_0 +)-x(t_0 -)$.

Grateful for any  help!
 A: Let $f:\Bbb R \to [0,\infty]$ an $\mathcal{A}$-measurable function such that $\mathcal{B}\subset \mathcal{A}$ (where $\mathcal{B}$ is the standard Borel $\sigma $-algebra on $\Bbb R $) and a sequence $(a_k)\uparrow a$. Then each $\mathbf{1}_{H}f$ is $\mathcal{A}$-measurable for any Borel set $H$ and $(\mathbf{1}_{[0,a_k]}f)\uparrow \mathbf{1}_{[0,a)}f$ point-wise, then by the monotone convergence theorem we find that
$$
\lim_{k\to\infty}\int_{\Bbb R }\mathbf{1}_{[0,a_k]}\,f\, d\mu=\int_{\Bbb R }\mathbf{1}_{[0,a)}\, f\, d\mu\tag1
$$
Now note that for any Lebesgue-Stieltjes measure any interval is measurable, then by the characterization of functional limits by sequential limits and $\rm(1)$ we find that
$$
\lim_{y\to x^-}\int_{[0,y]}f\, dF=\int_{[0,x)}f\, dF\tag2
$$
for any Lebesgue-Stieltjes measure and measurable function $f:\Bbb R \to [0,\infty]$.
If $f:\Bbb R \to \overline{\Bbb R }$ is measurable and $\int_{[0,x)}|f|\, dF<\infty $ for any chosen $x> 0$ then $\rm (2)$ still holds. 
Aside note: observe that if $\mu_F(\{x\})>0$ and $f(x)\neq 0$ then
$$
\lim_{y\to x^-}\int_{[0,y]}f\, dF=\int_{[0,x)}f\, dF\neq\int_{[0,x]}f\, dF
$$
