\documentclass[reqno]{amsart}
\usepackage{hyperref}
\usepackage{cite}

\AtBeginDocument{{\noindent\small
\emph{Electronic Journal of Differential Equations},
Vol. 2015 (2015), No. 107, pp. 1--13.\newline
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
\newline ftp ejde.math.txstate.edu}
\thanks{\copyright 2015 Texas State University - San Marcos.}
\vspace{9mm}}

\begin{document}
\title[\hfilneg EJDE-2015/107\hfil Measure integral inclusions]
{Measure integral inclusions with \\ fast oscillating data}

\author[B.-R. Satco \hfil EJDE-2015/107\hfilneg]
{Bianca-Renata Satco}

\address{Bianca-Renata Satco \newline
Stefan cel Mare University,
Faculty of Electrical Engineering and Computer Science,
Universitatii 13 - 720229 Suceava,  Romania}
\email{bianca.satco@eed.usv.ro Phone/Fax +40 230 524 801}

\thanks{Submitted April 29, 2014. Published April 21, 2015.}
\subjclass[2000]{34A60, 93C30, 26A42, 26A39}
\keywords{Measure integral inclusion;
 Kurzweil-Stieltjes integral; \hfill\break\indent  
regulated function; bounded variation}

\begin{abstract}
 We prove the existence of regulated or bounded variation solutions,
 via a nonlinear alternative of Leray-Schauder type, for the measure
 integral inclusion
 \[
 x(t) \in \int_0^t F(s, x(s)) \,du(s),
 \]
 under the assumptions of regularity, respectively bounded variation,
 on the function $u$. Our approach is based on the properties of
 Kurzweil-Stieltjes integral that, unlike the classical integrals,
 can be used for fast oscillating multifunctions on the right hand
 side and the results allow one to study (by taking the function $u$
 of a particular form) continuous or discrete problems, as well as
 impulsive or retarded problems.
\end{abstract}

\maketitle
\numberwithin{equation}{section}
\newtheorem{theorem}{Theorem}[section]
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{corollary}[theorem]{Corollary}
\allowdisplaybreaks


\section{Introduction}

Motivated by problems occurring in  fields such as mechanics, electrical 
engineering, automatic control and biology 
(see \cite{aub,miller,sesza}), an increasing attention has been given to
 measure-driven differential equations in the theory of differential equations: 
equations of the form
$$
dx(t) =g(t, x(t)) d\mu(t),
$$
where $\mu$ is a positive regular Borel measure. 
An equal interest has been shown to the related integral problems and, 
more recently, for practical reasons (arising, e.g., 
from the theory of optimal control), to the set-valued associated problems. 
Such studies cover some classical cases like usual differential inclusions 
(when $\mu$ is absolutely continuous with respect to the Lebesgue measure), 
difference inclusions (for discrete measure $\mu$) or impulsive problems
(when the measure $\mu$ is a combination of the two types of measures).

We shall approach the matter of existence of solutions of measure integral inclusion
\begin{equation} \label{e1}
x(t)\in \int_0^t F(s,x(s))\,du(s)
\end{equation}
via Stieltjes integration theory. As the function $u$ will not be assumed
absolutely continuous, we will not be able to find classical solutions.
More precisely, we will work with regulated or bounded variation function $u$
and the obtained solutions will be of the same kind.

Let us remark right from the start that the use of Riemann-Stieltjes integration 
theory is not possible when the function to integrate and the function $u$ are 
both discontinuous. Also, the Lebesgue-Stieltjes integral will not be appropriate 
(as in \cite{dalmasorampazzo,quincampoix,silva,cs}, to cite only a few) 
when the function under the integral sign is allowed to be highly oscillating. 
In the given situation, the most natural notion of integral is the 
Kuzweil-Stieltjes integral.

In the single-valued framework there is a series of existence results for 
Stieltjes integral equations using Kurzweil integral in the linear or nonlinear 
case (we refer to  \cite{tvrdy,schw,kurzweil} or to the more recent 
\cite{federsonmesqslavik,federsonmesqslavik2,afonsoronto}). 
As for the set-valued case, as far as the author knows, there are existence 
results via Lebesgue-Stieljes integral (such as \cite{cs}), but the problem 
has not been investigated yet in the setting of Kurzweil-Stieltjes integral.

In the first part, we will provide existence results for inclusion \eqref{e1} 
under the assumption that $u$ is regulated, using the notion of equi-regularity 
(introduced in \cite{frankova}). In the second part, the function $u$ will be 
assumed of bounded variation and the existence of solutions will be studied. 
Finally, in view of practical applications, the existence of bounded variation 
solutions will be obtained in a particular case: by considering 
Kurzweil-Stieltjes integrals of regulated functions with respect to a 
bounded variation function.
As the theory of measure driven problems cover many well-known situations 
(see \cite{afonsoronto} for a discussion in this sense), for a particular 
function $u$, new existence results can be deduced for usual integral 
inclusions, difference inclusions, impulsive or retarded problems for 
systems with fast oscillating data.

\section{Definitions and notation}

Let $(X,\|\cdot\|)$ be a separable Banach space (the separability allows 
us to apply the classical measurable selection theorems, see \cite{CV}). 
A function $u : [0,1] \to X$ is said to be regulated if there exist the 
limits $u(t+)$ and $u(s-)$ for all points $t \in [0,1)$ and
 $s \in (0,1]$. It is well-known (\cite{honig}) that the set of 
 discontinuities of a regulated function is at most countable, that 
 regulated functions are bounded and the space
 $G([0,1],X)$ of regulated functions $u:[0,1]\to X$ is a Banach space when 
endowed with the sup-norm $\|u\|_C={\sup_{t\in[0,1]}}\|u(t)\|$.

For a function $u : [0,1] \to X$ the total variation will be denoted by $\operatorname{var}_0^1(u)$
and if it is finite then $u$ will be said to have bounded variation 
(or to be a bounded variation function). Any bounded variation function is regulated.

Let us now recall some basic facts from the theory of Kurzweil-Stieltjes integration
in Banach spaces, which  is a particular case of Kurzweil integration 
\cite{kurzweil}.

Let $u:[0,1]\to \mathbb{R}$. A partition of $[0,1]$ is a finite collection of 
pairs $\{(I_i,\xi_i):  i=1,\dots ,p\}$, where $I_1, \dots,I_p$ are
non-overlapping subintervals of $[0,1]$,   $\xi_i \in I_i$,
$i=1,\dots,p$ and  $\cup^p_{i=1}I_i=[0,1]$. A gauge $\delta$
on $[0,1]$ is a positive function on $[0,1]$. For a given gauge
$\delta$  we say that a partition $\{(I_i,\xi_i): i=1,\dots ,p\}$ is
$\delta$-fine if $I_i \subset
(\xi_i-\delta(\xi_i),\xi_i+\delta(\xi_i))$, $i=1,\dots,p$.

\begin{definition}\label{henst} \rm
A function $f:[0,1]\to X$ is said to
be Kurzweil-Stieltjes-integrable with respect to $u:[0,1]\to \mathbb{R}$ 
on $[0,1]$ (shortly, KS-integrable) if there exists a
function denoted by
${(KS)\int_0^{\cdot}}f(s)du(s):[0,1]\to X$ such that, for every 
$\varepsilon > 0$, there is a gauge $\delta_{\varepsilon} $ on $[0,1]$ with
$$
\sum_{i=1}^p \big\|f(\xi_i)(u(t_i)-u(t_{i-1}))
- \Big({\rm (KS)}\int_{0}^{t_{i}} f(s)du(s)
- {\rm (KS)}\int_{0}^{t_{i-1}} f(s)du(s) \Big) \big\|
<\varepsilon
$$
for every $\delta_{\varepsilon}$-fine partition
$\{([t_{i-1},t_i],\xi_i): \ i=1,\dots ,p\}$ of $[0,1]$.
\end{definition}

The KS-integrability is preserved on all
sub-intervals of $[0,1]$. The function $t\mapsto (KS)\int_0^t f(s)du(s)$ is
called the KS-primitive of $f$ with respect to $u$ on $[0,1]$
(we refer to \cite{tvrdy} or \cite{schw} for the case where $X$
 is finite dimensional).

\begin{remark}{\rm 
When $u(s)=s$, this definition gives the concept of Henstock-Lebesgue-integrable 
function (\cite{cao}) or variational Henstock-integral \cite{marraffa}. 
If moreover $X$ is finite dimensional, in
the preceding definition the norm can be put outside the sum, giving the 
equivalent concept of Henstock integral 
(see \cite{cao,marraffa,schwye} for a comparison between the two notions 
in general Banach spaces).}
\end{remark}

\begin{definition} \rm
A collection $\mathcal{A}$ of KS-integrable functions is said to be KS 
equi-integrable if for every $\varepsilon>0$ there exists a gauge 
$\delta_{\varepsilon}$ (the same for all elements of $\mathcal{A}$) s
uch that all $f\in \mathcal{A}$ satisfy the condition in Definition \ref{henst}.
\end{definition}

As the KS-integral satisfies the Saks-Henstock Lemma \cite[Lemma 1.13]{schw}, 
the proof of \cite[Theorem 1.16]{schw} works in our
setting and gives:

\begin{proposition} \label{stie}
Let $u:[0,1]\to \mathbb{R}$ and $f:[0,1]\to X$ be KS-integrable with respect to
 $u$.
\begin{itemize}
\item[(i)] If $u$ is regulated, then so is the primitive $h:[0,1]\to X$,
 \[
 h(t)=(KS)\int_0^t f(s)\,du(s)
\] 
and for every $t \in [0,1]$,
\[
h(t^+)-h(t)=f(t)[u(t^+)-u(t) ],  \quad
h(t)-h(t^-)=f(t)[u(t)-u(t^-) ].
\]

\item[(ii)] If $u$ is of bounded variation and $f$ is bounded, 
then $h$ is of bounded variation.
\end{itemize}
\end{proposition}  
 
 For the rest of this article, unless otherwise stated, the function $u$ 
will be supposed to be regulated. 
The space of all functions that are KS-integrable with respect to $u$ will be
denoted by $\mathcal{KS}(u)$ and endowed  
with the supremum norm of the primitive 
(that is regulated, see Proposition \ref{stie} (i), namely the Alexiewicz 
norm with respect to $u$: 
$$
\| f\|^u_A=\sup_{t\in [0,1]}\| {\rm (KS)}\int_0^t f(s)du(s) \|.
$$
A  compact convex-valued multifunction $\Gamma:[0,1]\to \mathcal{P}_{ck}(X)$ 
is said to be upper semi-continuous at a point $t_0\in [0,1]$ if for 
every $\varepsilon>0$ there exists $\delta_{\varepsilon} > 0$ 
such that the excess of $\Gamma(t)$ over $\Gamma(t_0)$ 
(in the sense of Pompeiu-Hausdorff metric) is less than $\varepsilon$ whenever 
$|t-t_0|<\delta_{\varepsilon}$. Otherwise stated,
$$
\Gamma(t) \subset \Gamma(t_0) + \varepsilon B,
$$
where $B$ is the unit ball of $X$. A multifunction is upper semi-continuous 
when it is upper semi-continuous at each point $t_0 \in [0,1]$. 
Moreover, it is completely continuous if it is totally bounded and upper 
semi-continuous. The symbol $S_{\Gamma}$ stands for the family of measurable 
selections of $\Gamma$. We refer to 
\cite{ac,CV,hup,pgc,petr} for any aspect (classical or not) related 
to multivalued analysis.

A technical result will be used (see \cite{satco}).

\begin{lemma}\label{sirconv}
For any sequence $(\overline{y}_n)_n$ of measurable selections of
a $\mathcal{P}_{ck}(X)$-valued measurable multifunction $\Gamma$,
there exists $z_n \in \operatorname{conv} \{\overline{y}_m,m{\geq} n\}$ 
a.e. convergent to a measurable $\overline{y}$.
\end{lemma}

\section{Existence results - regulated case}

In this section, we  prove an existence result for measure integral inclusions 
considering Kurzweil-Stieltjes integrability with respect to a regulated 
function $u$, the main tool being the following concept:

\begin{definition}[\cite{frankova}] \rm
A set $\mathcal{A}\subset G([0,1],X)$ is said to be equi-regulated if 
for every $\varepsilon>0$ and every $t_0\in [0,1]$ there exists $\delta>0$ 
such that:\begin{itemize}
\item[(i)] for any $t_0-\delta<t'<t_0$: $\|x(t')-x(t_0^-)\|<\varepsilon$;
\item[(ii)] for any $t_0<t''<t_0+\delta$: $\|x(t'')-x(t_0^+)\|<\varepsilon$
for all $x\in \mathcal{A}$.
\end{itemize}
\end{definition}

A useful version of Ascoli's Theorem for regulated functions was proved 
in \cite{mesq} (see also \cite{frankova} in finite dimensional setting).

\begin{lemma}\label{ascoli}
Let $\mathcal{A}\subset G([0,1],X)$ be equi-regulated and, for every 
$t\in [0,1]$, $\mathcal{A}(t)=\{x(t),x\in \mathcal{A} \}$ be relatively compact. 
Then $\mathcal{A}$ is relatively compact in $G([0,1],X)$.
\end{lemma}

Moreover, as in the case of equi-continuous functions, one can prove the following
result.

\begin{lemma}\label{unifbounded}
An equi-regulated family $\mathcal{A}\subset G([0,1],X)$ which is pointwise 
bounded is uniformly bounded.
\end{lemma}

\begin{proof}
\cite[Theorem 1.2]{mesq} states that for every $\varepsilon>0$ one can 
find a finite collection $0=t_0<t_1<\dots <t_{n_{\varepsilon}}=1$ such that
$$
\|x(t')-x(t'')\|\leq \varepsilon
$$
for any $x\in \mathcal{A}$ and $[t',t'']\subset (t_{j-1},t_j)$, 
$j=1,\dots ,n_{\varepsilon}$.
Take now $\varepsilon=1$. There exist $0=t_0<t_1<\dots <t_{n_{1}}=1$ 
such that
$$
\|x(t')-x(t'')\|\leq 1
$$
for any $x\in \mathcal{A}$ and $[t',t'']\subset (t_{j-1},t_j)$, $j=1,\dots ,n_{1}$.
 If we note by $M_j=\sup \{ \|x\big(\frac{t_{j-1}+t_j}{2}\big)\|, 
x\in \mathcal{A}\}$ and by $N_j=\sup\{ \|x(t_j)\|, x\in \mathcal{A} \}$, 
$j=0,\dots ,n_{1}$, then for every $t\in [0,1]$ and any $x\in \mathcal{A}$ one gets
$$
\|x(t)\|\leq \max \big(\{M_j+1,j=1,\dots ,n_{1}\} \cup \{N_j,j=0,\dots ,n_{1}\}
\big)
$$
and the uniform  boundedness property is achieved.
\end{proof}

Bearing in mind the fact that the primitive of a function which is 
KS-integrable with respect to a regulated function is regulated as well,
 we prove the following result.

\begin{proposition}\label{equireg}
Let $u:[0,1]\to \mathbb{R}$ be regulated and $\mathcal{K}$ be pointwise
 bounded and KS equi-integrable with respect to $u$. Then the set 
$\{(KS)\int_0^{\cdot}f(s)du(s), f\in \mathcal{K}\}$ is equi-regulated.
\end{proposition}

\begin{proof}
Fix $t_0\in [0,1]$ and let $\varepsilon>0$. There exists $M>0$ such 
that $\|f(t_0)\|\leq M$ for every $f\in \mathcal{K}$. One can also find 
a gauge $\delta_{\varepsilon}$ with
$$
\sum_{i=1}^n \|f(\xi_i)(u(\tilde{t}_{i+1})-u(\tilde{t}_i))
-(KS)\int_{\tilde{t}_i}^{\tilde{t}_{i+1}}f(s)du(s) \|
\leq \frac{\varepsilon}{2}, \quad \forall f\in \mathcal{K}
$$
for any $\delta_{\varepsilon}$-fine partition 
$\left\{(\tilde{t}_i,\tilde{t}_{i+1}),\xi_i,0=1,\dots ,n\right\}$.
On the other hand, as $u$ is regulated, there exist 
$\overline{\delta}_{\varepsilon}>0$ such that
$$
\|u(t')-u(t_0^-)\|\leq \frac{\varepsilon}{2M}
$$
whenever $t_0-\overline{\delta}_{\varepsilon}<t'<t_0$ 
and the similar for the limit at the right. 

We will prove that 
$\delta'_{\varepsilon}=\min(\delta_\varepsilon(t_0),
\overline{\delta}_{\varepsilon})$ satisfies that for every 
$t_0-\delta'_{\varepsilon}<t'<t_0$:
$$
\|(KS)\int_0^{t'}f(s)du(s)-(KS)\int_0^{t_0^-}f(s)du(s)\|<\varepsilon, \quad
 \forall  f\in \mathcal{K}
$$
(and, obviously, the same for the right limit).
Indeed, as in the proof of \cite[Theorem 1.16]{schw}:
\begin{align*}
&(KS)\int_0^{t'}f(s)du(s)-(KS)\int_0^{t_0}f(s)du(s)\\
&=f(t_0)(u(t')-u(t_0))+\Big( (KS)\int_0^{t'}f(s)du(s)
 -(KS)\int_0^{t_0}f(s)du(s)\\
&\quad -f(t_0)(u(t')-u(t_0))\Big)
\end{align*}
where the last term can be made, by Saks-Henstock Lemma, (in norm) 
less that $\varepsilon/2$ and, from here:
$$
(KS)\int_0^{t_0^-}f(s)du(s)-(KS)\int_0^{t_0}f(s)du(s)=f(t_0)(u(t_0^-)-u(t_0)).
$$
It follows that
\begin{align*}
&(KS)\int_0^{t'}f(s)du(s)-(KS)\int_0^{t_0^-}f(s)du(s)\\
&= f(t_0)(u(t')-u(t_0^-))+\Big( (KS)\int_0^{t'}f(s)du(s)\\
&\quad -(KS)\int_0^{t_0}f(s)du(s)-f(t_0)(u(t')-u(t_0))\Big)
\end{align*}
and so,
$$
\|(KS)\int_0^{t'}f(s)du(s)-(KS)\int_0^{t_0^-}f(s)du(s)\|
\leq M\frac{\varepsilon}{2M}+\frac{\varepsilon}{2}=\varepsilon
$$
for any $f\in \mathcal{K}$ and $t'$ with $t_0-\delta'_{\varepsilon}<t'<t_0$.
\end{proof}

Let us recall a nonlinear alternative of Leray-Schauder type that will be 
applied below.

\begin{theorem}[\cite{oregan98}] \label{fp}
Let $D$ and $\overline{D}$ be open and closed subsets of a normed linear 
space $E$ such that $0\in D$ and let 
$T:\overline{D}\to \mathcal{P}_{ck}(E)$ be completely continuous. Then either
\begin{itemize}
\item[(i)] the inclusion $x\in T(x)$ has a solution, or
\item[(ii)] there exists $x\in \partial D$  such that 
$\lambda x \in T(x)$ for some $\lambda>1$.
\end{itemize}
\end{theorem}

Applying this theorem  will necessitate a convergence result, such as 

\begin{lemma}[{\cite[Theorem 6.1]{bpiazza}}] \label{conv1} 
Let $u$ be $ACG^{**}$ and $(f_n)_n$ a sequence KS equi-integrable with
 respect to $u$ which pointwise converges to $f$. Then $f$ is KS-integrable
 with respect to $u$ and 
$$
(KS)\int_0^1 f_n(s)du(s)\to(KS)\int_0^1 f(s)du(s).
$$
\end{lemma}

It works for functions $u$ that are more than regulated 
(but not necessarily of bounded variation), namely:

\begin{definition}[\cite{bpiazza}] \rm
(i) $u:[0,1]\to \mathbb{R}$ is said to be $ACG^{**}$ if it is continuous 
and the unit interval can be
written as a countable union of closed sets on each of which $F$ is
$AC^{**}$;

(ii) A function $u:[0,1]\to \mathbb{R}$ is $AC^{**}$ on
$E\subset [0,1]$ if, for any $\varepsilon>0$, there exists
$\eta_{\varepsilon}>0$ and a gauge $\delta:E\to\mathbb{R}_+$ such that, 
whenever $D1,D2$ are $\delta$-fine partitions of $E$ with
$\sum_{D1\setminus D2}|t'-t''| <
\eta_{\varepsilon}$, one has $$\sum_{D1\setminus D2}
|u(t')-u(t'')|<\varepsilon;$$ here $D1\setminus D2$ denotes the
collection of all connected components of $\cup D1\setminus \cup D2$.
\end{definition}

We give now the main result of this section. 
Notice that in \cite{tvrdy} it was explained why the space of regulated 
functions is the best choice for the space of solutions.

\begin{definition} \rm
A solution of measure driven inclusion \eqref{e1} is a regulated function 
$x : [0,1] \to X$ for which there exists
$g\in S_{F(\cdot,x(\cdot))}$ such that
\[
x(t) = (KS)\int_0^t g(s) \,d u(s), \quad  \forall  t \in [0,1].
\]
\end{definition}

\begin{theorem}\label{mainreg}
Let $u:[0,1]\to \mathbb{R}$ be $ACG^{**}$ and 
$F:[0,1]\times X\to \mathcal{P}_{ck}(X)$ satisfy:
\begin{itemize}
\item[(i)] for every $x\in X$, $F(\cdot,x)$ is measurable;
\item[(ii)] for every $R>0$:
 \begin{itemize}
\item[(ii1)] the family 
 $$
\cup \{ S_{F(\cdot,x(\cdot))}, x\in G([0,1],X), \|x\|_C\leq R\}
$$
 is pointwise bounded and KS equi-integrable with respect to $u$;
\item[(ii2)] the map $x\in G([0,1],X), \|x\|_C\leq R \to S_{F(\cdot,x(\cdot))}$ 
is upper semi-con\-tinuous with respect to the $\|\cdot\|_A^u$-topology on the 
space $\mathcal{KS}(u)$;
\item[(ii3)]  for each $t\in [0,1]$ ,
$$
 \big\{(KS)\int_0^t f(s)du(s), f \in S_{F(\cdot,x(\cdot))} \big\}
$$
is relatively compact for every $x\in G([0,1],X), \|x\|_C\leq R $ and
$$
\big\{(KS)\int_0^t f(s)du(s), f \in S_{F(\cdot,x(\cdot))}, 
x\in G([0,1],X), \|x\|_C\leq R  \big\}
$$ 
is bounded.
\end{itemize}
\end{itemize}
If moreover there exists $R_0$ such that $\|x\|_C\neq R_0$ for any regulated 
solution $x$ of
$$
x(t)\in \lambda \Big( x_0+\int_0^t F(s,x(s))du(s) \Big)
$$
for all $\lambda\in (0,1)$, then our integral inclusion possess regulated 
solutions with $\|x\|_C\leq R_0$.
\end{theorem}

\begin{proof}
Let $N:\overline{B_{R_0}}\to G([0,1],X)$ be the operator defined on the ball 
centered at the origin of radius $R_0$ of $G([0,1],X)$ by
$$
N(x)(t)=\big\{(KS)\int_0^t f(s)du(s), f\in S_{F(\cdot,x(\cdot))} \big\}.
$$
Obviously, the fixed points of this operator will be solutions to our inclusion.

We will check the hypothesis of Theorem \ref{fp}.
Let us note first that the values of $N$ are convex and non-empty; indeed,
 hypothesis $ii2)$ implies that for any $t\in[0,1]$ the map $F(t,\cdot)$ 
is upper semi-continuous and, thanks to hypothesis $i)$, this yields the existence 
of measurable selections for the superpositional map $F(\cdot, x(\cdot))$.

Let us prove that the values are compact.
We will get the relative compactness by Lemma \ref{ascoli}. From hypotheses (ii1), 
we are able to apply Proposition \ref{equireg} to obtain the equi-regularity,
 while the second condition in Lemma \ref{ascoli} is stated by hypotheses (ii3).

 It remains thus to prove that the values are closed. Fix then $x$ and consider 
a sequence $((KS)\int_0^{\cdot}f_n(s)du(s))_n\subset N(x)$ convergent to 
$g\in G([0,1],X)$ and show that there exists $f\in S_{F(\cdot,x(\cdot))}$ 
with $g(t)=(KS)\int_0^t f(s)du(s)$ for any $t\in [0,1]$. As $F$ is compact 
convex-valued, one can find a sequence of convex combinations 
$\tilde{f}_n \in \operatorname{co}\{f_m,m\geq n \}$ that pointwise converges to some selection 
$f$ of $F(\cdot,x(\cdot))$. Lemma \ref{conv1} implies that 
$$
(KS)\int_0^t \tilde{f}_n(s)du(s)\to (KS)\int_0^t f(s)du(s)
$$ 
and so, $g(t)=(KS)\int_0^t f(s)du(s)$ for any $t\in [0,1]$.

In the sequel, let us prove that $N$ is completely continuous. 
The total boundedness comes from Proposition \ref{equireg} and the 
pointwise boundedness hypothesis (ii3) since we can apply Lemma \ref{unifbounded}.

Let us now check that it is upper semi-continuous. To this aim, 
fix $x_0\in \overline{B_{R_0}}$ and consider an arbitrary $\varepsilon>0$. 
Hypothesis (ii2) says that there exists $\delta_{\varepsilon,x_0}>0$ 
such that for any $x\in G([0,1],X)$ with $\|x-x_0\|_C<\delta_{\varepsilon,x_0}$:
$$
S_{F(\cdot,x(\cdot))}\subset S_{F(\cdot,x_0(\cdot))}+\varepsilon B_A,
$$
where $B_A$ is the unit open ball of $\mathcal{KS}(u)$ endowed with the
 $\|\cdot\|_A^u$-topology.
By the definition of $\|\cdot\|_A^u$, it follows that for every 
$f\in S_{F(\cdot,x(\cdot))}$ one can find $f_0\in S_{F(\cdot,x_0(\cdot))}$ 
such that 
$\|(KS)\int_0^{\cdot}f(s)du(s)- (KS)\int_0^{\cdot}f_0(s)du(s)\|_C<\varepsilon$ 
which means that
$$
N(x)\subset N(x_0)+\varepsilon B_G,
$$
$B_G$ being the open unit ball of $G([0,1],X)$ and thus, the upper 
semi-continuity of $N$ is verified.

The conditions of Theorem \ref{fp} are satisfied and, as the alternative 
is excluded by hypothesis, it follows that the operator $N$ has fixed 
points and our inclusion has solutions.
\end{proof}

Another version of this result could be obtained in a similar manner.

\begin{theorem}
Let $u:[0,1]\to \mathbb{R}$ be $ACG^{**}$ and 
$F:[0,1]\times X\to \mathcal{P}_{ck}(X)$ satisfy:
\begin{itemize}
\item[(i)] for every $x\in X$, $F(\cdot,x)$ is measurable;
\item[(ii)] for every $x\in G([0,1],X)$, the family $S_{F(\cdot,x(\cdot))}$ 
is KS equi-integrable with respect to $u$;
\item[(iii)] for every $R>0$:
\begin{itemize}
\item[(iii1)] the map $x\in G([0,1],X), \|x\|_C\leq R 
 \to S_{F(\cdot,x(\cdot))}$ is upper semi-con\-tinuous with respect to the 
 $\|\cdot\|_A^u$-topology on the space $\mathcal{KS}(u)$;
\item[(iii2)]  for each $t\in [0,1]$ ,
$$
 \{(KS)\int_0^t f(s)du(s), f \in S_{F(\cdot,x(\cdot))} \}
$$
is relatively compact for every $x\in G([0,1],X), \|x\|_C\leq R $ and
$$
\big\{(KS)\int_0^t f(s)du(s), f \in S_{F(\cdot,x(\cdot))}, 
x\in G([0,1],X), \|x\|_C\leq R  \big\}
$$ 
is uniformly bounded.
\end{itemize}
\end{itemize}
If there exists $R_0$ as in Theorem \ref{mainreg}, then the integral 
inclusion possess regulated solutions.
\end{theorem}

\begin{proof}
Following the same line as in the preceding result, the operator $N$ 
has relatively compact values: they are equi-regulated by hypothesis (ii)
 and Proposition \ref{equireg} and they are pointwisely contained in a compact 
set by hypothesis (iii2). The values are closed (this can be proved as 
in Theorem \ref{mainreg}) and convex.
Besides, $N$ is totally bounded by (iii2) and upper semi-continuous by (iii1). 
Thus, the conditions of fixed point theorem are checked and so, the existence 
of solutions is obtained.
\end{proof}

\section{Existence results - bounded variation case}

When $u$ is of bounded variation, 
instead of \cite[Theorem 6.1]{bpiazza} we can use another convergence result.

\begin{lemma}\label{conv}
Let $u:[0,1]\to \mathbb{R}$ be of bounded variation and $f_n:[0,1]\to X$
 be a sequence of functions KS equi-integrable with respect to $u$ that  
converges pointwise to $f:[0,1]\to X$. Then $f$ is KS-integrable with 
respect to $u$ and
$$
(KS)\int_0^1 f(s)du(s)=\lim_{n\to\infty}(KS)\int_0^1 f_n(s)du(s).
$$
\end{lemma}

\begin{proof}
Let $\varepsilon>0$. There exists a partition 
$\mathcal{P}_0= \{(t_{i-1},t_i),\xi_i\}_{i=1}^{n_0}$ of $[0,1]$ such that
$$
\sum_{i=1}^{n_0} \big\| f_n(\xi_i)(u(t_i)-u(t_{i-1}))
-\Big((KS)\int_0^{t_i} f_n(s)du(s) - (KS)\int_0^{t_{i-1}} f_n(s)du(s)\Big)\big\|
<\varepsilon,
$$
for all $n\in \mathbb{N}$.
At the same time, one can find $n_{\varepsilon}\in \mathbb{N}$ such that 
$\|f_n(\xi_i)-f_m(\xi_i)\|<\frac{\varepsilon}{\operatorname{var}_0^1(u)}$ for every
$i=1,\dots ,n_0$ and every $m,n\geq n_{\varepsilon}$. It follows that
$$
 \sum_{i=1}^{n_0} \Big\|f_n(\xi_i)(u(t_i)-u(t_{i-1}))
-\sum_{i=1}^{n_0} f_m(\xi_i)(u(t_i)-u(t_{i-1}))\Big\|<\varepsilon,
\quad \forall m,n\geq n_{\varepsilon},
$$
whence
$$
\big\| (KS)\int_0^1 f_n(s)du(s) -(KS)\int_0^1 f_m(s)du(s)\big\|<3\varepsilon,
\quad \forall m,n\geq n_{\varepsilon}
$$
and the same for each $t\in [0,1]$: the sequence $((KS)\int_0^t f_n(s)du(s))_n$ 
is Cauchy. As in the proof of \cite[Theorem 3.6.18]{schw ye} it can be 
proved that its limit $L(t)$
equals the KS-integral of $f$ with respect to $u$ on $[0,t]$.
\end{proof}

We thus obtain, this time for a bounded variation function $u$ 
(instead of $ACG^{**}$):

\begin{theorem} \label{mainbv}
Let $u:[0,1]\to \mathbb{R}$ be of bounded variation and 
$F:[0,1]\times X\to \mathcal{P}_{ck}(X)$ satisfy the  hypothesis of 
Theorem \ref{mainreg}, except $(ii2)$, instead of which we impose:
\begin{itemize}
\item[(ii2')] the map $x\in G([0,1],X), \|x\|_C\leq R \to F(t,x(t))$ 
is upper semi-continuous uniformly in $t$.
\end{itemize}
 Then our integral inclusion possess regulated solutions with $\|x\|_C\leq R_0$.
\end{theorem}

\begin{proof}
Only the proof of the upper semi-continuity of $N$ has to be changed.
By (ii2') for each $x_0$ and $\varepsilon>0$ there exists 
$\delta_{\varepsilon,x_0}>0$ such that for any $x\in G([0,1],X)$ with 
$\|x-x_0\|_C<\delta_{\varepsilon,x_0}$:
$$
F(t,x(t))\subset F(t,x_0(t))+\varepsilon B, \quad \forall \; t\in [0,1],
$$
where $B$ is the unit open ball of $X$.
It follows that for every $f\in S^G_{F(\cdot,x(\cdot))}$ one can find 
$f_0\in S^G_{F(\cdot,x_0(\cdot))}$ such that $\|f(t)-f_0(t)\|\leq \varepsilon$ 
for every $t\in [0,1]$, whence (see \cite{stv})
$$
\|(KS)\int_0^{t}f(s)du(s)- (KS)\int_0^{t}f_0(s)du(s)\|
\leq \|f-f_0\|_C \operatorname{var}_0^1(u) \leq \varepsilon \operatorname{var}_0^1(u)
$$
 which means that
$$
N(x)\subset N(x_0)+\varepsilon \operatorname{var}_0^1(u) B_G,
$$
$B_G$ being the open unit ball of $G([0,1],X)$ and thus, the upper 
semi-continuity of $N$ is verified.
\end{proof}

For an alternative existence theorem, remark that in this setting a 
mean value result is available.

\begin{lemma}\label{mean}
Let $u:[0,1]\to \mathbb{R}$ be of bounded variation and 
$f:[0,1]\to X$ be KS-integrable with respect to $u$. 
\begin{itemize}
\item[(i)] If $u$ is nondecreasing, then
$$
(KS)\int_0^t f(s)du(s)\in (u(t)-u(0))\overline{\operatorname{co}}(f([0,t])), \quad
 \forall t\in [0,1].
$$
\item[(ii)] If $u$ is of bounded variation, then
$$
(KS)\int_0^t f(s)du(s)\in \operatorname{var}_0^t(u)
\overline{\operatorname{co}}(\{0\}\cup f([0,t]))
-\operatorname{var}_0^t(u)\overline{\operatorname{co}}(\{0\}\cup f([0,t])),
$$ 
 for all $t\in [0,1]$.
\end{itemize}
\end{lemma}

\begin{proof}
When $u$ is nondecreasing, the assertion is a consequence of the definition 
of KS-integral, since for any partition of $[0,t]$:
$$
\sum_{i=1}^p f(\xi_i)(u(t_i)-u(t_{i-1}))
=(u(t)-u(0))\sum_{i=1}^p f(\xi_i)\frac{u(t_i)-u(t_{i-1})}{u(t)-u(0)}.
$$
When $u$ is of bounded variation, it can be written as the difference 
of two non-decreasing functions $u_1$ and $u_2$ and so, by the first step,
\begin{align*}
(KS)\int_0^t f(s)du(s) 
&\in (u_1(t)-u_1(0))\overline{\operatorname{co}}(f([0,t]))-(u_2(t)-u_2(0))
 \overline{\operatorname{co}}(f([0,t]))\\
& \subset \operatorname{var}_0^t(u)\overline{\operatorname{co}}(\{0\}\cup f([0,t]))
-\operatorname{var}_0^t(u)\overline{\operatorname{co}}(\{0\}\cup f([0,t])).
\end{align*}
\end{proof}

From Theorem \ref{mainbv} we then get the existence of bounded variation solutions.

\begin{corollary}
Let $u:[0,1]\to \mathbb{R}$ be of bounded variation and 
$F:[0,1]\times X\to \mathcal{P}_{ck}(X)$ satisfy the hypothesis {\rm (i)}
 and {\rm (ii2')} of Theorem \ref{mainbv} together with:
\begin{itemize}
\item[(ii1')] the family 
$$
\cup \{ S_{F(\cdot,x(\cdot))}, x\in G([0,1],X), \|x\|_C\leq R \}
$$ 
is KS equi-integrable with respect to $u$;

\item[(ii3')] for each $t\in [0,1]$ ,
$$
 \big\{(KS)\int_0^t f(s)du(s), f \in S_{F(\cdot,x(\cdot))} \big\}
$$
is relatively compact for every $x\in G([0,1],X)$, $\|x\|_C\leq R $ 
and for any bounded $A\subset X$, 
$$
F([0,1]\times A)\; \text{  is  bounded.}
$$
\end{itemize}
If there exists $R_0$ as in Theorem \ref{mainreg}, then our integral 
inclusion possess bounded variation solutions with $\|x\|_C\leq R_0$.
\end{corollary}

\begin{proof}
The only modification to be made in the proof of Theorem \ref{mainbv} 
is at the step where the total boundedness of the operator $N$ must be verified, 
more precisely the pointwise boundedness of $N(\overline{B_{R_0}})$; 
at that point, under our assumptions, the property easily comes from 
Lemma \ref{mean} and hypothesis $ii3')$. Besides, as the found solution 
is the primitive of a bounded function with respect to a bounded variation 
function, by Proposition \ref{stie}, it is of bounded variation.
\end{proof}

In concrete situations, the Kurzweil-Stieltjes integral is mostly used 
in the case where the integrand is regulated and the function with respect 
to one integrates is of bounded variation (or viceversa); therefore, 
it could be more convenient to have an existence result for this case.
For this purpose, let us recall the following convergence result.

\begin{lemma}[{\cite[Theorem I.4.17]{stv}}] \label{convu}
 Let $u:[0,1]\to \mathbb{R}$ be of bounded variation and $f_n:[0,1]\to X$ 
be KS-integrable with respect to $u$ with $\|f_n-f\|_C\to 0$. 
Then $f$ is KS-integrable with respect to $u$ and 
$(KS)\int_0^1 f_n(s)du(s)\to (KS)\int_0^1 f(s)du(s)$.
\end{lemma}

Applying it will be possible by using another result.

\begin{lemma}[{\cite[Lemma 1.14]{mesq}}] \label{unif}
If an equi-regulated sequence of functions converges pointwise, 
then it converges uniformly towards the limit.
\end{lemma}

\begin{theorem}
Let $u:[0,1]\to \mathbb{R}$ be of bounded variation and 
$F:[0,1]\times X\to \mathcal{P}_{ck}(X)$ satisfy:
\begin{itemize}
\item[(i)] for every $x\in G([0,1],X)$, the family 
$S^G_{F(\cdot,x(\cdot))}$ of regulated selections of 
$F(\cdot,x(\cdot))$ is non-empty;

\item[(ii)] for every $R>0$:
\begin{itemize}
\item[(ii1)] the family 
$$
\cup \big\{ S^G_{F(\cdot,x(\cdot))},  x\in G([0,1],X), \|x\|_C\leq R \big\}
$$ 
is equi-regulated;
\item[(ii2')] the map $x\in G([0,1],X), \|x\|_C\leq R \to F(t,x(t))$ 
is upper semi-continuous uniformly in $t$;
\item[(ii3)]  for each $t\in [0,1]$,
$$
 \big\{(KS)\int_0^t f(s)du(s), f \in S^G_{F(\cdot,x(\cdot))} \big\}
$$
is relatively compact for every $x\in G([0,1],X), \|x\|_C\leq R $ and the 
family 
$$
\big\{(KS)\int_0^{\cdot} f(s)du(s), f \in S^G_{F(\cdot,x(\cdot))}, x\in G([0,1],X), 
\|x\|_C\leq R  \big\}
$$ 
is equi-regulated and pointwise bounded.
\end{itemize}
\end{itemize}
If moreover there exists $R_0$ such that $\|x\|_C\neq R_0$ for any
 regulated solution $x$ of
$$
x(t)\in \lambda \Big( x_0+\int_0^t F(s,x(s))du(s) \Big)
$$
for all $\lambda \in (0,1)$, then our integral inclusion possess bounded
 variation solutions with $\|x\|_C\leq R_0$.
\end{theorem}

\begin{proof}
Consider now the modified operator $N:\overline{B_{R_0}}\to G([0,1],X)$ 
defined on the ball centered at the origin of radius $R_0$ of $G([0,1],X)$ by
$$
N(x)(t)=\big\{ (KS)\int_0^t f(s)du(s), f\in S^G_{F(\cdot,x(\cdot))} \big\}.
$$
The proof of the fact that $N$ has fixed points is essentially that of 
Theorem \ref{mainreg}, except the point where it must be proved that 
the values of operator $N$ are closed; here this comes from the fact that 
the sequence $\tilde{f}_n$ pointwise converges to $f$ and it is equi-regulated so, 
by \cite[Lemma 1.14]{mesq}, $\|f_n-f\|_C\to 0$. 
Moreover, \cite[Corollary 3.2]{honig} states that $f$ is regulated. 
Now applying Lemma \ref{convu} gives the convergence of the integrals of
 $f_n$ towards the integral of $f$ and thus the closedness of the values of $N$.

Let us now check that $N$ is upper semi-continuous. Fix $x_0\in \overline{B_{R_0}}$ 
and consider an arbitrary $\varepsilon>0$. 
Hypothesis (ii2') yields that there exists $\delta_{\varepsilon,x_0}>0$ 
such that for any $x\in G([0,1],X)$ with $\|x-x_0\|_C<\delta_{\varepsilon,x_0}$:
$$
F(t,x(t))\subset F(t,x_0(t))+\varepsilon B, \quad \forall  t\in [0,1],
$$
where $B$ is the unit open ball of $X$.
It follows that for every $f\in S^G_{F(\cdot,x(\cdot))}$ one can find 
$f_0\in S^G_{F(\cdot,x_0(\cdot))}$ such that 
$\|f(t)-f_0(t)\|\leq \varepsilon$ for every $t\in [0,1]$, whence (see \cite{stv}):
$$
\|(KS)\int_0^{t}f(s)du(s)- (KS)\int_0^{t}f_0(s)du(s)\|
\leq \|f-f_0\|_C \operatorname{var}_0^1(u) \leq \varepsilon \operatorname{var}_0^1(u)
$$ 
which means that
$$
N(x)\subset N(x_0)+\varepsilon \operatorname{var}_0^1(u) B_G,
$$
$B_G$ being the open unit ball of $G([0,1],X)$ and thus, the upper 
semi-continuity of $N$ is verified.

Finally, as any solution is the KS-primitive of a regulated function 
(therefore bounded) with respect to the bounded variation function $u$, 
Proposition \ref{stie}.
 (ii) asserts that it is more than regulated: it is of bounded variation.
\end{proof}

\begin{remark} {\rm
The imposition of assumption (ii3) (equi-regularity of primitives) together with 
(ii1) (equi-regularity of selections) might look artificial but, 
in fact, the equi-regularity of primitives follows from the equi-regularity 
of selections only if we impose a pointwise boundedness condition on 
the family of selections. This pointwise boundedness condition would be 
very strong since, by Lemma \ref{unifbounded}, it would imply the uniform 
boundedness and many of the properties given above would then be obtained 
in a much simpler manner.}
\end{remark}

\begin{remark}{\rm
New existence results can be deduced in particular cases, namely when $u$ 
is absolutely continuous (leading to usual continuous problems), 
the sum of step functions (leading to discrete problems) or a sum between an 
absolutely continuous function and a sum of step functions 
(in which case one gets impulsive problems), as well as for retarded problems 
(see \cite{afonsoronto}).}
\end{remark}

\subsection*{Acknowledgements} 
This work was supported by a grant of the Romanian National Authority 
for Scientific Research, CNCS - UEFISCDI, project number
PN-II-RU-TE-2012-3-0336.


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\end{document}