I am working on an article based mainly on the notion of [Measure of non-compactness](https://en.wikipedia.org/wiki/Measure_of_non-compactness), to study a particular type of fixed point theorems.


Let $\mathcal M $ to be the family of all nonempty bounded
subsets of $E$.



> 
> **Definition:**
> 
> 
> The function $\nu : \mathcal M \rightarrow \mathbb R^+$
>  is said to be a measure of noncompactnees if:
> 
> 
> 1)$\nu (A)=\nu (\bar{A)}, \forall A \in \mathcal M$.
> 
> 
> 2)If $A\_n \in \mathcal M$, and $A\_{n+1}\subset A\_{n}$ and if $\lim\_{n \to +\infty} \nu (A\_n)=0$, then: $A\_{\infty}= \bigcap\_{n=1}^{+\infty}A\_n\neq \emptyset $.
> 
> 
> 3)If $A\in \mathcal M,$ closed and $\nu(A)=0$ Then, $A$ is compact. 
> 
> 
> 


Now, let $x\in BC(\mathbb R^+)$ *(EDIT: $BC(\mathbb R^+)$ furnished with the standard supremum norm*), i.e $x:\mathbb R ^+\rightarrow \mathbb{R}$ bounded, continuous.


Let $$\omega(x,r)=\sup\{|x(t)-x(s)| \colon t,s \in \mathbb R ^+ , \ |t-s|<r\}$$


(called [modulus of continuity](https://en.wikipedia.org/wiki/Modulus_of_continuity) of $x$ ).


Let us fix $X$ a nonempty bounded subset of $BC(\mathbb R^+)$, and $$\begin{align\*}
 \omega(X,r)&= \sup\{\omega(x,r), x\in X\}\\ 
 \omega\_0(X)&= \lim\_{r\to 0} \omega(X,r)\}
\end{align\*}
$$


I need to prove that $\mu$ is a measure of noncompactness, such that:


$$\mu(X)=\omega\_0(X)+\lim\_{t\to +\infty } sup\;diamX(t)$$


where $$diamX(t)=sup\{|x(t)-y(t)|:\: x,y \in X\}$$




---


**Its Okay with 1) and 2) of the definiton, How can I get 3)? I guess we have to apply [Arzelà–Ascoli](https://en.wikipedia.org/wiki/Arzel%C3%A0%E2%80%93Ascoli_theorem) somewhere..**



A particular measure of noncompactness?
$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$ 


Assume that the topology on $BC(\mathbb R^+)$ is induced by the $\sup$ norm and that $\lim\limits\_{t\to +\infty } sup$ in the definition of $\mu(X)$ means $\limsup\_{t\to\infty}$. 


Let us now show that condition 3) in the definition of a measure of noncompactness holds. Accordingly, suppose that $X$ a nonempty closed bounded subset of $BC(\mathbb R^+)$ with $\mu(X)=0$. We need to show that then $X$ is compact. 


Take any real $\ep>0$. Then the condition $\mu(X)=0$ implies that for some real $T>0$ one has 
\begin{equation}
 \sup\{|x(t)-y(t)|\colon x,y \in X, t\ge T\}<\ep. \tag{1}
\end{equation}
On the other hand, in view of the Arzelà--Ascoli theorem, the condition $\mu(X)=0$ implies that the set 
\begin{equation\*}
 X\_T:=\{x|\_{[0,T]}\colon x\in X\}
\end{equation\*}
of the restrictions $x|\_{[0,T]}$ of the functions $x\in X$ to the interval $[0,T]$ is compact with respect to the $\sup$ norm on $[0,T]$. So, $X\_T$ is totally bounded and thus has a finite $\ep$-net $\{x\_1|\_{[0,T]},\dots,x\_k|\_{[0,T]}\}$. Therefore and in view of (1), $\{x\_1,\dots,x\_k\}$ is a finite $\ep$-net for $X$. Thus, $X$ is totally bounded. Since $X$ is closed, it is compact, as desired. 


