A particular measure of noncompactness? I am working on an article based mainly on the notion of 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 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 somewhere..
 A: $\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}$ 
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. 
