I am working on an article based with the notion of Measure of noncompactness.
Let $ \mathcal {M}_E $ be the family of all nonempty bounded subsets of $ E $ (a Banach space)
Definition:
The function $ \nu: \mathcal{M}_E \rightarrow \mathbb R^ + $ is a measure of noncompactness if, for all $ X, Y \in \mathcal {M}_E $:
1) $ \nu (X) = \nu (\bar {X}) $.
2) $ \nu (X) = \nu (Conv (X)) $.
3) $ X \subset Y \Rightarrow \nu (X) \leq \nu (Y)$.
4) $ \nu \big(\lambda X + (1- \lambda) Y \big) \leq \lambda \nu (X) + (1- \lambda) \nu (Y) $, for all $ \lambda \in [0,1] $.
5) If $ A \in \mathcal {M}_E, $ closed and $ \nu (A) = 0 $ Then, $ A $ is compact.
6) If $ A_n \in \mathcal {M}_E $, and $ A_{n+1} \subset A_{n} $ and if $ \lim_{n \to + \infty} \nu (A_n) = 0 $, then: $ A_{\infty} = \bigcap_{n =}^ {+\infty} A_n \neq \emptyset $.
Now, let $x\in BC(\mathbb R^+)$ ( 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 $$diamX(t)=sup\{|x(t)-y(t)|:\: x,y \in X\}$$
I showed 1) ... 5) of the definition; How can I get 6) please?