I can't understand the following example of measure of non-compactness, which was given in this article.

Definition:A non-negative function $\phi$ defined on the bounded subsets of $X$ will be called Sadovskij functional if it satisfies the following for bounded subsets $M,N\subset X$ and $\alpha\in K$.

- $\phi(M\cup N)=\max\{\phi(M),\phi(N)\}$
- $\phi(M+N)\leq \phi(M)+\phi(N)$
- $\phi(\lambda M)=|\lambda|\phi(M)$
- $\phi(M) \leq \phi(N)$ for $M\subseteq N$
- $\phi([0, 1] ·M) = (M)$
- $\phi(\overline{co}M) = (M)$.

A Sadovskij functional is called **measure of noncompactness** if it satisfies

$\phi(M) = 0$ if and only if $M\subset X$ is pre-compact

.

**Example:**
Let $X = L^p[0, 1]$ $(1 \leq p < \infty)$ be the Lebesgue space of
all (classes of) p-integrable real functions on $[0, 1]$ with the usual norm, and denote by $\chi_D$ the characteristic function of a measurable subset $D\subset[0,1]$.

Can someone kindly explain me the following;

-How can we prove that $$\phi(M) := \limsup_{\textrm{mes}(D)\rightarrow 0} ~~\sup_{u\in M}||\chi_Du||,\quad (M\subset X)$$ is a Sadovskij functional?

-How is it not a measure of non-compactness?

-How to show that any set $M$ which is bounded in $L^q[0, 1]$ for some $q > p$ satisfies $\phi(M) = 0$ in $X$, by the Hölder inequality. How such a set need not be pre-compact in $X$?

Rather I'd be very thankful if someone can give very easy and simple examples of measure of non-compactness.

P.S. I couldn't even understand what $\limsup_{\textrm{mes}(D)\rightarrow 0}$ means.