I can't understand the following example of measure of non-compactness, which was given in [this](https://www.google.com.pk/url?sa=t&rct=j&q=&esrc=s&source=web&cd=3&cad=rja&uact=8&ved=0CCgQFjACahUKEwiphKzw7tfHAhXB1RQKHdgSAXg&url=http%3A%2F%2Fwww.math.ubbcluj.ro%2F~nodeacj%2Fdownload.php%3Ff%3D052appell.pdf&usg=AFQjCNEit9Ts2QkpmpCk_ZUr-4i-vNLT8A&sig2=6eetnRUc6j98ecyBT_vF0w&bvm=bv.101800829,d.bGQ) 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.



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


The set of Rademacher functions $R:=\{r\_n(t) : n \in N\}$ is bounded but not pre-compact (the sequence $(r\_n)$ is equivalent to the unit vector basis of $\ell\_2$) in $X$, but $\phi(R)=0$ because $|r\_n(t)|\leq 1$. 


-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 $X$?


Note that $\phi(M)>0$ in $X$ implies the existence of a constant $C>0$, a sequence $(f\_n)$ in $M$, and a disjoint sequence of measurable subsets of $[0,1]$ such that $\|f\_n\chi\_{A\_n}\|\_p\geq C$. However, if $(f\_n)$ is bounded in $L\_q[0,1]$, the Hölder inequality implies $\|f\_n\chi\_{A\_n}\|\_p\to 0$ as $n\to\infty$. Note also that $R$ is bounded but not precompact in each $L\_q[0,1]$.


For arguments and results of this kind, see the book Albiac, Fernando; Kalton, Nigel J. Topics in Banach space theory. Graduate Texts in Mathematics, 233. Springer, New York, 2006.


