A function $f$ is increasing on the closed internal $[a,b]$ Let the function $f$ be increasing on the closed internal $[a,b]$. If $a \le f(a)$ and $f(b)\le b$, prove that:


  $\exists x_0\in [a,b]$, such that $f(x_0)=x_0$.


Thanks for your help.

Note that $f$ need not be continuous.
Assume otherwise. Especially, $f(a)&gt;a, f(b)&lt;b$. Let $u_0=a,v_0=b$.
Given an interval $[u_n,v_n]\subseteq [a,b]$ with $f(u_n)&gt;u_n, f(v_n)&lt;v_n$ we can find a subinterval $[u_{n+1},v_{n+1}]\subset [u_n,v_n]$ with 
$$f(u_{n+1})&gt;u_{n+1}, f(v_{n+1})&lt;v_{n+1}\ \text{ and }\ |v_{n+1}-u_{n+1}|=\frac12|u_n-v_n|.$$ 
To do so let $w_n=\frac{u_n+v_n}{2}$. If $f(w_n)&gt;w_n$, let $u_{n+1}=w$, $v_{n+1}=v_n$; otherwise let $u_{n+1}=u_n$, $v_{n+1}=w_n$.
Then these nested intervals converge towards a number $x_0\in[a,b]$.
Then $u_n&lt;f(u_n)\le f(x_0)\le f(v_n)&lt;v_n$ for all $n$ implies $f(x_0)=x_0$.
Let $g(x)=f(x)-x$.
(*)Prove that if $x\in[a,b]$ and $g(x)&gt;0$ then there is an $\epsilon&gt;0 $ such that $g(x)&gt;0$ in $[x,x+\epsilon]$. For example for $x=a$ you could take 
$\epsilon=f(a)-a$. Prove a similar property for $x\in[a,b]$ with $g(x)&lt;0$.

Now if $r=\sup\{x\in[a,b]:g(x)&gt;0\}$ (or $\inf\{x\in[a,b]:g(x)&lt;0\}$) use (*) to show that $g(r)=0$.