Monotonic function in closed interval I am reading a proof that a monotonic function in closed interval is integrable.
The proof uses the assumption that "Let assume that $f$ is increasing, therefore for all $x\in[a,b]$: $f(a)\le f(x)\le f(b)$  and $f$ is bounded"
Why can we assume that $f$ is bounded? maybe $\lim_{x\to b}f(x)=\infty$?
Moreover if every monotonic function in closed interval is bounded then it also get it maximum and minimum values there? 

Certainly $f(a)$ is a lower bound and $f(b)$ is an upper bound.  If $f$ is real-valued, then that makes $f$ bounded on the interval.
If $f$ is weakly increasing and $f(x)\uparrow\infty$ as $x\uparrow b$, then $f(b)=\infty$.  If $f$ is allowed to take values in $[-\infty,\infty]$, then that can happen.  However, it is conventional not to call a function "bounded" when it's bounded only by $+\infty$ or or $-\infty$.
