Is there a value of the function $g$ that is greater than the length of the interval? I need to prove or disprove the following:
Suppose that $g$ is a positive real valued function of a real number. If $a < b$ are real numbers, then there is a finite sequence $a = t_0 < t_1 \dots < t_n = b$ of real numbers such that in each interval $[t_k,t_{k+1}]$ there is a point where the value of the function $g$ is greater than the length of the interval. 
With out the assumption of continuity, I'm not sure how to proceed. Any advice?
 A: Let $g$ be strictly positive, not necessarily continuous real valued funtion and suppose there are $a<b$ such that there is no finite sequence $a=t_{0}<t_{1}<...<t_{n}=b$ such that for all $0\leq k\leq n-1$ there is an $x\in[t_{k},t_{k+1}]$ such that $g(x)\geq t_{k+1}-t_{k}$. For $r\in[a,b]$ and $n\in\mathbb{N}$ we call a strictly increasing sequence $(t_{k})_{k=0}^{n}$ admissible if $t_{0}=a$, $t_{n}=r$ and for all $0\leq k\leq n-1$ there exists an $x\in [t_{k},t_{k+1}]$ such that $g(x)\geq t_{k+1}-t_{k}$.
Consider
$$c=\sup\{r\in[a,b]:\exists n\in\mathbb{N}\text{ and }(t_{k})_{k=1}^{n}\text{ admissible}\}.$$
Let $(t_{k})_{k=1}^{n}$ be an admissible sequence such that $c-t_{n}<g(c)$. Then we can pick $t_{n+1}=g(c)+t_{n}>c$ and $(t_{k})_{k=1}^{n+1}$ will be an admissible sequence which gives a contradiction.
A: For every $x\in [a,b]$ let $U_x$ be the open interval $(x-\frac 12g(x), x+\frac 12g(x))$. The $U_x$ form an open cover of the compact interval $[a,b]$, so there is a finite subcover $U_{x_1}, \ldots,U_{x_n}$. Without loss of generality it can be assumed that $x_1\lt x_2\lt \ldots\lt x_n$ and that the subcover is minimal in the sense that no real subcollection of the $U_{x_i}$ forms a cover.
Since the successive $U_{x_i}$ must overlap, there are $t_i\in U_{x_i}\cap U_{x_{i+1}}$ with $x_i<t_i<x_{i+1}$ for $i=1,\ldots, n-1$. From $x_i\in [t_{i-1},t_i]\subset U_{x_i}$ it follows that $\max_{t\in [t_{i-1},t_i]}g(t)\ge g(x_i)=\rm{length} (U_{x_i})>t_i-t_{i-1}$.
