Convex and concave functions This maybe a silly question... So mercy me.
Let $m,v:[0,S]\to \mathbb{R}$ be two Lebesgue integrable, monotone functions, say $m$ decreasing and $v$ increasing and set:
$$M(s):=\int_0^s m(\sigma)\ \text{d} \sigma \quad \text{and}\quad V(s):=\int_0^s v(\sigma)\ \text{d} \sigma\; .$$
Obviously $M$ is concave and $V$ is convex.
Now assume there exists a subinterval $[s_1,s_2]\subset ]0,S[$ such that $M(s)=V(s)$: in this case there exists $a,b\in \mathbb{R}$ s.t. $M(s)=as+b=V(s)$, because linear functions are the only functions which are simoultaneously convex and concave.
May I say that $m(s)=a=v(s)$ a.e. in $[s_1,s_2]$?
 A: Getting rid of some irrelevant details, the question sems to be as follows:

For every $t$ in $I=[t_0,t_1]$, let $F(t)=\int_{t_0}^tf(s)\mathrm ds$. Assume that $F(t)=\alpha\cdot(t-t_0)$ for every $t$ in $I$, for some constant $\alpha$. Can one deduce that $f(t)=\alpha$ for almost every $t$ in $I$?

The answer is: Yes we can. 
To see this, consider $g(t)=f(t)-\alpha$, note that $\int\limits_Ag(s)\mathrm ds=0$ for every $A=[t_0,t]$ with $t$ in $I$, and let $C$ denote the class of Borel subsets $A$ of  $I$ such that $\int\limits_Ag(s)\mathrm ds=0$. 
Then $C$ contains the intervals $[t_0,t]$ with $t$ in $I$ and $C$ is a sigma-algebra. Thus $C$ contains the sigma-algebra on $I$ generated by these intervals. This sigma-algebra is the Borel sigma-algebra on $I$ hence $\int\limits_Ag(s)\mathrm ds=0$ for every Borel subset $A$ of $I$.
The last step is to show that any function $h$ such that $\int\limits_Ah(s)\mathrm ds=0$ for every Borel subset $A$ is zero almost everywhere. There are several ways to do that--but maybe you know one of them?
