# fundamental theorem of calculus

Let $f: [a, b] \to \mathbb R$ be integrable function and $F: [a,b] \to \mathbb R$, $F(x) = \int_a^x f(t) dt$

a) Show that if $f(x) \ge 0$ for all $x \in [a, b]$ then $F$ is increasing

b) If $F$ is increasing, can you conclude that $f(x) \ge 0$ for all $x \in [a, b]$?

So, I can write down the proof of fundamental theorem of calculus, and that will prove it, but is there a neater way to prove it, just for the increasing part?

• The fundamental theorem of calculus supposes the function $f$ is continuous. Here it is only supposed integrable. – Bernard Feb 24 '17 at 11:29
• @Bernard: you are not right. One part of the fundamental theorem of calculus reads as follows: if $f$ is Riemann-integrable, then $F$ is Lipschitz-continuous. – Fred Feb 24 '17 at 11:36
• @Fred: I'm quite sorry, but the function $f(x)=1$ if $x\ne0$, $f(0)=0$ is Riemann-integrable, but certainly not continuous (whether Lipschitz- or not). More generally, any step function is integrable, and not continuous. Unless we don't have the same definition of (Lipschitz-)continuity (?). – Bernard Feb 24 '17 at 11:46
• @Bernard: the function $F(x) = \int_a^x f(t) dt$ is Lip. -continuous, if $f$ is Riemann- integrable. What is your problem ? – Fred Feb 24 '17 at 11:49
• The fundamental theorem asserts $F$ is differentiable if $f$ is continuous. My problem was mentioning the fundamental theorem about a possibly non-continuous function. – Bernard Feb 24 '17 at 11:53

a) If $a \le x <y \le b$, then $F(y)-F(x)=\int_x^y f(t) dt \ge 0$, hence $F(x) \le F(y)$
b) Let $a=0,b=1$ and
$f(0)=-1$ and $f(x)=0$ for $x \in (0,1]$. Then $F$ is constant, hence increasing, but $f(0)<0$.