If $F$ is increasing on $\mathbb{R}$ then show that $F(b)-F(a)\geq \int_a^b F'(t)dt$.
My work:
Since $F$ is increasing on $\mathbb{R}$, $F'$ exists a.e. on $\mathbb{R}$. So $F'$ is integrable on $[a,b]$. Now by $L-R-N$ theorem there exists a complex measure $\lambda \bot m $ ($m$ is the lebesgue measure) such that, $F(b)-F(a)=\mu_F([a,b])=\lambda ([a,b])+\int_a^b F'(t)dt$. Then there is a decomposition of $[a,b]=A\cup B, $ where $A\cap B = \phi$ such that $m(A)=0$ and $\lambda (B)=0$. Now I must prove that $\lambda ([a,b])\geq 0$. I am stuck now. How do I prove it? can anyone please help?
In the proof of Folland's "Real Analysis" (Theorem 3.8) you can see that whenever $\mu$ and $m$ are positive measures, $\lambda$ is positive too. That is clearly your case.