Real Analysis, Folland problem 3.5.33 Functions of Bounded Variation Problem 33 - If $F$ is increasing on $\mathbb{R}$, then $F(b) - F(a) \geq \int_{a}^{b}F'(t)dt$.
My question: I want to know what facts arise when we know $F$ is increasing on $\mathbb{R}$? 
There must be something within those facts that lead to this conclusion I am just lost on what would lead to this result without making a plethora of assumptions or constructing a function or something where this works out.
Also we can assume $F$ is right-continuous as well.
Any suggestions is greatly appreciated. 
 A: To prove this you need some background in measure theory including Fatou's lemma and the fact that a monotone increasing function is almost everywhere differentiable.
Consider the sequence of functions for $x \in [a,b],$
$$F_n(x) = n\left[F(x + 1/n)-F(x)\right].$$
This sequence is well-defined by extending $F$ with $F(x) = F(b)$ for $x > b.$
If $F$ is monotone increasing then it can be shown that the derivative $F'$ exists almost everywhere.  Furthermore, $F_n(x) \to F'(x)$ almost everywhere as $n \to \infty$ and $F'$ is measurable.
Integrating we get
$$\int_a^b F_n(x) \, dx= n \int_a^b \left[F(x + 1/n)-F(x)\right] \, dx \\ = n \int_{a + 1/n}^{b + 1/n} F(x) \, dx -n\int_{a}^{b} F(x) \, dx \\ =n \int_{b}^{b + 1/n} F(x) \, dx -n\int_{a}^{a + 1/n} F(x) \, dx \\ \leqslant F(b) - F(a).$$
Applying Fatou's lemma we get
$$\int_a^b F'(x) \, dx \leqslant  \liminf_{n \to \infty}\int_a^b F_n(x) \, dx \leqslant F(b) - F(a).$$
A: Another approach (only sketched): There is a Borel measure $\mu$, finite on compact subsets of $\Bbb R$, such that $\mu\{(a,b]\}=F(b)-F(a)$ for all real $a\le b$. Let $\mu_s+\mu_a$ be the Lebesgue decomposition of $\mu$ into singular and absolutely continuous components. The measure $\mu_a$ admits a representation $\mu_a(B)=\int_B f(x)\,dx$ for all Borel $B\subset\Bbb R$, where $f\ge 0$ is Borel measurable. In fact, $f(x) = F'(x)$ for Lebesgue a.e. $x$. Finally, $F(b)-F(a)=\mu\{(a,b]\}\ge\mu_a\{(a,b]\}=\int_a^b F'(x)\,dx$.
