$\alpha(b) - \alpha(a) \geq \int_a^b \alpha'(x) dx$ for $\alpha$ monotone and differentiable a.e. This is a problem from Reed and Simon's book on functional analysis, where you are asked to prove $\alpha(b) - \alpha(a) \geq \int_a^b \alpha'(x) dx$ for a monotone function $\alpha$ that is differentiable a.e.. Based on the comments, I assume the answer if equality always holds is no, and that is when $\alpha$ is not differentiable everywhere.
This seems to be a direct application of the fundamental theorem of calculus, and so the proof is almost trivial and equality would hold everywhere. But this does not require the monotone assumption and seems too simple. Am I missing something?
 A: Not exactly sure of the formulation you mean, but the basic result here is that if $\alpha$ monotonic on $[a,b]$ then $\alpha$ is differentiable ae and $\int_a^b\alpha(x)dx \le \alpha(b)-\alpha(a)$ with the famous Cantor function a typical example when $\alpha'=0$ ae so the integral is zero but $\alpha(1)=1, \alpha(0)=0$ so RHS is $1$
(if $x$ is in the Cantor set so it has a decimal representation $x=0.x_1..x_n.., x_k =0,2$, then $\alpha(x)=0.y_1..y_n..$ where $y_k=x_k/2$, while if $x$ is in an excluded interval, first, it's easy to show that at the two ends which are in the Cantor set, $\alpha$ defined above takes the same value and then $\alpha(x)$ is defined as the common value at the two ends)
The differentiability ae of $\alpha$ monotonic is a difficult result but the integral inequality follows easily from Fatou's lemma.
We assume wlog that $\alpha$ is non decreasing and we extend it to the right of $[a,b]$ a little as the constant $\alpha(b)$ respectively; then since for each $h>0$ small $\frac{\alpha(x+h)-\alpha(x)}{h} \ge 0$ while the limit as $h \to 0$ is $\alpha'(x)$ almost everywhere (namely at the points of differentiability of $\alpha$), Fatou's lemma says that $$\liminf_{h \to 0^+}\int_a^b\frac{\alpha(x+h)-\alpha(x)}{h}dx \ge \int_a^b\alpha'(x)$$
But now for small enough $h>0$, $$\int_a^b\frac{\alpha(x+h)-\alpha(x)}{h}dx=\frac{\int_{a+h}^{b+h}\alpha(x)dx-\int_a^b \alpha(x)dx}{h}=$$ $$=\alpha(b)-\frac{1}{h}\int_a^{a+h}\alpha(x)dx$$
Since $\alpha$ is nondecreasing $\frac{1}{h}\int_a^{a+h}\alpha(x)dx \ge \frac{1}{h}\int_a^{a+h}\alpha(a)dx=\alpha(a)$,
so $\alpha(b)-\alpha(a) \ge \alpha(b)-\frac{1}{h}\int_a^{a+h}\alpha(x)dx$ and letting $h \to 0$ we are done!
Edit later: just to make the above clear, it may happen that $\alpha$ is monotone and continuous and the inequality is still strict as in the Cantor function case; we have equality (a non trivial result though) if $\alpha'$ exists everywhere (though it may not be continuous) and is integrable (which here holds automatically since by monotonicity $\alpha'$ has constant sign where it exists and by Fatou, its integral is finite on any finite interval) which then implies $\alpha$ absolutely continuous, and of course if $\alpha$ is absolutely continuous regardless of monotonicity (when $\alpha'$ may exist only ae)
