I'm reading "An introduction to measure theory" from Terry Tao and I'm stuck understanding part of the proof of the following theorem:(whole argument can be found in his weblog Theorem 53.)
(Monotone differentiation theorem) Any function ${F: {\bf R} \rightarrow {\bf R}}$ which is monotone (either monotone non-decreasing or monotone non-increasing) is differentiable almost everywhere.
The idea to prove this theorem, is considering Dini Derivatives of a function $F$.namely:
$\bullet$ The upper right derivative
$${\overline{D^+} F(x) := \limsup_{h \rightarrow 0^+} \frac{F(x+h)-F(x)}{h}}$$
$\bullet$The lower right derivative
$${\underline{D^+} F(x) := \liminf_{h \rightarrow 0^+} \frac{F(x+h)-F(x)}{h}}$$
$\bullet$ The upper left derivative
$${\overline{D^-} F(x) := \limsup_{h \rightarrow 0^-} \frac{F(x+h)-F(x)}{h}}$$
$\bullet$ The lower right derivative
$${\underline{D^-} F(x) := \liminf_{h \rightarrow 0^-} \frac{F(x+h)-F(x)}{h}}$$
now to show a.e differentiability we should show these 4 quantities agree a.e but it's easy to see that the only thing that is needed to be proved is ${\overline{D^+} F(x)} = {\underline{D^-} F(x)}$ a.e.
to prove this the idea is to show that $E_{r,R} := \{x \in R: {\overline{D^+} F(x)} > R > r > {\underline{D^-} F(x)}\}$ is a null set and then by letting $R>r$ both iterate over all rational numbers, conclusion follows. Tao proceeds with showing that $E_{r,R}$ has no Lebesgue point and it would give the desired conclusion.
Hence the only thing one need to prove is following lemma:
For any interval ${[a,b]}$ and any ${0 < r < R}$, one has ${m( E_{r,R} \cap [a,b] ) \leq \frac{r}{R} |b-a|}$.
The proof is as follows:
We begin by applying the rising sun lemma to the function ${G(x) := r x + F(-x)}$ on ${[-b,-a]}$; the large number of negative signs present here is needed in order to properly deal with the lower left Dini derivative ${\underline{D_-} F}$. This gives an at most countable family of disjoint intervals ${-I_n = (-b_n,-a_n)}$ in ${(-b,-a)}$, such that ${G(-a_n) \geq G(-b_n)}$ for all ${n}$, and such that ${G(-x) \leq G(-y)}$ whenever ${-x \leq -y \leq -a}$ and ${-x \in (-b,-a)}$ lies outside of all of the ${-I_n}$. Observe that if ${x \in (a,b)}$, and ${G(-x) \leq G(-y)}$ for all ${-x \leq -y \leq -a}$, then ${\underline{D_-} F(x) \geq r}$. Thus we see that ${E_{r,R} \cap (a,b)}$ is contained inside the union of the intervals ${I_n = (a_n,b_n)}$…
(full proof is not given here and can be found in the first link given above)
I don't understand how rising sun lemma is applied to $G$ since I think applying rising sun lemma to $G$ gives an at most countable family of disjoint intervals ${-I_n = (-b_n,-a_n)}$ in ${(-b,-a)}$, such that ${G(-a_n) \geq G(-b_n)}$ for all ${n}$, and such that ${G(-x) \geq G(-y)}$ whenever ${-x \leq -y \leq -a}$ and ${-x \in (-b,-a)}$ lies outside of all of the ${-I_n}$.I don' understand why in proof given by tao, it's said that ${G(-x) \leq G(-y)}$ whenever ${-x \leq -y \leq -a}$ and ${-x \in (-b,-a)}$ lies outside of all of the ${-I_n}$.
Thanks.
Best Answer
You are right. It should be $G(-x)\ge G(-y)$ and $G(-a_n)\ge G(-b_n)$.
Since $G(-a_n)\ge G(-b_n)$ and $G(x)=rx+F(-x)$, you have that $G(-a_n)=-ra_n+F(a_n)\ge G(-b_n)=-rb_n+F(b_n)$, which gives $F(b_n)-F(a_n)\le r( b_n-a_n)$, which is the inequality he needs.
Now using $G(-x)= -rx+F(x)\ge -ry+F(y)=G(-y)$ whenever $-x\le -y\le -a$, you get $$F(y)-F(x)\le r(y-x)$$ whenever $a\le y\le x$. If you divide by $y-x<0$, you get $$\frac{F(y)-F(x)}{y-x}\ge r,$$ which gives $D_{-}f(x)\ge r$.
PS If you write on his blog, he will correct the misprint. Since you are at it, when he writes "So we can arrange the inequality $G(-a_n)\le G(-b_n)$", he should write $G(-a_n)\ge G(-b_n)$.