This is Exercise $58$ in Tao's blog: https://terrytao.wordpress.com/2010/10/16/245a-notes-5-differentiation-theorems/comment-page-2/#comment-678096
(One-sided Hardy-Littlewood inequality) Let ${F: [a,b] \rightarrow {\mathbb R}}$ be a monotone non-decreasing function, and let ${\lambda > 0}$. Show that
$\displaystyle m( \{ x \in [a,b]: \overline{D^+} F(x) \geq \lambda \} ) \leq C\frac{F(b)-F(a)}{\lambda}$
for some absolute constant ${C>0}$. Here $\overline{D^+} F(x)$ denotes the upper right derivative ${\overline{D^+} F(x) := \limsup_{h \rightarrow 0^+} \frac{F(x+h)-F(x)}{h}}$.(Hint: the rising sun lemma is no longer available, but one can use either the Vitali-type covering lemma (which will give ${C=3}$) or the Besicovitch lemma (which will give ${C=2}$)).
Lemma $1$: Let $E \subset \mathbb{R}^d$ be a bounded set, then $E$ is Jordan measurable if and only if the topological boundary $\partial E$ has Jordan outer measure zero.
Lemma $2$: Let $f$ be continuous on the interval $[a,b]$, then if any of the four dini derivates is nonnegative on this interval $[a,b]$, then $f(b) > f(a)$.
Proof: By modifying ${\lambda}$ by an epsilon, and dropping the endpoints from ${[a,b]}$ as they have measure zero, it suffices to show that
$\displaystyle m( \{ x \in (a,b): \overline{D^+} F(x) > \lambda \} ) \leq 3\frac{F(b)-F(a)}{\lambda}$.
By inner regularity, it suffices to show that
$\displaystyle m( K )\leq 3\frac{F(b)-F(a)}{\lambda}$ whenever ${K}$ is a compact set that is contained in $\{ x \in (a,b): \overline{D^+} F(x) > \lambda \}$. By lemma $1$, $K - \partial K$ is open and $m(K - \partial K) = m(K)$. Since $F$ is monotone, it has at most countably many discontinuities, which is a null set. Hence by inner regularity again, it suffices to show that $m(J) \leq 3\frac{F(b)-F(a)}{\lambda}$ whenever $J$ is a compact set that is contained in the set $A = \{x \in K - \partial K: F \text{ is continuous at $x$} \}$. By openness and continuity, $\forall x \in A$, there exists some $\varepsilon_x > 0$ sufficiently small such that the interval $I_x := (x - \varepsilon_x, x + \varepsilon_x) \subset A$. By compactness of $J$, we can then cover $J$ by a finite number $I_1, I_2, \ldots ,I_n$ of such intervals. Moreover, on each of these intervals $I_k = (a_k, b_k)$, we have:
$\displaystyle {b_k - a_k \leq \frac{F(b_k)-F(a_k)}{\lambda}}$.
As can be seen by using lemma $2$ on the function ${G(x) := F(x) - \lambda x}$.
Applying the Vitali-type covering lemma, we can find a subcollection ${I'_1,\ldots,I'_m}$ of disjoint intervals such that
$\displaystyle m( \bigcup_{i=1}^n I_i ) \leq 3 \sum_{j=1}^m m(I'_j) \leq 3\sum_{j=1}^m \frac{F(b'_j)-F(a'_j)}{\lambda}.$
From telescoping series and the monotone nature of ${F}$ we have ${\sum_{j=1}^m F(b'_j)-F(a'_j) \leq F(b)-F(a)}$, and the claim follows.