First, context: I'm doing a course on measure/integration theory following the book by Donald L. Cohn. In section 6.2, he defines the (upper and lower) derivates of a finite Borel measure $\mu$ on $\mathbb{R}^d$ as $$ (\overline{D}\mu)(x) = \lim_{\varepsilon\downarrow 0}\sup\left\{ \frac{\mu(C)}{\lambda(C)} : C\in\mathscr{C},x\in C, e(C) < \varepsilon \right\} $$ and $$ (\underline{D}\mu)(x) = \lim_{\varepsilon\downarrow 0}\inf\left\{ \frac{\mu(C)}{\lambda(C)} : C\in\mathscr{C},x\in C, e(C) < \varepsilon \right\}, $$ where $\mathscr{C}$ is the collection of all cubes in $\mathbb{R}^d$ (i.e. products of closed intervals of common length), $e(C)$ is the length of the edges of the cube $C$, and $\lambda$ is the Lebesgue measure.
In exercise 6.2.4, Cohn asks to confirm that certain assumptions in the following lemma are required:
Lemma 6.2.5. Let $\mu$ be a finite Borel measure on $\mathbb{R}^d$ such that $\mu \ll \lambda$, let $a$ be a positive real number, and let $A$ be a Borel set of $\mathbb{R}^d$ such that $(\underline{D}\mu)(x)\leq a$ holds at each $x\in A$. Then $\mu(A)\leq a\lambda(A)$.
In particular, the exercise asks to show that the requirement that $\mu\ll\lambda$ is indeed necessary, so there should be some $\mu$ such that $\mu\not\ll\lambda$, but $\underline{D}\mu$ is bounded on some set $A$ where the statement $\mu(A) \leq \text{constant}\cdot\lambda(A)$ fails.
I have an extremely meager attempt at a solution to this: I know that any "continuous" looking measure cannot work, since any "naturally" constructed one will be absolutely continuous w.r.t. the Lebesgue measure, and any discrete measure will have either zero derivative (where it does not produce a counter example), or it'll have infinite derivative, which will also not work. Therefore, one needs to use some very strange measure, probably singular. Therefore, I suspect that the Cantor measure works, i.e. the measure induced by integrating next to the Cantor function. However, I have no idea how to calculate the upper and lower derivatives of this at the points where it would be non-zero (i.e. points of the Cantor set), nor can I even really show that the derivatives ought to be bounded. So I am stuck.
Is this the correct approach, and if so, how should I proceed? If it isn't, what is?
It appears like there isn't a counterexample because the statement does seem to still hold even if you remove the absolute continuity requirement. I'll write up a self-contained exposition here but the details can be found in Functions of Bounded Variation and Free Discontinuity Problems by Ambrosio, Fusco and Pallara.
$\textbf{Definition:}$ Let $(X,d)$ be a locally compact, separable metric space. Let $\mu: \mathcal{B}(X) \to [0,\infty]$ be a measure. The set of all points $x \in X$ such that $\mu(U) > 0$ for every nbhd $U$ of $x$ is the support $\mu$, denoted by $\text{supp}(\mu)$. It can be shown that this is a closed set.
$\textbf{Definition:}$ Let $(X,d)$ be a locally compact, separable metric space. Let $\mu: \mathcal{B}(X) \to [0,\infty]$ be a measure. If $\mu$ is finite on compact sets, then it is called a Radon measure.
It can be shown that the Lebesgue measure is finite on compact sets. So, it is a Radon measure.
$\textbf{Definition:}$ Let $(X,d)$ be a locally compact, separable metric space. Let $\mu,\nu$ be Radon measures on the Borel $\sigma$-algebra on $X$. Let $x \in \text{supp}(\mu)$. Then, the upper and lower derivatives are defined by: $$D^+_{\mu} \nu(x) = \limsup_{\rho \downarrow 0} \frac{\nu(B_{\rho}(x))}{\mu(B_{\rho}(x))}$$ $$D^-_{\mu} \nu(x) = \liminf_{\rho \downarrow 0} \frac{\nu(B_{\rho}(x))}{\mu(B_{\rho}(x))}$$
Okay, so the difference here is that you've defined these in terms of cubes and so on. But because cubes in $\mathbb{R}^n$ can be approximated by balls, there's actually no real issue here. The three definitions above provide sufficient background for the following proposition.
$\textbf{Proposition:}$ Let $\mu,\nu$ be Radon measures on $\mathbb{R}^n$ and let $t \geq 0$ be a constant. For any Borel set $E \subseteq \text{supp}(\mu)$, the following hold:
If for all $x \in E$ we have that $D_{\mu}^- \nu(x) \leq t$, then $\nu(E) \leq t \mu(E)$.
If for all $x \in E$, we have that $D^+_{\mu} \nu(x) \geq t$, then $\nu(E) \geq t \mu(E)$.
For us, we are merely concerned with the first statement. In particular, let $\mu$ be the Lebesgue measure restricted to the Borel $\sigma$-algebra. Next, we note that $\text{supp}(\mu) = \mathbb{R}^n$. If $\nu$ is a finite Borel measure, then $\nu$ is obviously a Radon measure. Hence, your statement actually does seem to hold. In the book, the proof actually proceeds via the Vitali-Besicovitch Covering Theorem.