In Cohn's measure theory, second edition, p166, there is written:
Let $\mu$ be a finite Borel measure on $\mathbb{R}^d$. Then the upper derivate $(\overline{D}\mu)(x)$ of $\mu$ at $x$ is defined by
$$(\overline{D}\mu)(x) = \lim_{\epsilon \to 0}\sup \left\{\frac{\mu(C)}{\lambda(C)}: C \in \mathcal{C}, x \in C, e(C) < \epsilon\right\}$$
where $\mathcal{C}$ is the set of all cubes in $\mathbb{R}^d$ and $e(C)$ denotes the length of a side of a cube.
Can someone explain what the notation $(\overline{D}\mu)(x) = \lim_{\epsilon \to 0}\sup \left\{\frac{\mu(C)}{\lambda(C)}: C \in \mathcal{C}, x \in C, e(C) < \epsilon\right\}$ means?
I am only familiar with $\limsup$ of sequences, for example $\limsup_{n \to \infty} x_n = \inf_{n} \sup_{k \geq n} x_k$
Are you sure they really mean the $\limsup$? To me, it looks more like one is supposed to take the supremum of the set and then pass to the limit, as it says $\lim_{\epsilon\to0}\sup$ and not $\limsup_{\epsilon\to0}$.
In very small steps: Look at the set $$ A_\epsilon:= \left\{\frac{\mu(C)}{\lambda(C)}: C \in \mathcal{C}, x \in C, e(C) < \epsilon\right\}, $$ take its supremum $$ S_\epsilon:= \sup A_\epsilon \in\mathbb [0,\infty],$$ and then pass to the limit $$ \lim_{\epsilon \to 0}S_\epsilon, $$ which has to exist in $[0,\infty]$, as by the definition of $A_\epsilon$, the quantity $S_\epsilon$ is decreasing for $\epsilon\to0$.