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Let $\mu: \mathcal{B}(\mathbb{R}^N)\to [0,\infty)$ be a measure such that $\mu$ is absolutely continuous with respect to the Lebesgue measure $\lambda_N$, and let $A\in \mathcal{B}(\mathbb{R}^N)$ be such that the lower derivative $(\underline{D}\mu)(x)\leq r$ holds for all $x\in A$. Then $\mu(A)\leq r\lambda(A)$ holds.

Here $\mathcal{B}(\mathbb{R}^N)$ denotes the Borel sigma-algebra.

Why cannot we drop the condition "$\mu$ is absolutely continuous with respect to $\lambda_N$"?

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You can drop it: $\mu$ is a finite Borel-measure on a Polish space, so it is regular. Using the Vitali-Besicovitch covering theorem instead of the Vitali covering theorem the implication can be proved, see for instance Proposition 2.21 in Functions of Bounded Variation and Free Discontinuity Problems by Ambrosio, Fusco and Pallara.

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