Lemma:
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 $\lambda_N$"?