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Let A be such that $\mu(A) = 1, \lambda(A) = 0$. There is an open set $O$ with $ A \subset O, \lambda(A) < \epsilon$ Suppose for convenience that $$\liminf\frac{\mu(B(x,r))}{\lambda(B(x,r))}< a $$ on $A$, otherwise replace $A$ with the set on which it is true. For each x in A pick $r_x$ so that $$B(x, r_x) \subset O$ \text{ and } $\frac {\mu(B(x,r_{3x}))}{\lambda(B(x,r_{3x}))}< a $$ . By the Vitale covering lemma there is a disjoint subcollection $J$ with $$A \subset \cup_J B(x, r_{3x})$$ But then $$1 = \mu(a) \le \sum_J\mu(B(x,r_{3x}) \le a \sum_J \lambda(B(x,r_{3x} )) \le a*3*\epsilon $$ and since $\epsilon$ is at your disposal, this is a contradiction. I've written this for 1 dimension but I think it applies mutatis mutandis to all.