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.

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 $$B(x, r_x) \subset O \quad \text{ and } \quad \frac {\mu(B(x,r_{3x}))}{\lambda(B(x,r_{3x}))}< a .$$ By the Vitali covering lemma there is a disjoint subcollection $J$ with $$A \subset \cup_J B(x, r_{3x})$$ But then $$1 = \mu(A) \le \sum_J\mu(B(x,r_{3x})) \le a \sum_J \lambda(B(x,r_{3x} )) \le a \cdot 3 \cdot \epsilon $$ 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.

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 $B(x, r_x) \subset O$ 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 $. I've written this for 1 dimension but I think it applies mutatis mutandis to all.

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.