I encountered a problem in measure theory recently which asked that for any set $A \in B_{\mathbb{R}}$, if $\Phi(A)=0$ does that imply that $\lambda(A)=0$, where $\Phi$ is the standard normal law and $\lambda$ is the Lebesgue measure on $(\mathbb{R},B_{\mathbb{R}})$.
I found that all countably infinite sets $A$ have $\Phi(A)=0$ and also $\lambda(A)=0$. But is there a way to prove this or is there a counter example lurking somewhere?
Well $\Phi$ is absolutely continuous with respect to $\lambda$: $$ \Phi(dx) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2} \,\lambda(dx).$$ Since the density is positive, we can invert this $$\lambda(dx) = \sqrt{2\pi} e^{x^2/2}\, \Phi(dx),$$ which implies that $\lambda \ll \Phi$.