If ν is a signed measure and µ a positive measure, we say that
ν is absolutely continuous w.r.t. µ if µ(E) = 0 ⇒ ν(E) = 0.
If |ν| is a finite measure then this last condition is equivalent to the assertion that
for each ϵ > 0 there exists δ > 0 such that µ(E) < δ ⇒ |ν|(E) < ϵ.
while in general this is a strictly stronger assertion.
Give an example where |ν| is not finite, and the first assertion does not imply the second.
I think the problem is just about how you define $\infty$ times $\epsilon$? If the product is $\infty$ then we can not get the second(stronger) assertion.
But can someone give me an example? I am quite confused about signed measure right now. Thanks in advance~
Let $f\in L^1$ be a positive function, $\mu:=f\cdot\mathcal{L}$ ($\mathcal{L}$= Lebesgue measure on $\mathbb{R}$), $\nu:=\mathcal{L}$.
Clearly $\mu(E)=0$ implies $\nu(E)=0$, but $\mu([n,+\infty))\to 0$ as $n\to +\infty$, while $\nu([n,+\infty))\equiv +\infty$.