Given the following proposition :
Let $P$ be a probability and $\mu$ a $\sigma$ finite measure. Then $P \ll \mu \iff \forall \hspace{0.1cm} \varepsilon > 0 \hspace{0.1cm} \exists\hspace{0.1cm} \delta > 0 : \forall A$ with $\mu(A) < \delta$ we have $P(A) \leq \epsilon$
I do understand the proof but I'd like to see the necessity of $P$ being a probability, so I'd like to prove that the result doesn't hold if we take a $\sigma-$finite measure instead of $P$, i.e $P \not\ll \mu$.
Here I'm taking $P = \lambda$ the lebesgue measure on $\mathbb{R}$ and $\mu = f.\lambda$ where $f(x) = \frac{1}{1+x^2}$.
In this way we have $\lambda(A) = \int_{A} 1+x^2 \frac{1}{1+x^2}d\lambda(x)$, i.e $\lambda \ll \mu$.
The idea is that $\mu(]x_1,x_2[) = \int_{x_1}^{x_2} \frac{1}{1+x^2}dx = \text{arctg}(x_2)-\text{arctg}(x_1)$ and $\lim\limits_{x \to +\infty}\text{arctg}(x) = \frac{\pi}{2}$.
This is where I got stuck, trying to prove that $\exists \hspace{0.1cm} \varepsilon > 0 \hspace{0.1cm} \forall\hspace{0.1cm} \delta > 0$ we have $\mu(]x_{1},x_{2}[) < \delta$ but $\lambda(]x_1,x_2[) \geq \varepsilon$.
Any help or hint would be appreciated.
What about taking the sets $A_n=[n,\infty)$ instead. Since $\frac{1}{1+x^2}$ is integrable then, $\mu(A_n)\to 0$ as $n\to \infty$. But $\lambda(A_n)=\infty$. Hence, for example if you pick $\epsilon_0=1$. Then, for every $\delta>0$ you can find some $n_\delta$ with $\mu(A_{n_{\delta}})<\delta$ but $\lambda(A_{n_\delta})\geq 1$.
You can also take the sets $[n,n+1]$. Then again $\mu([n,n+1])=\arctan(n+1)-\arctan(n)\to 0$. But $\lambda([n,n+1])=1$, so again for $\epsilon_0=1$ you can get what you want.