Does $\int_{\Gamma}\frac{x^2}{1+x^2}\,\Lambda_1(dx)=\int_{\Gamma}\frac{x^2}{1+x^2}\,\Lambda_2(dx)$ suffices to prove $\Lambda_1=\Lambda_2$? Suppose we have two measure $\Lambda_1$ and $\Lambda_2$ on $\mathcal{B}\left(\mathbb{R}\right)$.
They may be infinite measure but for all $\Gamma\in \mathcal{B}\left(\mathbb{R}\right)$, and $\Lambda_1\left(\left\{0\right\}\right)=\Lambda_2\left(\left\{0\right\}\right)=0$, we have
$$
\int_{\Gamma}\frac{x^2}{1+x^2}\,\Lambda_1(dx)=\int_{\Gamma}\frac{x^2}{1+x^2}\,\Lambda_2(dx)<\infty.
$$
Does it follows that $\Lambda_1=\Lambda_2$?
 A: Yes, it does.
First note that
$$ \infty>\int_{\mathbb{R}\setminus\left(\frac{-1}{n},\frac{1}{n}\right)}\frac{x^2}{1+x^2}\Lambda_i(dx)\ge\frac{\Lambda_i\left(\mathbb{R}\setminus\left(\frac{-1}{n},\frac{1}{n}\right)\right)}{n^2+1}\ ,
$$
so both $\ \Lambda_i\ $ must be $\sigma$-finite.
If $\ \Gamma\ $ is any Borel set with $\ \Lambda_2(\Gamma)=0\ $, then
\begin{align}
0&=\int_\Gamma\frac{x^2}{1+x^2}\Lambda_2(dx)\\
&=\int_\Gamma\frac{x^2}{1+x^2}\Lambda_1(dx)\\
&\ge\int_{\Gamma\setminus(-a,a)}\frac{x^2}{1+x^2}\Lambda_1(dx)\\
&\ge\frac{a^2\Lambda_1\big(\Gamma\setminus(-a,a)\big)}{1+a^2}\ ,
\end{align}
for any $ a>0\ $. Therefore $\ \Lambda_1\big(\Gamma\setminus(-a,a)\big)=0\ $ for any $\ a>0 $, and
\begin{align}
\Lambda_1(\Gamma)&=\Lambda_1\big(\Gamma\cap\{0\}\big)+\Lambda_1\left(\bigcup_{n=1}^\infty\Gamma\setminus\left(\frac{-1}{n},\frac{1}{n}\right)\right)\\
&=0
\end{align}
From Radon-Nikodym it follows that $\ \Lambda_1\ $ has a density $\ \varphi\ $ with respect to $\ \Lambda_2\ $:
$$
\Lambda_1(\Gamma)=\int_\Gamma\varphi(x)\Lambda_2(dx)
$$
for any $\ \Gamma\in\mathcal{B}(\mathbb{R})\ $.  Then
\begin{align}
\int_\Gamma\frac{x^2}{1+x^2}\Lambda_1(dx)&=\int_\Gamma\frac{x^2\varphi(x)}{1+x^2}\Lambda_2(dx)\\
&=\int_\Gamma\frac{x^2}{1+x^2}\Lambda_2(dx)
\end{align}
for any $\ \Gamma\in\mathcal{B}(\mathbb{R})\ $, from which it follows that $\ \varphi(x)=1\ $ for $\ \Lambda_2$-a.e. $\ x\ $. Therefore
\begin{align}
\Lambda_1(\Gamma)&=\int_\Gamma\Lambda_2(dx)\\
&=\Lambda_2(\Gamma)
\end{align}
for all $\ \Gamma\in\mathcal{B}(\mathbb{R})\ $.
