Why does $\lim_{\lambda\to+\infty}\frac{\mathcal{L}\{v(x)\leq \lambda\}}{\mathcal{L}\{v(x)+|x|\leq \lambda\}}=1$? ($|x|^2\leq v(x)\leq 2|x|^2$) 
Let $v(x)$ be a  real-valued function and $\mathcal{L}$ be the Lebesgue measure on $\Bbb R$.
  Suppose that $|x|^2\leq v(x)\leq 2|x|^2$ for all $x\in\Bbb R$. How can we prove that
  $$
\lim\limits_{\lambda\to+\infty}\frac{\mathcal{L}\{x\in\Bbb R:v(x)\leq \lambda\}}{\mathcal{L}\{x\in\Bbb R:v(x)+|x|\leq \lambda\}}=1.
$$

We observe that $|x|^2\leq v(x)\leq 2|x|^2$. Then for any lager $\lambda>0$, we have
$$
2^{\frac 12}\lambda^{\frac 12}=\mathcal{L}\{x\in\Bbb R:2|x|^2\leq \lambda\}\leq\mathcal{L}\{x\in\Bbb R:v(x)\leq \lambda\}\leq\mathcal{L}\{x\in\Bbb R:|x|^2\leq \lambda\}=2\lambda^{\frac 12},
$$
so there exist $p_\lambda\in[2^{\frac 12},2]$ such that
$$
\mathcal{L}\{x\in\Bbb R:v(x)\leq \lambda\}=p_\lambda \lambda^{\frac 12}.
$$
Similarity, there exist $q_\lambda\in[2^{\frac 12},2]$ such that
$$
\mathcal{L}\{x\in\Bbb R:v(x)+|x|\leq \lambda\}=q_\lambda\lambda^{\frac 12}+O(1).
$$
Hence, it is equivalent to prove 
$$
\lim\limits_{\lambda\to\infty}\frac{p_\lambda}{q_\lambda}=1.
$$
 A: This is false. Fix a large $N$ and define $v(x)=N^2$ for $N/\sqrt{2}\le |x|\le N$. Note that this is consistent with the bounds that are imposed. On $|x|<N/\sqrt{2}$, we could set $v=2x^2$, and $v$ will be $>N^2$ for $|x|>N$. Now consider the quotient for $\lambda=N^2$. The set in the numerator is $[-N,N]$, but in the denominator we have at most $[-N/\sqrt{2},N/\sqrt{2}]$ in the set. Thus the quotient is $\ge \sqrt{2}$.
We can give $v$ this behavior for a sequence $N_k\to\infty$, so the quotient need not go to $1$.
A: Christian Remling is right. It is false. What you can prove is that
$$
\liminf_{\lambda\rightarrow\infty}\frac{\mathcal{L}(\{x\in\mathbb{R}
:\,v(x)\leq\lambda\})}{\mathcal{L}(\{x\in\mathbb{R}:v(x)+|x|\leq\lambda\})}=1.
$$
If $v(x)+|x|\leq\lambda$, then $v(x)\leq\lambda$ and so $\{x\in\mathbb{R}
:v(x)+|x|\leq\lambda\}\subset\{x\in\mathbb{R}:v(x)\leq\lambda\}$. It follows
that
$$
1\leq\frac{\mathcal{L}(\{x\in\mathbb{R}:\,v(x)\leq\lambda\})}{\mathcal{L}
(\{x\in\mathbb{R}:v(x)+|x|\leq\lambda\})}$$
and so
$$
\liminf_{\lambda\rightarrow\infty}\frac{\mathcal{L}(\{x\in\mathbb{R}
:\,v(x)\leq\lambda\})}{\mathcal{L}(\{x\in\mathbb{R}:v(x)+|x|\leq\lambda
\})}\geq1.
$$
On the other hand, if $v(x)\leq\lambda(1-\varepsilon)$, then $|x|^{2}
\leq\lambda(1-\varepsilon)$ and so
$$
v(x)+|x|\leq\lambda(1-\varepsilon)+\lambda^{1/2}(1-\varepsilon)^{1/2}%
=\lambda\left(  1-\varepsilon+\frac{(1-\varepsilon)^{1/2}}{\lambda^{1/2}%
}\right)  \leq\lambda
$$
provided $\lambda$ is large enough. Hence, $\{x\in\mathbb{R}:v(x)\leq
\lambda(1-\varepsilon)\}\subset\{x\in\mathbb{R}:v(x)+|x|\leq\lambda\}$ and so
$$
\mathcal{L}(\{x\in\mathbb{R}:v(x)\leq\lambda(1-\varepsilon)\})\leq
\mathcal{L}(\{x\in\mathbb{R}:v(x)+|x|\leq\lambda\}).
$$
Next let
$$
\ell=\limsup_{\lambda\rightarrow\infty}\frac{\mathcal{L}(\{x\in\mathbb{R}
:\,v(x)\leq\lambda\})}{\lambda^{1/2}}.
$$
Then
$$
\frac{\mathcal{L}(\{x\in\mathbb{R}:\,v(x)\leq\lambda\})}{\lambda^{1/2}}
\leq\ell+\varepsilon
$$
for all $\lambda$ large and
$$
\ell-\varepsilon\leq\frac{\mathcal{L}(\{x\in\mathbb{R}:\,v(x)\leq\lambda
_{n}\})}{\lambda_{n}^{1/2}}
$$
for a sequence $\lambda_{n}\rightarrow\infty$. Then
\begin{align*}
\mathcal{L}(\{x  & \in\mathbb{R}:v(x)+|x|\leq\lambda_{n}(1-\varepsilon
)^{-1}\})\geq\mathcal{L}(\{x\in\mathbb{R}:\,v(x)\leq\lambda_{n}\})\\
& \geq(\ell-\varepsilon)\lambda_{n}^{1/2}=\frac{\ell-\varepsilon}
{\ell+\varepsilon}(1-\varepsilon)^{1/2}(\ell+\varepsilon)\frac{\lambda
_{n}^{1/2}}{(1-\varepsilon)^{1/2}}\\
& \geq\frac{\ell-\varepsilon}{\ell+\varepsilon}(1-\varepsilon)^{1/2}
\mathcal{L}(\{x\in\mathbb{R}:\,v(x)\leq\lambda_{n}(1-\varepsilon)^{-1}\})
\end{align*}
and so
$$
\frac{\ell-\varepsilon}{\ell+\varepsilon}(1-\varepsilon)^{1/2}\frac
{\mathcal{L}(\{x\in\mathbb{R}:\,v(x)\leq\lambda_{n}(1-\varepsilon)^{-1}
\})}{\mathcal{L}(\{x\in\mathbb{R}:v(x)+|x|\leq\lambda_{n}(1-\varepsilon
)^{-1}\})}\leq1.
$$
In turn,
$$
\frac{\ell-\varepsilon}{\ell+\varepsilon}(1-\varepsilon)^{1/2}\liminf
_{\lambda\rightarrow\infty}\frac{\mathcal{L}(\{x\in\mathbb{R}:\,v(x)\leq
\lambda\})}{\mathcal{L}(\{x\in\mathbb{R}:v(x)+|x|\leq\lambda\})}\leq1.
$$
Letting $\varepsilon\rightarrow0$ gives
$$
\liminf_{\lambda\rightarrow\infty}\frac{\mathcal{L}(\{x\in\mathbb{R}
:\,v(x)\leq\lambda\})}{\mathcal{L}(\{x\in\mathbb{R}:v(x)+|x|\leq\lambda
\})}\leq1.
$$
I guess that if you assume that there exists 
$$
\ell=\lim_{\lambda\rightarrow\infty}\frac{\mathcal{L}(\{x\in\mathbb{R}
:\,v(x)\leq\lambda\})}{\lambda^{1/2}}.
$$
then you can prove that what you want.
