Convergence of measures

Consider the measurable space $(\mathbb R,\mathscr B_{\mathbb R})$. Suppose that $\{\mu_n\}_{n=1}^{\infty}$ is a sequence of finite Borel measures and $\mu$ is a finite Borel measure on $\mathbb R$. Suppose also that $$\lim_{n\to\infty}\mu_n(\mathbb R)=\mu(\mathbb R).$$

I am trying to prove the following:
$$\lim_{N\to\infty}\left\{\limsup_{n\to\infty}\left[\mu((-N,\infty))-\mu_n((-N,\infty))\right]\right\}=0.\quad(\spadesuit)$$


Since measures are continuous from below, it is clear that
\begin{align*}\lim_{N\to\infty}\mu((-N,\infty))=&\,\mu(\mathbb R),\\\lim_{N\to\infty}\mu_n((-N,\infty))=&\,\mu_n(\mathbb R)\quad\forall n\in\mathbb N,\end{align*}
but since the convergence in the second line may not be uniform across $\{\mu_n\}_{n=1}^{\infty}$, I am not sure how to proceed to prove $(\spadesuit)$ (provided that it is true in the first place). Note that it is the order of the lim and limsup in $(\spadesuit)$ that makes this problem tricky.


Any hint or counterexample would be appreciated.

Solutions Collecting From Web of "Convergence of measures"

Take $\mu = \delta_0$ and $\mu_n = \delta_{-n}$.
Then, for a fixed $N > 0$, and for any $n > N$,
$$
\mu((-N, \infty))

\mu_n((-N, \infty))
=
1 – 0 = 1.
$$
This means that, for any fixed $N > 0$,
$$
\limsup_{n \rightarrow \infty}
\left[
\mu((-N, \infty))

\mu_n((-N, \infty))
\right]
=
1.
$$

All you need is that $\mu_n$ “escapes” to $-\infty$.

Let $\mu_n$ be a dirac delta concentrated in $-n$ (or, if you prefere, you can choose $\mu_n$ as the characteristic function of $[-n-1,-n]$, it will work the same). And set $\mu=\mu_0$

Then $\mu(\mathbb R)=\mu_n(\mathbb R)=1$ for any $n$.

Also $\mu((-N,\infty))=1$.

On the other hand, for every fixed $N$, $\lim_{n\to\infty}\mu_n((-N,\infty))=0$

Thus $\lim_N(\lim_n(\mu((-N,\infty))-\mu_n((-N,\infty))))=0$