I am trying to prove that if $\{\mu_n\}_{n=1}^\infty$ is a sequence of probability measures on $(\mathbb{R},\mathcal{B})$ such that $\mu_n$ has distribution $N(m_n,\sigma_n^2)$ for each $n \in \mathbb{N}$, then $\{\mu_n\}_{n=1}^\infty$ is tight if and only if $\{m_n\}_{n=1}^\infty$ and $\{\sigma_n\}_{n=1}^\infty$ are bounded. I know this has already been posted here, but I'd like to prove it with a different approach.
The part I'm struggling with is proving that if $\{m_n\}_{n=1}^\infty$ is not bounded, then $\{\mu_n\}_{n=1}^\infty$ is not tight, that is, there exists $\varepsilon > 0$ such that for every $a > 0$ there exists $n \in \mathbb{N}$ verifying $\mu_n([-a,a]) \leq 1-\varepsilon$.
As $\{m_n\}_{n=1}^\infty$ is not bounded, for every $a >0$ there exists $n \in \mathbb{N}$ such that $|m_n| > a$. Then
$$ \mu_n([-a,a]) = \int_{-a}^a \frac{1}{\sqrt{2\pi\sigma_n^2}}e^{-\frac{(t-m_n)^2}{2\sigma_n^2}} \, dt = \frac{1}{\sqrt{2\pi\sigma_n^2}}\int_{\frac{a-m_n}{\sigma_n}}^{-\frac{a+m_n}{\sigma_n}}e^{-\frac{s^2}{2}} \sigma_n ds = \frac{1}{\sqrt{2\pi}}\int_{\frac{a-m_n}{\sigma_n}}^{-\frac{a+m_n}{\sigma_n}}e^{-\frac{s^2}{2}} ds. \tag{$\ast$} $$
where in the second equality I've done the substitution $\frac{t-m_n}{\sigma_n} = s$, $dt = \sigma_n ds$ (this substitution transforms $[-a,a]$ into $[\frac{a-m_n}{\sigma_n}, -\frac{a+m_n}{\sigma_n}]$ because $|m_n| > a$ implies $a-m_n < 0$ and $a+m_n < 0$). If it could be proven something like
$$ \int_{\frac{a-m_n}{\sigma_n}}^{-\frac{a+m_n}{\sigma_n}}e^{-\frac{s^2}{2}} ds \leq \int_0^\infty e^{-\frac{s^2}{2}} ds = \sqrt{\frac{\pi}{2}}, $$
then we would have
$$ \mu_n([-a,a]) \leq \frac{1}{\sqrt{2\pi}}\sqrt{\frac{\pi}{2}} = \frac{1}{2}, $$
which would prove that $\{\mu_n\}_{n=1}^\infty$ is not tight. So the part in which I'm stuck is bounding the integral in $(\ast)$.
