Let $(X, \mathcal{A}, \mu)$ be a measure space such that $\mu(X) = 1$. Let $(f_n)_{n}$ be a sequence of measurable functions $$f_n : (X, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R})) $$ such that for every $y \in \mathbb{R}$, $$\mu(f_n^{-1}((-\infty, y])) = \frac{\sqrt{n}}{\sqrt{2\pi}} \int_{-\infty}^y e^{-n\frac{x^2}{2}}dx, \forall n \in \mathbb{Z}_+. $$
How can we prove that there is some $c \in \mathbb{R}, c>0$ such that, for every $n \in \mathbb{Z}_+$ and for every $\varepsilon > 0$, we have that $$\mu (\{x \in \Omega \ \mid \ |f_n(x)| > \ \varepsilon \}) \leq c \cdot e^{-n\frac{\varepsilon^2}{2}} ?$$
I have tried writing that \begin{align*} \mu (\{x \in \Omega \ \mid \ |f_n(x)| > \ \varepsilon \}) &= \mu (\{x \in \Omega \ \mid \ f_n(x) > \ \varepsilon \}) + \mu (\{x \in \Omega \ \mid \ f_n(x) < \ -\varepsilon \}) = \\ &= 1 - \mu (\{x \in \Omega \ \mid \ f_n(x) \leq \ \varepsilon \}) + \mu (\{x \in \Omega \ \mid \ f_n(x) < -\ \varepsilon \}) = \\ &= 1 - \frac{\sqrt{n}}{\sqrt{2\pi}}\int_{-\varepsilon}^\varepsilon e^{-n\frac{x^2}{2}}dx, \end{align*} but I don't know how to proceed from here, because the function $x \mapsto e^{-n\frac{x^2}{2}}$ is decreasing, so I can't simply derive an upper bound from the above integral.
Let $g_n=\sqrt n f_n$. Then $\mu (g_n \leq y)=\frac 1 {\sqrt {2\pi}} \int_{-\infty} ^{y} e^{-x^{2}/2}dx$. The required inequality reduces to the fact that $\int_t^{\infty} \frac 1 {\sqrt {2\pi}} e^{-x^{2}/2}dx \leq ce^{-t^{2}/2}$ for some $c$ where $t=\sqrt n /\epsilon$. This is equivalent to saying that $e^{t^{2}/2} \int_t^{\infty} \frac 1 {\sqrt {2\pi}} e^{-x^{2}/2}dx$ is a bounded function. Bringing $e^{t^{2}/2}$ to the denominator and using L'Hopital's Rule we see that above expression tends to $0$ as $ t \to \infty$. By continuity it is bounded for bounded $t$. Hence the function is bounded and we can take $c$ as its supremum.