In my probability class, we had a proof of tightness (for Levy's theorem) and I'm confused by a step. So let $(\mu_n)_{n \geq 1}$ such that $\hat{\mu}_n(t) \to f(t)$ (where $\hat{\mu}$ is the characteristic function of $\mu$). We want to show that the $\mu_n$ are tight. Here is what the professor did: for some small $\alpha$:
\begin{align} \mu_n\left(\left[\frac{-2}{\alpha}, \frac{2}{\alpha}\right]^C\right) & = \int_{-\alpha}^{\alpha}{\mu_n(dx)} \\ & \leq 2\int_{\mathbb{R}}{1 - \frac{sin(\alpha x)}{\alpha x}\mu_n(dx)} \\ & = 2 - \int_{-\alpha}^{\alpha}{\int_{\mathbb{R}}{e^{itx}\mu_n(dx)}dt} \\ & = \frac{1}{\alpha}\int_{-\alpha}^{\alpha}{1 - \hat{\mu}_n(t)dt} \\ & \to 0 \end{align}
I don't understand how this got from the second line to the third line. The professor vaguely said "we are undoing the Fourier transform" but I'm not clear on how we got this. Also, I'm not clear on how we get form the third line to the fourth line either. Any help is appreciated!
$\int_{-\alpha}^{\alpha} e^{itx}dt=\frac {e^{ix\alpha}-e^{-ix\alpha}} {ix}=\frac {2i\sin x\alpha} {i x}=\frac {2\sin x\alpha} {x}$. So $\frac {\sin x\alpha} {\alpha x}=\frac 1 {2\alpha} \int_{-\alpha}^{\alpha} e^{itx}dt $.