P3571
By definition, we have $\mathfrak{F}_X (t) = \mathbb{E} (e^{i t X})$. Therefore
\begin{equation}
\mathfrak{F}_{X + \mu} (t)
= \mathbb{E} (e^{i t (X + \mu)})
= \mathbb{E} (e^{i t X} e^{i t \mu})
= e^{i t \mu} \mathbb{E} (e^{i t X})
\end{equation}
$\square$
