Integration of $e^{it}$ I am pretty sure that
$$\bigg|\int_{A}e^{it}\,dt\bigg|\leq2$$
for every measurable set $A\subseteq[-\pi,\pi]$,
but I cannot prove this...

We have
$$ I=\int_a^be^{it}dt=\frac{1}{i}(e^{ib}-e^{ia}) $$
Then
$$ |I|=|-i||e^{ib}-e^{ia}|=|e^{ib}||1-e^{i(b-a)}|\le |1|+|e^{i(b-a)}|\le 1+1=2$$
