I wanted to compute the characteristic function of the distribution in question here:
How to multiply a standard normal RV times a uniform{-1.1} RV?
Let $X$ be standard $N(0,1)$, $Y$ be Uniform $[-1,1]$, consider $Z=XY$, then, for $t\neq 0$
$E[e^{itXY}] = E[E[e^{itXY}|Y]]= E\phi(itX)$
where $\phi$ is the uniform distribution's MGF, that is
$E[e^{itXY}] = E[\frac{e^{itX}-e^{-itX}}{2it} ]=0$
where is the mistake? This is extremely alarming, because the moment generating function is 0 except at the origin and this argument hold for all value of $\sigma$! surely they are not the same distribution!
I think that, in your second equality, you are stating that:
$$ \mathbb{E}\left[\mathrm e^{itXY}|Y \right] $$
is some function of $X$.
As you condition on the $\sigma$-algebra $\sigma(Y)$, this conditional expectation must be a measurable function of $Y$.
For example, in the convenient case where $X$ and $Y$ are independent, we have: $$ \mathbb{E}\left[\mathrm e^{itXY}|Y \right] = f(Y),$$ where $f(y) = \mathbb{E}\left[ \mathrm e^{itXy}\right] = \phi_X(ty)$. Under the same independence assumption, we also have: $$ \mathbb{E}\left[\mathrm e^{itXY}|X \right] = g(X),$$ where $g(x) = \mathbb{E}\left[ \mathrm e^{itxY}\right] = \phi_Y(tx)$.
Now, $$\phi_Y(s) = \frac{\mathrm e^{is} - \mathrm e^{-is}}{2is} .$$
And, as noted by Byron Schmuland (see comments),
$$g(X) = \phi_Y(tX) = \frac{\mathrm e^{itX} - \mathrm e^{-itX}}{2itX}, $$
not $$ \frac{\mathrm e^{itX} - \mathrm e^{-itX}}{2it}.$$