Characteristic function of a random variable with symmetric distribution is real 
Show characteristic function of a distribution is real iff the a probability distribution is symmetric.

I am trying to show the first direction, in which a symmetric distribution implies a real characteristic function. We have that
$$\varphi(t)=\int_{-\infty}^\infty e^{itx}d\mu(x) = \int_{-\infty}^\infty \cos(tx)  d\mu(x) + i\int_{-\infty}^\infty \sin(tx)  d\mu(x)$$
I need to show that
$$\int_{-\infty}^\infty \sin(tx)  d\mu(x) = 0$$
This would be trivial under the Lebesgue measure, but I need to do it for an arbitrary symmetirc measure over the line. I think one may be able to do this using a convergence theorem and a sequence of simple functions and the dominated convergence theorem, but is there a better way to do this?
Is there a way to show this without resorting to a discretization of the $\sin$ function?
 A: Let me just first address that the integral
$$\int_{-\infty}^\infty \sin(tx) \, d\mu(x)$$
does not exist in general, especially not in the case of the Lebesgue measure as you mentioned. This is because
$$\int_{-\infty}^\infty \sin(tx)^+\, d\mu(x)=\int_{-\infty}^\infty \sin(tx)^-\, d\mu(x)=\infty$$
In the case of a probability measure, however, the integral does exist. In any case, the theorem you mentioned at the beginning is still true and there are at least two ways of proving it, both of which are equivalent. The first way addresses your question most directly, namely that we will show that for a symmetric probability measure $\mu$,
$$\int_{-\infty}^\infty \sin(tx) \, d\mu(x)=0$$
Firstly, see that both
$$\int_{-\infty}^\infty \sin(tx)^+\, d\mu(x) \qquad \int_{-\infty}^\infty \sin(tx)^-\, d\mu(x)$$
are finite since $\sin(tx)^+, \sin(tx)^-$ are bounded functions and $\mu$ is a finite measure. Thus, $\sin(tx)$ is integrable and we can write
$$\begin{align*}
\int_{-\infty}^\infty \sin(tx) \, d\mu(x) 
&=\int_{-\infty}^0 \sin(tx) \, d\mu(x)+\int_0^\infty \sin(tx) \, d\mu(x) \\
&=-\int_{-\infty}^0 \sin(-tx) \, d\mu(x)+\int_0^\infty \sin(tx) \, d\mu(x) \\
&=-\int_{0}^\infty \sin(tx) \, d\nu(x)+\int_0^\infty \sin(tx) \, d\mu(x) \\
&=-\int_0^\infty \sin(tx) \, d\mu(x)+\int_0^\infty \sin(tx) \, d\mu(x) \\
&= 0
\end{align*}$$
where we have used linearity of the integral, $\sin$ is an odd function, the change of variable $x\mapsto -x$, $\nu$ is the measure defined by $\nu(A)=\mu(-A)$ and symmetry of $\mu$.
For the second method, we need the use the fact that complex conjugation and integrals can be interchanged. The rest uses the same ingredients
$$\begin{align*}
\overline{\varphi(t)} &= \overline{\int_{-\infty}^\infty e^{itx}\, d\mu(x)} \\
&= \int_{-\infty}^\infty \overline{e^{itx}}\, d\mu(x) \\
&= \int_{-\infty}^\infty e^{-itx}\, d\mu(x) \\
&= \int_{-\infty}^\infty e^{itx}\, d\nu(x) \\
&= \int_{-\infty}^\infty e^{itx}\, d\mu(x) \\
&= \varphi(t)
\end{align*}$$
