Quick summary: I am stuck on the implication: $\phi_X$ real-valued $\rightarrow$ $X$ symmetric.
Assume you have a probability space $(\Omega, \mathcal{F},P)$, and a random varaiable $X: \Omega \rightarrow \mathbb{R^d}$, it is measurable with respect to $\mathcal{B}(\mathbb{R^d})$.
We define the characteristic function $\phi_X: \mathbb{R^d}\rightarrow\mathbb{C}$. With $\phi_X(u)=\int_\Omega e^{i<u,X(\omega)>}dP=\int_{\mathbb{R^d}}e^{i<u,x>}d\,u_x$, the measure $u_x$ is the distribution function:$u_x(B)=P(X^{-1}(B)), B \in \mathcal{B}(\mathbb{R^d}))$.
We say that X is symmetric iff $P(X\in B)=P(X\in-B)), B \in \mathcal{B}(\mathbb{R^d})$. Or equivalently $\mu_X(B)=\mu_X(-B)$.
I am supposed to show that X is symmetric if and only if the characteristic function of X: $\phi_X$ is real valued.
One part I seem to have managed: If we start with that X is symmetric, by then splitting $\mathbb{R^d}$ in all its hyperdimensional "octants" or "quadrants" and $\{0\}$, the result follows(but there is some work involved, but I think it is ok this way).
The problem is the other way: If I start with the knowledge that the characteristic function is real valued, how should I then work to show that X is symmetric? I tried assumeing that there is one set B such that $P(X \in B)\ne P(X \in -B)$, and then try to arrive at a contradiction:
If I split $\mathbb{R}^d$ in: $\{0\}, L_1=(+,+,+..,+), L_2=(-,-,-,,-), L_3=(+,+,+,..,-)L_4=(-,-,-,..,+)$ etc.
So that I get $\{0\}$ and $2^d$ $L_k$s. Where $L_{2i-1}=-L_{2i}$. Then I must have that $P(X \in B)\ne P(X \in -B)$, implies that there is atleast one i, such that $P(X \in B\cap L_{2i-1})\ne P(X \in -B\cap L_{2i})$ or $P(X \in B\cap L_{2i})\ne P(X \in -B\cap L_{2i-1})$, because if all these were equal, and since they are a partition of $\mathbb{R}^d$(with $\{0\}$) we would end up with: $P(X \in B)= P(X \in -B)$.
But this is as far as I got, beacuse even if I have that: $P(X \in B\cap L_{2i-1})\ne P(X \in -B\cap L_{2i})$ I can't see how this would contradict the characteristic function beeing real valued. My only idea is to chose one u, such that we get a complex number when we integrate, my only idea is choosing u for instance 1 in those parts that is a + in $L_{2_i-1}$, but I can't really see this leading anywere, since 0 in a exponential funciton, doesnt lead to a 0 in the function itself.
Any tips or hints or help on how to solve this?