I am coming back to a well-known example: let $X$ follow $N(0,1)$ and $T$ follow a Rademacher distribution $(p(T=1)=p(T=-1)=1/2)$. Then, it can be easily demonstrated that $TX$ follows also $N(0,1)$.
My problem is about the demonstration that the random vector $(X,TX)$ is not gaussian.
Usually, the demonstration shows that the linear combination $X+TX$ doesn't follow a normal distribution (thus implying the desired result that $(X,TX)$ is not gaussian...).
This demonstration is usually explained by setting: $p(X+TX=0)=p((1+T)X=0)=p(1+T=0)=p(T=-1)=1/2$
which shows a concentration of probability in $0$ and contradicts the fact that $X+TX$, as normal, should be continuous.
This is perfectly clear... as far as the above demonstration is perfectly understood... But the problem is that I don't understand why the equality $p((1+T)X=0)=p(1+T=0)$ holds!
I know it must be obvious,but I'm just stuck on it.