$(X, Y)$ is not Bivariate Normal because $P(X + Y = 0) = P(S = −1) = 1/2$, which implies that $X + Y$ can’t be Normal

Question

My textbook, Introduction to Probability by Blitzstein and Hwang, gives the following example:

Example 7.5.2 (Non-example of MVN). Here is an example of two r.v.s whose marginal distributions are Normal but whose joint distribution is not Bivariate Normal. Let $X \sim \text{N}(0, 1)$, and let

$$S = \begin{cases} 1 \ \ \text{with probability} \ \dfrac{1}{2} \\ -1 \ \ \text{with probability} \ \dfrac{1}{2} \end{cases}$$

be a random sign independent of $X$. Then $Y = SX$ is a standard Normal r.v., due to the symmetry of the Normal distribution (see Exercise 32 from Chapter 5). However, $(X, Y)$ is not Bivariate Normal because $P(X + Y = 0) = P(S = −1) = 1/2$, which implies that $X + Y$ can’t be Normal (or, for that matter, have any continuous distribution). Since $X + Y$ is a linear combination of $X$ and $Y$ that is not Normally distributed, $(X, Y)$ is not Bivariate Normal.

I understood everything until the claim that what was shown implies that $X + Y$ can't be Normal (or have any continuous distribution). I would greatly appreciate it if someone could please take the time to clarify why this is the case.

Answer 1

If a random variable $Z$ is continuous, we have for every $c \in \mathbb{R}$ $$P(Z=c)=0.$$