Let $X\in\mathbb{R}^n$ be a random vector with continuous coordinates such that $\|X\|_2=1$ a.s. Define the random matrix $A=XX^T$. It is obvious that $A$ has rank 1 and its unique non-zero eigenvalue is 1. Therefore, $A$ is not positive definite since it has $n-1$ eigenvalues equal to zero.
On the other hand, for any $v\in\mathbb{R}^n$ such that $v\neq 0$, we have that $Y_v:=v^TAv=v^TXX^Tv=(v^TX)^2$, and $Y$ has to be a continuous random variable since $X$ has continuous coordinates. Therefore, $P(Y_v=0)=0$ and $P(Y_v>0)=1$, and this happens for any non-zero $v$. This is saying that $A$ is positive definite with probability one!!
What is wrong with this reasoning? Any help will be appreciated.
(I think the problem is with $P(Y_v>0)\neq P(Y_v>0\text{ for all }v)$. )
Positive Definite can be defined as $v^TAv >0 $ for all $v\neq 0$. So as you mentioned, to find the probability of $XX^T$ being positive definite we need to calculate $P(v^TXX^Tv >0 \text{ for all } v\neq 0)$. This probability is zero. So $XX^T$ is almost surely not positive definite.
If we fix $v$ then compute $P(v^TXX^Tv>0)$ we will find that the probability is $1$. This however, is not the probability we are interested in.