Let $X$ be a random vector in $\mathbb R^d$ and $Y$ be a binary random variable on the same probability space, and with values in $\{\pm 1\}$. Suppose that for every $y \in \{\pm 1\}$: conditioned on the event $Y=y$, (the distribution of) $X$ has density.
Question. Is it true that the random vector $Z:=YZ$ has density ?
Rough guess
For any $y \in \{\pm 1\}$, let $\pi_y := \mathbb P(Y=y)$ and let $f_y:\mathbb R^d \to \mathbb R_+$ be the density of $X$ conditioned on the event $Y=y$. For any measurable $A \subseteq \mathbb R^d$, one has
$$ \begin{split} \mathbb P(Z \in A) &= \sum_y \pi_y\mathbb P(yX \in A \mid Y = y) = \sum_y \pi_y\mathbb P( X \in yA \mid Y = y)\\ & = \sum_y\pi_y\int_{y A}f_y(b)\mathrm{d}b = \sum_y \pi_y \int_{A}f_y(ya)\mathrm{d}a, \end{split} $$ where we have used the change of variable $b=ya$, which has Jacobian determinant $1$. Thus, it would seem that $Z$ has density given by the explicit formula $f(a) := \sum_y \pi_y f_y(ya)$.
If $E$ has Lebesgue measure $0$ in $\mathbb R^{d}$ then $$ \mathbb P(YZ\in E)=\mathbb P(Z\in E \mid Y=1)\mathbb P(Y=1)+\mathbb P(Z\in E \mid Y=-1)\mathbb P(Y=-1)=0 $$ since $-E$ also has Lebesgue measure $0$. Hence, $YZ$ has a density w.r.t. Lebesgue measure.