I am confused about one trivial thing.
We have independent variables $X,Y,\beta$ where $X$ has a distribution function $F(x)$, $Y$ has $G(x)$ and $\beta$ is $Bin(1,p)$ distributed. Also $X,Y$ have densities $f(x)$ and $g(x)$ correspondingly. Clear enough $Z=\beta X+(1-\beta)Y$ has a distribution function $H(x)=pF(x)+(1-p)G(x)$. What is the density of $Z$?
If $f(x)$ and $g(x)$ are continuous, then clearly $h(x)=p*f(x)+(1-p)*g(x)$ is the density of $Z$ because $H'(x)=h(x)$ and by the fundamental theorem of calculus $\int_{-\infty}^{x}h(t)dt=H(x)$ for any $x$, which proves that $h(x)$ is a distribution function for $Z$.
But if $g(x)$ or $f(x)$ is discontinuous at $x^*$ then $H'(x^*)=h(x^*)$ is not necessarily true (not clear whether $H(x)$ is at all differentiable at $x^*$). So $H(x)$ is not an antiderivative for $h(x)$ on $R$. How then do I show that $\int_{-\infty}^{x}h(t)dt=H(x)$ for any $x$, i.e. $h(x)$ is a density function for $Z$?