This appeared as a throwaway statement in a proof – that a strictly monotonic (increasing) transformation of a continuously distributed random variable (I am assuming that this means that the distribution function is continuous, not that the random variable is absolutely continuous) is also a continuously distributed random variable.
So the setup $X:\left(\Omega, \mathscr{F}, \mathbb{P}\right) \longmapsto \left(\mathbb{R}, \mathscr{B}(\mathbb{R}), \mathbb{P}_X\right)$ and $h: \mathbb{R} \longmapsto \mathbb{R}$ and $h(X) = Y$, where clearly $h$ needs to be measurable. So the claim is that if $h$ is monotonic, then $Y$ is a continuously distributed random variable.
A proof or a reference to a textbook would be appreciated.
Best Answer
You are interested in what you call continuously distributed random variables, more commonly called random variables with atomless distributions. These are the random variables $X$ such that $\mathrm P(X=x)=0$ for every $x$.
Assume the function $h$ is (strictly) increasing and the distribution of $X$ is atomless. Then $h$ is measurable hence $Y=h(X)$ is a random variable and the task is to show that $\mathrm P(Y=y)=0$ for every $y$.
Fix some $y$, then $[Y=y]=[h(X)=y]=[X\in B]$, where $B=h^{-1}(y)=\{x\mid h(x)=y\}$. Let us study the set $B$. If $x'$ and $x''$ are both in $B$, then $h(x')=y=h(x'')$ hence, due to the strict monotonicity of $h$, both cases $x'\lt x''$ and $x''\lt x'$ are impossible. Thus, $B=\varnothing$ or $B$ is a singleton. In both cases, $\mathrm P(X\in B)=0$, hence $\mathrm P(Y=y)=0$ and you are done.