A continuous random variable I want to know whether the following statement is true. If $X$ is a continuous random variable and $f$ is 1-1 on the range of $X,$ then $f(X)$ is a continuous random variable. If it is true, could you tell me the way to prove it?
 A: The statement is false. For a counterexample, let $\Omega=[0,1]$, $\mathcal F=\mathcal B([0,1])$, $X(\omega)=\omega$, and $f(x) = x\chi_E -x\chi_{E^c}$ where $E\subset[0,1]$ is not (Borel)-measurable. Then $(0,1)$ is a measurable set but $f^{-1}(0,1)$ is not measurable.
In general, given a random variable $X$ and a real-valued function $f$, the composite function $f\circ X$ is a random variable if $f$ is measurable (although this is not a necessary condition, as @drhab pointed out in the comments). This follows from the measurability of the composition of measurable functions.
A: Under the extra condition that $f$ is measurable the answer is: "yes".
Let $A\subseteq\mathbb{R}$ denote the range of $X$.
It is to be shown that for each constant $c$ we have $\mathbb P\left(\left\{ f\left(X\right)=c\right\} \right)=0$.
Note that $\{f(X)=c\}=\{X\in f^{-1}(\{c\}\}=\{X\in A\cap f^{-1}(\{c\}\}$.
Assume that $\mathbb P\left(\left\{ f\left(X\right)=c\right\} \right)>0$
or equivalently that $\mathbb P\left(\left\{ X\in A\cap f^{-1}\left(\left\{ c\right\} \right)\right\} \right)>0$. 
Since $X$ is a continuous random variable this can only be true if
$A\cap f^{-1}\left(\left\{ c\right\} \right)$ contains more than
one element.
However, this contradicts that $f$ is one-to-one on $A$. 
