Sufficient conditions for continuous functions of continuous random variables to themselves be continuous random variables I've been trying to figure out nontrivial conditions for continuous functions of continuous random variables to themselves be continuous random variables without much success. Here's what I know so far:


*

*Continuous functions of continuous random variables are random variables, see this thread

*Continuous functions of continuous random variables needn't be continuous random variables in general, see this counterexample.

*Strictly monotonic functions of continuous random variables are continuous random variables


Are there more general conditions under which a continuous/smooth/analytic function of a continuous random variable is itself a continuous random variable?
Ultimately, what I am after is the following: if $\Omega$ is a continuous random variable with a bounded density function and $f$ is a continuous/smooth/analytic function, then what are some general conditions for the density function of $f(\Omega)$, if it exists, to be bounded?
Edit: As per @Malkin's comments, I want to clarify that by a continuous random variable, I mean a random variable that has a continuous cumulative distribution function (c.d.f.). I am also interested in the case when the c.d.f. is absolutely continuous, see the previous paragraph.
 A: Claim 1:
If $f$ is any function that is constant on some interval $I$ then there exists a continuous random variable $X$ such that $f(X)$ is not a continuous random variable.
Proof:
Suppose $f$ is constant on $I=[a,b]$ with $a \neq b$ and let $X \sim N(0,1)$. Put $$\varepsilon :=P(X \in I)>0$$ $$Y:=f(X)$$ $$F_Y(x)=P(Y \leq x)$$ $$x_0:=f(a)=f(b)$$
Then $\forall \, \delta>0$ we have:
$$
\begin{align}
\vert F_Y(x_0)-F_Y(x_0-\delta) \vert &= P \left( Y \in (x_0-\delta, x_0]\right)
\\ &= P \left( f(X) \in (x_0-\delta,x_0] \right)
\\ &\geq P \left( f(X) =x_0 \right)
\\ &\geq P \left( X \in I \right)
\\ &= \varepsilon
\end{align}
$$
Hence $F_Y$ is not continuous at $x_0$ and $f(X)$ is not a continuous random variable.
Claim 2:
If $f$ is any real analytic function that is not constant on any interval $I \subset \mathbb{R}$ then $f(X)$ is a continuous random variable for any continuous random variable $X$.
Proof:
Let $X$ be a continuous random variable with CDF $F_X$ and let $U\subset \mathbb{R}$ be the range of $f$. Define $Y := f(X)$ and let $F_Y$ be the CDF of $Y$, so that $F_Y$ has domain $U$. We will show that $F_Y$ is continuous.
Let $\varepsilon>0$ and $x_0 \in U$. 
By simple properties of random variables, $P(\vert X \vert > M) \rightarrow 0$ as $M \rightarrow \infty$. Pick $M$ such that $P(\vert X \vert > M) < \frac{\varepsilon}{2}$.
Now consider $S=f^{-1}(\{x_0\})$. Because $f$ is not constant on any interval, $S$ consists of countably many points: $S=\{s_i\}_{i \in J}$ for some $J \subset \mathbb{N}$. 
Define $S':=S \cap [-M,M]$. Suppose $S'$ contains infinitely many points. Then, since $S'$ is bounded, there exists a subsequence $(s_{i_n})_{n \in \mathbb{N}}$ such that $s_{i_n} \rightarrow c$ for some $c \in S'$. Since $f(s_{i_n})=x_0 \, \forall \, n$ by Rolle's Theorem we have a sequence $(r_n)_{n \in \mathbb{N}}$ with $s_{i_n} \leq r_n<s_{i_{n+1}}$ and $f'(r_n)=0 \, \forall \, n$. Also $s_{i_n} \rightarrow c \implies r_n \rightarrow c$. But by this reasoning, such a sequence $(r_n)$ cannot exist for an analytic function $f$. And so $S'$ must only contain finitely many points. Re-label them $S'=\{s'_i\}_{i=1}^N$.
$F_X$ continuous $\implies$ for each $s'_i \, \exists \, \delta_i>0$ s.t. $\vert F_X(x)-F_X(y) \vert < \frac{\varepsilon}{2N} \, \, \forall \, x,y \in (s'_i-\delta_i, s'_i+ \delta_i)$
Consider $f'(s'_i)$. Suppose $f'(s'_i)=0$. Since $f$ is not constant on any interval and since $f'$ is differentiable, $\exists \, \gamma_i>0$ s.t. $f$ is monotonic on $(s'_i,s'_i+\gamma_i)$ and monotonic on $(s'_i-\gamma_i,s'_i)$. If instead $f'(s'_i) \neq 0$ then again $\exists \, \gamma_i>0$ s.t. $f$ is monotonic on $(s'_i,s'_i+\gamma_i)$ and monotonic on $(s'_i-\gamma_i,s'_i)$. (See the answer here for a justification.) 
Define $k:=\frac{1}{2}\min\{\delta_i,\gamma_i \}_i$ and $t:=\frac{1}{2} \min\{\vert f(s'_i+k)-f(s'_i)\vert ,\vert f(s'_i-k)-f(s'_i)\vert \}_i$. 
Constructing $k$ and $t$ in this way gives us that $(s'_i-k, s'_i+ k) \subset (s'_i-\delta_i, s'_i+ \delta_i) \, \forall \, i$ ; that $f$ is monotonic on $(s'_i-k, s'_i) \, \forall \, i$ and separately on $(s'_i, s'_i+ k) \, \forall \, i $ ; and then that $(x_0-t,x_0+t]=(f(s'_i)-t,f(s'_i)+t] \subset f((s'_i-k,s'_i+k])\, \forall \, i$. These facts will be used in the working below.
Let $x \in (x_0-t,x_0 + t)$. Then:
$$
\begin{align}
\vert F_Y(x)-F_Y(x_0) \vert &\leq P \left( Y \in (x_0-t,x_0+t] \right)
\\ &= P \left( f(X) \in (x_0-t,x_0+t] \right)
\\ &= P \left( X \in f^{-1}((x_0-t,x_0+t]) \right)
\\ &\leq P \left( X \in f^{-1}((x_0-t,x_0+t]) \cap [-M,M] \right) + P(\vert X \vert > M)
\\ &= P \left( X \in f^{-1}((x_0-t,x_0+t]) \cap [-M,M] \right) + \frac{\varepsilon}{2}
\\ &\leq P \left( X \in \bigcup_i (s'_i-k,s'_i+k] \right) + \frac{\varepsilon}{2}
\\ &\leq \sum_i \vert F_X(s'_i+k)-F_X(s'_i-k) \vert + \frac{\varepsilon}{2}
\\ &\leq \sum_i \frac{\varepsilon}{2N}+ \frac{\varepsilon}{2}
\\ &= \frac{\varepsilon}{2} + \frac{\varepsilon}{2}
\\ &= \varepsilon
\end{align}
$$
Hence $F_Y$ is continuous.
We may conclude that $f(X)$ is a continuous random variable.
