I thought this was going to be straightforward but I've obviously been staring at it too long and got stuck!
I am considering a random variable $x \in \mathbb{R}$ drawn from a symmetric distribution $f(x)$ with mean $\mu$ and support in $(-\infty,+\infty)$. For some values of $x$, the inequality $x_0-\delta \leq x \leq x_0$, where $\delta>0$ is a real constant, will hold. I would like to derive the pdf for the random variable $y$ which satisfies this inequality.
I am thinking that the required density function should take the form $$ g(y) = \left\{ \begin{array}{lcl} 0 & , & y<x_0-\delta \\ ? & , & x_0-\delta \leq y \leq x_0 \\ 0 & , & y > x_0 \end{array} \right. $$ with $g(x_0-\delta)=\max_\mathbb{R} g(y)$ and $g(x_0)=0$, and clearly $\int_{-\infty}^{+\infty}g(y)\,\text{d}y=1$. But I am going round in circles figuring out what form $g(y)$ takes in $(x_0-\delta, x_0)$.
As a hint, notice that if $x \leq x_0 \leq x + \delta$, then $x_0 - \delta \leq x \leq x_0$.