Q: Is the boundary of an HDR a region of the sample space with equal density value? After reading this question, I read in the reference provided (Hyndman, 1996, The American Statistician) the following:

It follows inmediately from the definition that the boundary of an HDR consists of those values of the sample space with equal density.

How does it follow from the definition ?

Let $f(x)$ be the density function of a random variable $X$. Then the
  $100(1-\alpha)\%$ HDR is the subset $R(f_\alpha)$ of the sample space
  of $X$ such that 
  $$R(f_\alpha) = \{x\colon f(x)\geq f_\alpha\},$$
  where $f_\alpha$ is the largest constant such that
  $$P\big(X\in R(f_\alpha)\big)\geq 1-\alpha.$$


A: It doesn't, at least not in general. This only holds for continuous densities.
If the density $f$ is continuous, then the boundary of an HDR 
$$R(f_\alpha) = \{x\colon f(x)\geq f_\alpha\}$$
(which is the pre-image of the half-open interval $[f_\alpha,\infty[$ under $f$) is given by $\{x\colon f(x)=f_\alpha\}$. 
For a counterexample with a non-continuous density, define
$$ f(x) := \begin{cases}
0, & x<0 \\
\tau, & x=0 \\
\frac{1}{(x+1)^2}, & x>0
\end{cases}$$
This integrates to 1, so it is a density. Every HDR will be of the form $[0,x_\alpha]$, where $x_\alpha>0$ depends on $\alpha$, and obviously, we will typically not have $\tau=f(0)=f(x_\alpha)$.
Alternatively, use any discrete distribution, where this statement will usually also not hold.
