Consider a measure space $\left(S,\Sigma\right)$ where each state $s\in S$ can be expressed as $s=\left(x,c\right)$, where $x\in\mathbb R$ and $c\in\mathbb N$. E.g., suppose $s$ denotes the state of a particle, where $x$ is its position and $c$ is its colour (or some analogous discrete property).

Consider a probability measure $P$ on $\left(S,\Sigma\right)$. Is it safe to say that there exists a measure $\mu$ such that $P$ has density $p$ with respect to $\mu$? I'd say this measure is neither the Lebesgue nor the counting measure as the measure space is neither discrete nor continuous, but does it always exit?

  • $\begingroup$ My answer below explains the proper way of using the word "discrete" in this context (which differs from what your question seems to suggest) and says something about how "continuous" is to be understood as well. $\endgroup$ Dec 2, 2023 at 19:29
  • $\begingroup$ Can you just take $\mu = P$? Then $P$ has density identically $1$ with respect to $\mu$. $\endgroup$ Dec 2, 2023 at 20:22
  • $\begingroup$ I'd like to write a density with respect to a different measure, e.g. the product measure of the Lebesgue and counting measures. $\endgroup$ Dec 3, 2023 at 13:36
  • $\begingroup$ Well, then it depends on what "different measure" you have in mind. There's no universal choice. Indeed, given any $\sigma$-finite measure $\mu$ on an uncountable Polish space $S$, there exists a probability measure $P$ which is singular to $\mu$. $\endgroup$ Dec 4, 2023 at 7:59

2 Answers 2


A discrete probability distribution is one made up entirely of point masses; i.e. there is some set $S_1\subseteq S$ for which, for every $s\in S_1,$ the probability $P\{s\}$ assigned to the set $\{s\}$ is positive, and $\sum_{s\in S_1} P\{s\} = 1.$

A probability measure $P$ is a mixture – a weighted average – of a discrete probability distribution and a distribution with no discrete part precisely if $\sum_{s\in S_1} P\{s\}<1$ where $S_1 = \{ s\in S : P\{s\}>0\} \ne\varnothing. $

A probability distribution has no discrete part if for all $s\in S,$ $P\{s\}=0.$

In all of the above I avoided the word "continuous." More on that below.

There is nothing in your description that implies that a probability measure on your space has any discrete part.

If $X$ is a random variable taking values in the space you describe, certainly the $c$ component of $X$ – call it $c(X)$ – is a random variable whose probability distribution is discrete.

But that doesn't mean $X$ itself as a positive probability of being at any one point.

"Continuous" can mean having a density, which in many contexts is equivalent to being absolutely continuous with respect to some underlying measure. Having a density is always a sufficient condition for absolute continuity.

But sometimes "continuous" means having a continuous cumulative distribution function. That is a sufficient condition for having no discrete part, but does not entail that there is a density function.


Your notion of “has a density with respect to another measure” is essentially the notion of “absolute continuity”. Not all measures are absolutely continuous with respect to another, though one can often decompose a measure into an absolutely continuous part (which “admits a density”) and a part that does not, see Lebesgue’s decomposition theorem.

Trivially, any measure is absolutely continuous with respect to itself, but this is likely not what you’re interested in, and instead it sounds like you are interested if it is absolutely continuous with respect to something like the Lebesgue measure.


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