What is an example of the probability distribution function that does not have a density function?

  • 3
    $\begingroup$ Any discrete probability distribution, such as the one that picks an integer between 1 and 10 (inclusive) with equal probability. $\endgroup$ Commented Jan 13, 2012 at 16:32
  • 1
    $\begingroup$ Thank you. I should have said the probability distribution on a continuum set. $\endgroup$
    – user12586
    Commented Jan 13, 2012 at 16:33
  • 1
    $\begingroup$ Thank you. I should have said: do we have an example with atomless distributions? $\endgroup$
    – user12586
    Commented Jan 13, 2012 at 16:39
  • 2
    $\begingroup$ See the "devil's staircase" as in this answer: math.stackexchange.com/questions/4683/… $\endgroup$
    – user940
    Commented Jan 13, 2012 at 16:41
  • 1
    $\begingroup$ @user12586 Yes! Weird isn't it? Actually "non-decreasing" is a bit more accurate than "increasing" since the function is virtually always flat. $\endgroup$
    – user940
    Commented Jan 13, 2012 at 16:46

2 Answers 2


About genericity (see the comments), note that every probability distribution $\mu$ on the Borel line may be written uniquely as a sum $\mu=\mu_a+\mu_d+\mu_s$ of measures such that $\mu_a$ is absolutely continuous with respect to Lebesgue measure, $\mu_d$ is discrete and $\mu_s$ is... well, the remaining part.

Thus, for every Borel set $B$, $\mu_a(B)=\displaystyle\int_Bf(x)\mathrm dx$ for some nonnegative integrable density $f$, $\mu_d(B)=\displaystyle\sum\limits_{n}p_n\cdot[x_n\in B]$ for some finite or infinite sequence $(x_n)_n$ of points of the real line and some sequence $(p_n)_n$ of nonnegative weights. The measure $\mu_a$ is called the densitable part of $\mu$. The measure $\mu_d$ is called the discrete part of $\mu$. The third measure $\mu_s$ is called the singular part of $\mu$ and is somewhat the most mysterious part since $\mu_s$ is atomless AND has no density.

The measures $\mu_a$, $\mu_d$ and $\mu_s$ are mutually singular, in the sense that there exists some disjoint Borel sets $A$, $D$ and $S$ such that $\mu_a(\mathbb R\setminus A)=\mu_d(\mathbb R\setminus D)=\mu_s(\mathbb R\setminus S)=0$. The set $D$ is always discrete, hence at most countable. The set $S$ might be a Cantor set with Lebesgue measure zero.

One sees that, in a sense, probability distribution functions with a density are the opposite of generic, since they correspond to measures $\mu$ such that $\mu_d=\mu_s=0$. And asking that $\mu=\mu_a$ is a bit like asking that a point $(x,y,z)$ in $\mathbb R^3$ is in fact located on the first coordinate axis $y=z=0$...

  • $\begingroup$ +1 Actually, my questions there were sparkled from your reply here: 1. Can singular continuous measures be generalized to a more general measure space than Lebesgue measure space R? 2. The purpose of knowing it is that to what extent the decomposition of a singular measure into a discrete measure and a singular continuous measure exist, wrt some reference measure? $\endgroup$
    – Tim
    Commented Jan 14, 2012 at 20:08
  • $\begingroup$ This is a wonderful answer. Where could one look for some examples of this decomposition $\mu=\mu_a+\mu_d+\mu_s$? I remember seeing that this decomposition always exists in a measure theory course, but at the time I did not have time to build intuition for what kinds of measures induce nonzero components $\mu_a, \mu_d, \mu_s$. $\endgroup$
    – kdbanman
    Commented Sep 17, 2020 at 19:00
  • $\begingroup$ For those wondering, this decomposition is called Lebesgue's decomposition theorem : en.wikipedia.org/wiki/Lebesgue%27s_decomposition_theorem $\endgroup$
    – Voidt
    Commented Apr 1, 2023 at 11:39

Take $f$ to be the Cantor function, then it has no density, but is continuous.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .