Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In uniform distribution, a continuous distribution, for example where $A = -1$ and $B = 1$, $P(X < 0)$ is said to be the same as $P(X \le 0)$. Why is this?

share|cite|improve this question
Because the set $\{X = 0\}$ is a set of measure zero. Or easier $\int_0^0 f(x) \, dx = 0$ for any $f$. – Jonas Teuwen Sep 23 '11 at 10:56
Hmm, I still don't really get it. What do you mean by "Measure" zero? – Arvin Sep 23 '11 at 10:57
more generally, $\int_{a}^{b}f(x)dx=0$ if $a=b$ – sigma.z.1980 Sep 23 '11 at 11:03
up vote 6 down vote accepted

Generally, continuous distribution means $P(X=c)=0$ for every $c$. Thus in your case $P(X=0)=0$, and this is the difference between the two probabilities you mention.

Note, though, that continuous does not imply absolutely continuous, so the distribution may not be of the form $P(X \in A) = \int_A f(x) dx$ for any function $f$. The word continuous comes from writing the distribution as $P(X \in A) = \int_A \,dF(x)$ in Stieltjes form, and requiring $F$ to be continuous.

share|cite|improve this answer
+1: I supposed the absolute continuity so your answer more related to the question – Ilya Sep 23 '11 at 12:42

I don't know what you mean with $A,B$ but I suppose that with a continuous distribution you mean a distribution on the real line which has a density, i.e. $$ P(X\in [a,b]) = \int\limits_a^bf(x)\,dx. $$ Note that $$ P(X\leq 0) = P(X<0)+P(X=0) = P(X<0)+\int\limits_{0}^0f(x)dx $$ and the latter integral is zero as Jonas has written.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.