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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?

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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
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2 Answers

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.

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

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