# Limit/Infinitesimal Expression for Probability of a Continuous Random Variable at Any Single Value in Its Support

The following is a lecture slide from a machine learning class:

I already have basic understanding of probability, including continuous random variables. And I'm familiar with the typical explanations of why the probability of a continuous random variable at any single value in its support is "almost surely" $0$. However, I'd like further explanation of the limit expression used in the above slide. I've never encountered such an expression (with limits and $dx$), and I think it would increase my understanding by understanding a new way of thinking about the concept.

I would greatly appreciate it if people could please take the time to explain this (in a generalised way).

EDIT: I found a section in my textbook Introduction to Probability, by Blitzstein and Hwang, that indicates that the use of $dx$ in the slide is incorrect:

• The probability isn't almost surely $0$. It is $0$. The phrase "almost surely" appears in the context "The variable almost surely doesn't take this particular value", which is another way of saying that the probability of the variable taking that particular value is $0$. – joriki Jul 7 '18 at 13:11
• @joriki So is it used incorrectly in the accepted answer to this question, or did I just misstate it? math.stackexchange.com/questions/1259928/… – The Pointer Jul 7 '18 at 13:13
• I'm afraid it is, though I hesitate to say that, because Graham is usually very precise and the fact that he wrote it makes me wonder whether I might be wrong :-) But e.g. the Wikipedia article on "almost surely" consistently uses the term as I've described and not how it's used in that answer. – joriki Jul 7 '18 at 13:18
• @joriki Ahh, ok. Thank you for the clarification. :) – The Pointer Jul 7 '18 at 13:20
• @joriki - To me, the issue with the slide is that he declares it to be a density, and then (informally?) uses the form of a distribution when writing the limit - shouldn't it be something like an integral from 20.5-dx/2 to 20.5+dx/2? – John Polcari Jul 7 '18 at 14:04

For continuous RV’s, I separately define a probability density ${p_x}\left( x \right)$ and a distribution function ${P_x}\left( x \right)$, where $${p_x}\left( x \right) = \frac{{d{P_x}\left( x \right)}}{{dx}}$$ and $$P\left( x \right) = \int\limits_{ - \infty }^x {dy\,{p_x}\left( y \right)}$$ Now, the density has a nonzero value at any single value in the support. My definition of “probability” is $$Prob\left( {{x_L} \le x \le {x_U}} \right) = \int\limits_{{x_L}}^{{x_U}} {dy\,{p_x}\left( y \right)} = P\left( {{x_U}} \right) - P\left( {{x_L}} \right)$$ so that, almost surely $$Prob\left( {{x_B} \le x \le {x_B}} \right) = 0$$ I add the almost surely as a caveat against the possibility that the density includes a delta function, in which case (in my vernacular) the probability at that particular value is indeed nonzero.
Now the slide declares $P\left( X \right)$ to be a density with nonzero value 0.125 at $X = 20.5$. If that is the case, my rendering of the appropriate limit would be $$\mathop {\lim }\limits_{dx \to 0} \frac{{\int\limits_{20.5 - {{dx} \mathord{\left/ {\vphantom {{dx} 2}} \right.} 2}}^{20.5 + {{dx} \mathord{\left/ {\vphantom {{dx} 2}} \right.} 2}} {dYP\left( Y \right)} }}{{dx}} \approx \mathop {\lim }\limits_{dx \to 0} \frac{{P\left( {20.5} \right)dx}}{{dx}} = P\left( {20.5} \right) = 0.125$$ All in all, a bit of a tempest in a teapot, but it helps to keep the nomenclature precise.