Q:

Area under a point

My question relates to continuous functions. I understand that the integral, over an interval, is the area under the curve over that interval. However, if you are calculating, say, the probability of a single point in a continuous probability distribution function, why is the answer zero? Doesn't the width of that single point, while infinitely small, have to be greater than zero? If not, it would lead me to believe that the sum of the areas of every single point on the interval would be zero. I understand that it's mathematically not possible to calculate the area under a single point, but why do we say it's zero?

A:

This is actually a very common misconception about what Riemann integration does. The idea that many people have is that we're adding up the signed areas of infinitely-many vertical line segments, but this is patently not the case. We can always add up a finite multi-set (meaning repeats are allowed) of real numbers, and in "nice enough" cases, we can "add up" a countably-infinite multi-set of real numbers. However, if we want to add up an uncountable multiset, then we have to have only countably-many of them be non-$0$, and that countable collection has to be "nice enough." Indeed, adding up uncountably-many $0$s does get us $0$, but that isn't what we're doing.
If a function $f$ is "nice enough" to be integrable on a real interval $[a,b]$ (which can be made precise, but I don't want to digress more than necessary), then we can proceed as follows. For each integer $n\ge0$ define a real number $S_n$ by: $$S_n=\sum_{k=1}^{2^n}f\left(a+\frac{2k-1}{2}\cdot\frac{b-a}{2^n}\right)\cdot\frac{b-a}{2^n}$$ Visually, what we're doing is dividing $[a,b]$ up into $2^n$ sub-intervals of equal width--namely, $\frac{b-a}{2^n}$--and adding up the signed areas of the rectangles with height given by the function values at the center points of the subintervals. (See the picture here labeled "Middle" for the idea.) We then define $$\int_a^bf(x)\,dx:=\lim_{n\to\infty}S_n.$$ At every stage, we were only adding up finitely-many signed areas of rectangles with positive width. At no point did we add an area of a vertical line segment.

