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Recalling from my days in Calc BC that if we have some integrable function $\ f\colon \mathbb R \to \mathbb R, x \mapsto f(x)$, three numbers $\ a,b,c \in \mathbb R,a \lt b \lt c$ and we know $\int_a^c f(x)\, dx=A$, then it follows that $\int_a^b f(x)\, dx +\int_b^c f(x)\, dx =A$, too. This always bothered me, because it seems like we're double counting at b.

To take the above property to its extreme, I wondered what would happen if we start dividing up the interval $[a,c]$ into smaller and smaller chunks and then summing up the integrals over said smaller and smaller chunks. In mathematical notation, I believe the above would be expressable as follows – define some $N \in \mathbb N$ and let $c-a=N\epsilon$, and take $$\displaystyle\lim_{N \to \infty} \displaystyle \sum_{j=0}^{N-1} \int_{a+j\epsilon}^{a+(j+1)\epsilon} f(x)\, dx.$$

My question is, can this above expression ever $\ne A$? If not, why does double counting a point on an integral an infinite number of times not affect its net value? As a follow up, what could happen if we let the various limits of integration overlap on some small interval larger than a point?

Thanks!

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    $\begingroup$ Intuition: points have zero width. So "double-counting" a single point won't change your integral -- an integral is only sensitive to changes over intervals with positive width. 1/2 $\endgroup$
    – Neal
    Apr 25, 2018 at 2:24
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    $\begingroup$ If you recall the definition of the integral of an integral function, and recall criteria allowing one to interchange two limits, I think you'll be able to show that the limit expression is equal to $A$ for any function $f$ integrable on $[a,b]$. 2/2 $\endgroup$
    – Neal
    Apr 25, 2018 at 2:25

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One possible full answer to this question is to read a book on measure theoy and get a grip on the construction of the Lebesgue measure and the Lebesgue integral.

On a very informal level, the Lebesgue measure (on the real line) is an attemt to assign a length to as many subsets of the real line as possible. It starts from the basic assumption that an open interval $]a,b[$ should have length $b-a$. The further construction to measure the length of more complicated set is not so straightforward and I skip this. If you assume that the length of the set should be somehow continuous then what should be the length of the singleton $\{a\}$? Well $\bigcap_{\epsilon>0} ]a-\epsilon,a+\epsilon[\ = \{a\}$ and thus, it should be $\lim_{\epsilon\to 0} 2\epsilon = 0$. Another important thing in measure theory is that the measure is (should be) countably additive, i.e., if you take the union of countable many disjoint sets, you can add up (i.e. take the limit of the series) the measures. (Note that this would not be a good property without "countably", as every set is the union of its elements, so union of set of length zero…)

Now, in your question, you count a countable number of points twice, so (still intuitively) you count a set of measure zero twice.

The Lebesgue integral is in fact quite similar to the Riemann integral - its main difference is that you divide the range of the function (and not the domain) which gives you an adaptive partition of the domain of the function. To be more precise, you approximate the integral of a positive function by discretizing the range by $0=y_0 < y_1 <\cdots y_n$ and taking $$ \sum_i y_i\lambda(\{x\ :\ y_{i-1}<f(x)\leq y_i\}) $$ where $\lambda(A)$ is the (Lebesgue) measure of $A$. (Draw a sketch to convice yourself that this is indeed an approximation of the area under the curve.) An important consequence is, that changing a function on a set of measure zero, does not change its integral. So in your example, the (Lebesgue) integral of $f$ stays exactly the same if you set $$ \tilde f(x) = \begin{cases} 0, & x = a+j\epsilon,\ j=0,1,\dots\\ f(x), & \text{else}\end{cases}. $$ This new function $\tilde f$, has the same integral (area under the curve) but the values that you count twice are all zero…

A very short, intuitive answer: A point has zero length; coutalbly many points also have "zero total length", so counting them twice does not matter.

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    $\begingroup$ This is terrific - thanks for the measure theory primer! $\endgroup$
    – yungblud
    Apr 25, 2018 at 12:57

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