# Integration and area - why is integrating over a single point zero?

This is a naive question, but I'm only starting to learn calculus so please cut me some slack.

So we all know the integral from $a$ to $b$ of a function over an interval measures the area under the curve $(x, f(x))$ from $a, b$. For any point $c$ in this interval

$$\int_c^c f(x)dx = 0$$

So,

$$\int_a^b f(x)dx = \sum_{c \in [a, b]} \int_c^cf(x)dx = 0$$

But obviously this isn't true. So it cannot be the case that $\int_c^c f(x)dx = 0$. What is wrong with my reasoning?

• What do you think $\sum_{c \in [a,b]}$ means? – MartianInvader Jun 4 '13 at 1:02

A good question! The very important and interesting point is that integrals are only countably additive, so the identity you wrote would only work for a countable sum of "c"'s, but there are uncountably many in a (non-trivial) interval $[a,b]$.

The critical distinction between countable and uncountable operations in integration was only clarified around 1910 by Lebesgue, and subsequently by quite a few others.

In case the terms "countable" and "uncountable" are new to you: to say a set $S$ is "countable" means that there is a way (maybe not "constructive") to enumerate the elements of $S$, that is, to map from the positive integers onto $S$. Such maps can be a bit clever or complicated, as Cantor's map from the positive integers to the set of all rational numbers.

Further, it is not obvious that the points in a non-trivial interval $[a,b]$ are not countable. It was discovered by Cantor that this is the case. You might do a search on "Cantor's diagonal argument".

• If the question is about Riemann integrals, then the distinction is between finite additivity and infinite additivity, not countable and uncountable additivity, right? Someone just starting to learn calculus is probably not using Lebesgue integration – Trevor Wilson Jun 3 '13 at 22:26
• It doesn't have to be about Lebesgue integration versus Riemann integration. These are mere formalizations. Well, ok, if you want to play by restrictive rules... as is indeed unfortunately done in many calculus classes... many things are not allowed. I would not want to encourage anyone to declare countable sums off-limits just because the course number was too low... See what I mean? – paul garrett Jun 3 '13 at 22:31

It's worth examining how you got to your incorrect equation $$\int_a^b f(x) \, dx = \sum_{c \in [a,b]} \int_c^c f(x) \, dx.$$ There is a correct equation along these lines: $$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx, \qquad (1)$$ where $c \in [a,b]$ is any number (also, it doesn't have to be in $[a,b]$). You may have internalized this as "if you divide up an interval, then you divide up the corresponding integral as well". This is a good kind of motto to have but it is definitely a generalization of equation (1) rather than being exactly the same. All you can do with (1) is apply it repeatedly: $$\int_a^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^b f(x) \, dx = \int_a^c f(x) \, dx + \int_c^d f(x) \, dx + \int_d^b f(x) \, dx = \dots,$$ where each successive step adds one more point of division for the integral. Using induction you can prove that (1) works with finitely many terms, but normal induction does not allow you to proceed to the "infinite step", so you don't get anything like your proposed equation.

Now, if you wanted to go a little deeper into this question you might learn about transfinite induction and wonder why you can't induct over the ordinal $\omega$ to get to the "infinite step". In fact, you can try, but it involves a wholly different operation than simply "adding one more point", because $\omega$ is a limit ordinal rather than a successor, and has no immediate predecessor than which it is "one more". Consequently, the matching induction step will require some kind of limiting operation as well, and one fault of Riemann integrals is that they do not always interoperate with limits very well.

Here is an analogue of (1) for doing a countable subdivision:

Proposition: Let $c_1, c_2, \dots \in [a,b]$ be an increasing sequence of countably infinitely many points, and suppose that $f(x)$ is integrable on $[a,b]$. Then we have $$\int_a^b f(x) \, dx = \int_a^{c_1} f(x) \, dx + \sum_{n = 1}^\infty \int_{c_n}^{c_{n + 1}} f(x) \, dx + \int_{\sup c_n}^b f(x) \, dx.$$

Note that I have to introduce the supremum $\sum c_n$ into the equation simply to be able to write the formula, because the sequence of $c_n$'s does not have a maximal term but I need to start the last integral "right after" the end of the others.

To prove this you can use the following trick, common in real analysis: let $$\chi_{[s,t]}(x) = \begin{cases} 1 & x \in [s,t] \\ 0 & x \notin [s,t] \end{cases}$$ be the "indicator function" (or characteristic function, hence the letter) of an interval $[s,t]$. Then if, say, $[s,t] \subset [a,b]$ we have $$\int_s^t f(x) \, dx = \int_a^b f(x) \chi_{[s,t]}(x) \, dx$$ and $$\chi_{[a,c_1]}(x) + \sum_{n = 1}^m \chi_{[c_n,c_{n + 1}]}(x) = \chi_{[a,c_m]}.$$ The combination of these two is: \begin{align}\int_a^{c_m} f(x) \, dx &= \int_a^b f(x) \chi_{[a,c_m]}(x) \, dx \\ &= \int_a^b f(x) \left(\chi_{[a,c_1]}(x) + \sum_{n = 1}^m \chi_{[c_n,c_{n + 1}]}(x)\right) \, dx \\ &= \int_a^b f(x) \chi_{[a,c_1]}(x) \, dx + \sum_{n = 1}^m \int_a^b f(x) \chi_{[c_n,c_{n + 1}]}(x) \, dx \\ &= \int_a^{c_1} f(x) \, dx + \sum_{n = 1}^m \int_{c_n}^{c_{n + 1}} f(x) \, dx. \end{align} which is more or less the finitely-many-points version of equation (1). Passing from $m$ to $m + 1$ is "adding one more point". But how do you assemble all these results into a version with $m \to \infty$? Letting $m \to \infty$ requires the following manipulation: \begin{align}\int_a^b f(x) \, dx &= \int_a^b f(x) \chi_{[a,b]}(x) \, dx \\ &= \int_a^b f(x) \left(\chi_{[a,c_1]}(x) + \sum_{n = 1}^\infty \chi_{[c_n,c_{n + 1}]}(x) + \chi_{[\sup c_n,b]}(x) \right) \, dx \\ &\overset?= \int_a^b f(x) \chi_{[a,c_1]}(x) \, dx + \sum_{n = 1}^\infty \int_a^b f(x) \chi_{[c_n,c_{n + 1}]}(x) \, dx + \int_a^b f(x) \chi_{[\sup c_n,b]}(x) \, dx\\ &= \int_a^{c_1} f(x) \, dx + \sum_{n = 1}^\infty \int_{c_n}^{c_{n + 1}} f(x) \, dx + \int_{\sup c_n}^b f(x) \, dx. \end{align} The questionable step is: how do you break up an integral of a sum of infinitely many terms? Of course, we have $$\sum_{n = 1}^\infty = \lim_{m \to \infty} \sum_{n = 1}^m,$$ so the question is really "how do you exchange a limit with an integral?". In Riemann integration there is often no way; with the Lebesgue integral, a more sophisticated version, there are theorems such as the Dominated Convergence Theorem (which applies here) that will allow it.

However, your equation is even worse: it has uncountably many pieces. Even the Dominated Convergence Theorem will not help you there, since it applies only to countably many terms. And as you observed, if the equation worked then integration would not work, so this is truly something of a different character than a countable split. In sum, making the leap from finite sums to infinite sums is not as simple as simply replacing the index on the summation sign.

Your second line is incorrect...the sum you wrote is an uncountable one, and things don't work. It is certainly true that $\int_c^cf(x)\,dx=0$, and it probably agrees with your intuition, since the area of a line is $0$. It is true that $\int_a^bf(x)\,dx=\int_a^cf(x)\,dx+\int_c^bf(x)\,dx$ for $c\in(a,b)$, but not what you wrote.

• Well, the equality holds if you divide $[a, b]$ into intervals of equal length, right? So can't you shrink those intervals to a point? – user80838 Jun 3 '13 at 22:28
• @user80838 Your intervals can be points, but you can only subdivide into finitely many intervals (or countably many as Paul points out, if you want to get fancy.) So you cannot cover all of $[a,b]$ by such "degenerate" intervals and still expect the original integral to be the sum of all of these subdivided integrals. – Trevor Wilson Jun 3 '13 at 22:31