# Darboux's Integral vs. the “High School” Integral

The definition of the integral below is what I usually call the "High School definition," because that's usually where I've seen it in use.

Take a partition $\Delta = \{ x_0, x_1, x_2, \ldots, x_n\}$, where $$a = x_0 \leq x_1 \leq \cdots \leq x_{n-1} \leq x_n = b$$ of an interval $[a, b]$, such that $\Delta x_i = x_i - x_{i - 1}$ is constant, and define the norm of this partition $||\Delta||$ by $$||\Delta||=\frac{b-a}{n}.$$ Let $c_i$ be any point in $[x_{i-1}, x_i]$. If the following limit exists, then we call it the definite integral of a function $f$ on $[a, b]$. $$\lim_{||\Delta|| \to 0}\sum_{i=1}^nf(c_i)\Delta x_i = \int_{a}^bf(x)\ dx.$$

The integral often used in more advanced textbooks (that aren't advanced enough to into Riemann-Stieltjes integrals or Lebesgue integration), which I believe is called the "Darboux integral", I've seen defined roughly as follows.

Take a partition $\Delta$ of an interval $[a, b]$, in the same way as above, and define $$m_i=\inf\{f(x) : x_{i-1} \leq x \leq x_i \},$$ $$M_i = \sup\{ f(x) : x_{i - 1} \leq x \leq x_i \}.$$ Then, the lower and upper sums of $f$ for the partition $\Delta$ are defined as $$L(f, \Delta) = \sum_{i = 1}^n m_i(x_i - x_{i - 1}),$$ $$U(f, \Delta) = \sum_{i = 1}^n M_i(t_i - t_{i - 1}).$$

After the hassle of proving (I'm not going to type it up, as it's lengthy and irrelevant) that for any two partitions $\Delta_1, \Delta_2$, that we have $$L(f, \Delta_1) \leq U(f, \Delta_2),$$ it is stated that if we have $$\sup\{ L(f, \Delta) : \Delta\ \mathrm{is\ a\ partition\ of\ } [a, b] \} = \inf\{ U(f, \Delta) : \Delta\ \mathrm{is\ a\ partition\ of\ } [a, b] \}$$ then we define this common value to be $$\int_{a}^{b} f(x)\ dx.$$

It seems that the "High School definition" is significantly simpler than the Darboux integral, so my question is: why does the latter definition often appear in more advanced texts (e.g. Spivak's Calculus)? Are these two definitions equivalent, or does each have specific advantages over the other? I assume that the first definition is preferred in most lower-level classes because it avoids the need for $\sup$'s and $\inf$'s, but this still does not explain why the second definition is used in other texts (especially because it seems to take considerably more time to construct).

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Darboux integrals are equivalent to Riemann integrals, meaning that a function is Darboux-integrable if and only if it is Riemann-integrable, and the values of the two integrals, if they exist, are equal. link – Konrad Szałwiński Jan 22 '14 at 23:19
The description of the "high school" definition is someoff off kilter. There's nothing in it, for example, to prevent $x_0= x_1=\ldots= x_{n-1}=a$ and $x_n=b$. – Barry Cipra Jan 22 '14 at 23:19
@BarryCipra The "high school" definition does not need to demand that $\delta$ become small. Because the integral is the supremum of all lower sums, it is natural that the supremum is "reached" at the divisions with smaller $\delta$. – 5xum Jan 22 '14 at 23:22
@5xum, I don't understand your comment. There's no mention of suprema in the OP's description of the high school definition. His definition of "small" $\Delta$ just says there are lots of points in the partition. I think he should instead define the norm of $\Delta$ to be the max of the $\Delta x_i$'s. – Barry Cipra Jan 22 '14 at 23:32
I agree with Barry Cipra, if the distance between the abscissa is not forced to go to zero then the "high school" definition OP provided will not converge to the integral. There is no supremum mentioned in the definition given at this time. – Spencer Jan 23 '14 at 1:49

## 1 Answer

As you noted, Darboux's definition uses the infimum and the supremum throughout, which is not a concept one usually studies at highschool. Moreover, the definition using the mesh of partitions going to zero is somewhat messy when it comes to rigorous work, and it somewhat difficult to verify integrability using it versus Darboux's criterion. However, it is quite convincing for students to consider the integral as an approximation using sums of rectangles, so the "high school" definition may be more natural at first. It is sensible that in favour of good definitions, books such as Spivak's Calculus use Darboux's definition which gives a cleaner exposition.

One can look into more details that show why Darboux's definition is (maybe) better. For each $\delta >0$ one can define $U_\delta=\{P\text{ is a (tagged) partition of } [a,b]:\lVert P\rVert <\delta\}$. Observe that $\delta_1<\delta_2\implies U_{\delta_1}\subseteq U_{\delta_2}$, so every set is comparable with each other, and $\bigcap_{\delta >0}U_\delta=\varnothing$. In this sense, we say the collection $\{U_\delta:\delta >0\}$ defines a direction on the set of partitions of $[a,b]$. Observe how punctured balls $B'(x,\delta)$ define a direction in the set of nbhds of a point of a metric space, or how the tails $\{n,n+1,n+2,\ldots\}$ define a direction in the set of natural numbers.

With this in mind, we would say $I=\displaystyle\int\limits_a^b f$ if for each $\varepsilon >0$ there exists $U_\delta$ such that for any $P\in U_\delta$, $|R(f,P)-I|<\varepsilon$. Again, this is a bit problematic: we need to pick an arbitrary partition and an arbitrary set of tags and find a suitable candidate for $I$, and verify the inequality for any partition of this kind.

On the other hand, we know that $\sup_PL(f,P)$ and $\inf_P\{U(f,P)\}$ always exist, and we have a theorem which asserts that $L(f,P_n)\to \underline{\int_a^b f}$ for a sequence of partitions $P_1,P_2,\ldots$ provided only that $\lVert P_n\rVert \to 0$. This helps greatly in the evaluation of integrals, the same claim holding for $U(f,P_n)$. This is in fact the theorem one uses (maybe unknowingly) when evaluating the integral of say a continuous function by means of regular partitions and limits of Riemann sums.

Moreover, Darboux's idea translates to the (very useful) criterion of Riemann, namely, that for each $\varepsilon >0$ we can find two partitions $P,P'$ such that $$U(f,P)-L(f,P')<\varepsilon$$

This in turn translates to a more geometrical picture: that for each $\varepsilon >0$; we can find step functions $s,t$ with $s\leqslant f\leqslant t$ and $$\int_a^b t-\int_a^b s<\varepsilon$$

In fact, one can check that $$\overline{\int_a^b}f=\inf_{f\leqslant t}\int_a^b t$$ $$\underline{\int_a^b}f=\sup_{s\leqslant f}\int_a^b s$$

where $s,t$ are always step functions. This idea somewhat passes over to Lebesgue's integrals, which might also give Darboux's insight another point. Darboux's integral is indeed equivalent to Riemann's integral.

The morale is, maybe, it is always easier to work with suprema and infima rather than with any kind of limits. The former are usually much more well behaved and lead to simpler proofs, for example.

A comment on

"...especially because it seems to take considerably more time to construct."

I would say a reason why Darboux's definition is more useful is precisely because one has taken the time to carefully construct it!

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Great answer! This is exactly what I was looking for. – James Pirlman Jan 23 '14 at 15:59