nLab Cousin's theorem

Idea

Cousin's Theorem, attributed to Pierre Cousin, is a generalization of the Heine–Borel theorem (which Cousin also proved), and an important lemma to prove the nontriviality of the Henstock–Kurzweil integral.

Statements

The basic result (containing the hard part of the proof) can be distilled to the following:

Theorem

Let $a \leq b$ be real numbers, and let $\mathcal{U}$ be an open cover of the interval $[a, b]$. Then there exists a partition (in the sense of Riemann sum?s) $a = x_0 \leq \cdots \leq x_n = b$ such that each subinterval $[x_i, x_{i+1}]$ (for $0 \leq i \lt n$) is contained in some $U_i \in \mathcal{U}$.

We give a proof valid in both classical mathematics and intuitionistic mathematics. (See below for other forms of constructive mathematics.) This proof is based on one that appeared uncredited on Wikipedia.

Proof

Let $S$ be the set of real numbers $r \in [a, b]$ such that $[a, r]$ has a partition of the desired form; we will show that $S$ is all of $[a, b]$.

If $r \in S$, with a partition ending with $x_n = r$ and $[x_{n-1}, x_n]$ contained in some $U_{n-1} \in \mathcal{U}$, then we can (if $r \lt b$) increase $x_n$ slightly and keep $[x_{n-1}, x_n]$ within $U_{n-1}$, showing that $S$ is open on the right. Similarly, we can (if $r \gt a$) decrease $x_n$ slightly as long as $x_{n-1} \lt x_n$, to show that $S$ is open on the left; to deal with the possibility that $x_{n-1} = x_n$ and keep the argument intuitionistically valid, note that in any case $x_{n-1} \lt x_n$ or $x_{n-1} \gt a$, so we can either decrease $x_n$ as originally planned, or lower $n$ by $1$ and decrease the new $x_n$ (the old $x_{n-1}$) instead (which requires comparing with the old $x_{n-2}$ and so on, but since there are only finitely many subintervals, this process will eventually end). Therefore, $S$ is open.

Now suppose that $s \in [a, b]$ and $[a, s) \subseteq S$; we will show that $[a, s] \subseteq S$. So suppose $r \in [a, s]$, and let $U$ be a member of $\mathcal{U}$ such that $s \in U$; note that (even intuitionistically) $r \lt s$ or $r \in U$. If $r \lt s$, then we already have a partition of $[a, r]$; if $r \in U$, then we can (if $r \gt a$) take $t \lt r$ with $t \in U$, get a partition of $[a, t]$ and extend it with $[t, r]$ to get a $1$-step longer partition of $[a, r]$. To handle intuitionistically the possibility that $r = a$, let $V$ be a member of $\mathcal{U}$ such that $a \in V$, so that $a \lt r$ or $r \in V$; if $a \lt r$, then we can proceed as in the previous sentence, while if $r \in V$, then we can just use $[a, r]$ as a $1$-step partition.

Now we use the open induction principle?, a form of completeness of the real line that is both classically and intuitionistically valid, which states that if $S$ is an open subset of $[a, b]$, and $[a, s] \subseteq S$ whenever $[a, s) \subseteq S$, then $S = [a, b]$.

Sometimes it's desired that every subinterval have positive length, so that $x_i \lt x_{i+1}$ for $0 \leq i \lt n$. This is trivial classically; simply throw out any duplicates among the list of $x$s. Intuitionistically, we can't expect this to be possible in all situations; if $a \lt b$ then we obviously need $n \gt 0$, while if $a = b$ then we need $n = 0$ to have all subintervals of positive length, but we can't expect to be able to decide this. However, this is the only hang-up:

Corollary

If $a \lt b$, then there exists a partition in which each subinterval has positive length.

Proof

Form a partition as above; because $a \lt b$, we have $n \gt 0$. By performing a finite number of comparisons, we can verify $x_i \lt x_{i+1}$ for all $i \lt n$, or find some $i \lt n$ such that $x_i \in U_{i+1}$ (unless $i = n - 1$, since there is no $U_n$) and $x_{i+1} \in U_{i-1}$ (unless $i = 0$, since there is no $U_{-1}$). Since $n \gt 0$, we have $x_i \in U_{i+1}$ or $x_{i+1} \in U_{i-1}$ no matter what value $i$ takes, and this allows us to remove $x_{i+1}$ or $x_i$ (respectively) from the list of $x$s. Repeating this (for a maximum of $n - 1$ times), we eventually get a partition with only positive-length subintervals.

Cousin's Lemma

For purposes of integration theory, we want a form of Cousin's Theorem with tagged partitions; that is, each subinterval should be equipped with a point (its tag), generally required to belong to the subinterval. Since each subinterval is inhabited, it's easy enough to tag each one (even constructively since there are only finitely many), but Cousin's Lemma (so-called because it's a lemma in integration theory) gives us more: it lets us specify ahead of time which open neighbourhood of the tag its subinterval should be contained in.

We start with a relatively easy version fitted to the McShane integral (which is equivalent to the Lebesgue integral on the real line):

Theorem

Let $N$ be a function from $[a, b]$ to its powerset (or equivalently a binary relation on $[a, b]$) such that for each $t \in [a, b]$, $N(t)$ is a neighbourhood of $t$. Then there exists a freely tagged partition of $[a, b]$, that is a list $a = x_0 \leq \cdots \leq x_n = b$ (as before) together with a list $t_0, \ldots, t_{n-1} \in [a, b]$ (but without requiring $x_i \leq t_i \leq x_{i+1}$, which is why the partition is freely tagged), such that for each $i \lt n$, we have $[x_i, x_{i+1}] \subseteq N(t_i)$.

Proof

Let $\mathcal{U}$ consist of the interiors of the $N(t)$; then $\mathcal{U}$ is an open cover of $[a, b]$. Let $x_0, \ldots, x_n$ be a partition as given by Theorem . Each subinterval $[x_i, x_{i+1}]$ is contained in some member of $\mathcal{U}$, which in turn is contained in $N_t$ for some point $t$; pick one and call it $t_i$. (These choices are no problem constructively, since there are only finitely many to make.) Using these points as tags, we have a freely tagged partition as desired.

Using Theorem instead, we can guarantee that each subinterval has positive length if we wish (assuming that $a = b$ or $a \lt b$ to be constructive). It's common to start with a function $\delta\colon [a, b] \to {]0, \infty]}$ (a gauge) and let $N(t)$ be a ball around $t$ of radius $\delta(t)$. (Conversely, we can get $\delta$ from $N$ by taking a supremum, although constructively this means that $\delta$ is valued in the lower reals.)

For the Henstock–Kurzweil integral, we need a tagged partition that is not free, so that each tag belongs to the corresponding subinterval.

Theorem

As before, let $N$ be a function mapping each $t \in [a, b]$ to a neighbourhood $N(t)$. Then there exist $a = x_0 \leq t_0 \leq x_1 \leq \cdots \leq x_{n-1} \leq t_{n-1} \leq x_n = b$, such that for each $i \lt n$, we have $[x_i, x_{i+1}] \subseteq N(t_i)$.

Proof

…

We can again guarantee that each subinterval has positive length, or start with a gauge $\delta$, if we wish.

Constructive version

See at locale of real numbers for now.

References

• Wikipedia (English). Cousin's theorem. Accessed 2026-02-02.