Solving interval problems without outer measure

Question

Is it possible to solve the following two problems on intervals using elementary methods, without using the outer measure ?

Problem 1

If $(I_n)$ is a disjoint sequence of subintervals of interval $I$ with $\bigcup_{n=1}^{\infty} I_n = I$, then $\sum_{n=1}^{\infty} l(I_n) = l(I)$, where $l$ denotes the interval length function.

I can see how to solve it where the $(I_n)$ are orderable :

Theorem

If $(I_n)$ is a disjoint sequence of subintervals of interval $I$ with $\bigcup_{n=1}^{\infty} I_n = I$, and if the $I_n$ can be rearranged into L to R order or into R to L order, then $\sum_{n=1}^{\infty} l(I_n) = l(I)$.

Proof

Consider the L to R case. Since the series has non-negative terms its convergence and sum are unaffected by rearrangements of its terms. So WLOG assume $(I_n)$ are in L to R order.

Case $I$ unbounded. Then $l(I) = \infty$. If $I_1$ unbounded then the result follows. Otherwise if $I_1$ is bounded then $I$ must have form $[a, \infty)$ or $(a, \infty)$, and LH of $I_1$ must be at $a$, and there must be no gaps between successive $I_n$, so that $l(I_1 \cup \cdots \cup I_n) = l(I_1) + \cdots + l(I_n)\ \forall\ n$. Given any $L > a, \exists\ N \mbox{ with } I_N \cap (L, \infty) \neq \emptyset$. Then $l(I_1 \cup \cdots \cup I_N) > L - a$, ie. $l(I_1) + \cdots + l(I_N) > L - a$. Thus partial sums are unbounded, and the result follows.

Case $I$ bounded. Consider the non-trivial case $l(I) > 0$. Then the $I_n$ are bounded also, and LH of $I_1$ must align with LH $a$ of $I$, and there must be no gaps between successive $I_n$, so that $l(I_1 \cup \cdots \cup I_n) = l(I_1) + \cdots + l(I_n)\ \forall\ n$. Because the $(I_n)$ are disjoint every partial sum is $\leq l(I)$. For any $\epsilon > 0, \exists\ N \mbox{ with } I_N \cap (a + l(I) - \epsilon, \infty) \neq \emptyset$. Then $l(I_1 \cup \cdots \cup I_N) > l(I) - \epsilon$, ie. $l(I_1) + \cdots + l(I_N) > l(I) - \epsilon$. Thus $\forall n \geq N, l(I) - \epsilon < l(I_1) + \cdots + l(I_n) \leq l(I)$. Thus the partial sums converge to $l(I)$.

The case of R to L follows similarly.

$\blacksquare$

However we cannot always rearrange a sequence of intervals $(I_n)$ into L to R or R to L order, eg. take $I = (0, 2)$ and form a sequence from the subintervals $\left(\frac{1}{n + 1}, \frac{1}{n}\right]$ for $n \geq 1$, the interval $(1, 1\frac{1}{2})$, and $\left[2 - \frac{1}{n}, 2 - \frac{1}{n + 1}\right)$ for $n \geq 2$ - this sequence then has no leftmost nor rightmost element.

Is there an elementary means of proving the general case, where no ordering can be assumed, or is outer measure necessary to solve this problem ?

Problem 2

Prove that if $\delta > 0$ and if $(E_n)$ is a sequence of sets in $[0, 1]$ with each $E_n$ containing a finite number of non-overlapping closed intervals with a total length $\geq \delta$, then there exists a point belonging to infinitely many of the $E_n$.

This result is readily proved using Axler [1, §2.60, p44] on the measure of a decreasing intersection, applied to outer measure on $\mathbb{R}$ on the Borel sets, since we require to prove that the set $E = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} E_m$ is non-empty, and if we denote by $F_n$ the union of the above intervals within $E_n$ (so $F_n$ is a Borel set), and let $F = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} F_m \subseteq E$, then we have from that theorem $|F| = \lim_{n \rightarrow \infty} |\bigcup_{m=n}^{\infty} F_m|$, where $\forall\ n,\ |\bigcup_{m=n}^{\infty} F_m| \geq |F_n| \geq \delta$ (noting $F_n$ equals a finite union of disjoint intervals with a total length $\geq \delta$), so that $|F| \geq \delta > 0$ and hence $F \neq \emptyset$ and thus $E \neq \emptyset$, as required.

Problem 2 is part of a proof of the Bounded Convergence Theorem for Riemann Integrals in Bartle [2, §22.14, p288], but Bartle omits proof, though the rest of the proof of the BCT is provided. In Bartle & Sherbert [3, §7.2.5, p304], the BCT for Riemann Integrals is stated but the proof is omitted entirely and described as 'quite delicate'. Bartle's book [2] covers quite a bit of real analysis but does not anywhere mention the concept of outer measure. It would be interesting to know if an elementary solution exists for Problem 2 or not. I asked about Problem 1 on MSE but did not receive any answers.

[1] Sheldon Axler (2020), Measure, Integration & Real Analysis, Springer Graduate Texts in Mathematics, https://measure.axler.net/.

[2] Robert G. Bartle (1964), The Elements of Real Analysis, John Wiley & Sons.

[3] Robert G. Bartle & Donald R. Sherbert (1982), Introduction to Real Analysis, John Wiley & Sons.

Answer 1

I give an elementary solution to Problem 1 in $\mathbb{R}^n$ in my book Measure Theory and Functional Analysis (Proposition 2.16, p. 48). Here is the one-dimensional version.

I guess it's clear that the length function defined on finite disjoint unions of intervals is finitely additive. Thus for each $N \in \mathbb{N}$ we have $$\sum_{n=1}^N l(I_n) = l\left(\bigcup_{n=1}^N I_n\right) \leq l(I),$$ and taking $N \to \infty$ yields $\sum l(I_n) \leq l(I)$.

For the reverse inequality, let $\epsilon > 0$. Let $K$ be a closed interval contained in $I$ whose length is at least $l(I) - \epsilon$ and for each $n$ let $J_n$ be an open interval that contains $I_n$ and whose length is at most $l(I_n) + \frac{\epsilon}{2^n}$. Then the $J_n$ cover $K$, so by compactness finitely many of them cover $K$. Thus $K$ is contained in $\bigcup_{n=1}^N J_n$ for some $n$. Defining $K_n = (K \cap J_n) \setminus (J_1 \cup \cdots \cup J_{n-1})$ then yields finitely many disjoint intervals $K_1$, $\ldots$, $K_N$ whose union is $K$. Thus $$l(I) \leq l(K) + \epsilon = \sum_{n=1}^N l(K_n) + \epsilon \leq \sum_{n=1}^N l(J_n) + \epsilon \leq \sum_{n=1}^N l(I_n) + 2\epsilon.$$ Taking $\epsilon \to 0$ yields $l(I) \leq \sum l(I_n)$.

Answer 2

Concerning problem 2. I do not know whether this "elementary", but at least formally it does not use the notion of outer measure or Lebesgue measure.

Replace the closed intervals to open intervals. Now the sets $A_m:=\cup_{n\geqslant m} E_m$ are open, have measure (=the sum of lengths of the inclusion-maximal intervals) $\mu(A_k)\geqslant \delta$, and they are nested: $A_1\supset A_2\supset A_3\ldots$. Let $B_k\subset A_k$ be a finite union of closed segments with measure at least $\mu(A_k)-\delta/10^k$. Then $B_1\cap B_2\ldots\cap B_n$ is a finite union of closed segments with measure at least $\mu(A_n)-\sum_{k=1}^n \delta/10^n\geqslant\delta-\sum_{k=1}^n \delta/10^n$: otherwise $A_n$ is covered by the intervals of total length less than $\mu(A_k)$ (the sets $B_1\cap\ldots \cap B_n$ and the sets $A_i\setminus B_i$ for $i=1,2,\ldots,n$) which contradicts to the statement of Problem 1 (for not disjoint intervals, but this is not very important) applied to each inclusion-maximal interval in $A_n$.

Therefore $B_1\cap\ldots\cap B_n$ is a non-empty compact set, and the intersection of such nested sets is non-empty.

Answer 3

Further to Nik Weaver's answer, which proves the case of $I$ bounded in Problem 1, the unbounded case is proved below - it follows as a corollary of the bounded case. I will update this answer later if I manage to find an elementary solution to Problem 2.

Theorem

If $(I_n)$ is a disjoint sequence of subintervals of unbounded interval $I$ with $\bigcup_{n=1}^{\infty} I_n = I$, then $\sum_{n=1}^{\infty} l(I_n) = \infty$, where $l$ denotes the interval length function.

Proof

The result is clear if there is an $I_n$ which is unbounded. Thus assume every $l(I_n) < \infty$. Consider the case where $I$ has form $[a, \infty)$. Take $L > a$. Then : \begin{eqnarray*} [a, L] & \subseteq & \bigcup_{n=1}^{\infty} I_n \\ \Rightarrow \hspace{1em} [a, L] & = & \bigcup_{n=1}^{\infty} [a, L] \cap I_n, \hspace{1.5em} \mbox{ a union of disjoint intervals } \\ \Rightarrow \hspace{1em} \sum_{n=1}^{\infty} l( [a, L] \cap I_n ) & = & L - a, \hspace{1.5em} \mbox{ by the bounded case } \\ \Rightarrow \hspace{1em} \sum_{n=1}^{\infty} l(I_n) & \geq & L - a, \hspace{1.5em} [a, L] \cap I_n \mbox{ being a subinterval of } I_n \\ \Rightarrow \hspace{1em} \sum_{n=1}^{\infty} l(I_n) & = & \infty, \hspace{1.5em} \mbox{ since $L > a$ is arbitrary. } \end{eqnarray*}

The cases of $(a, \infty), (-\infty, a]$, and $(-\infty, a)$ follow similarly. In the case of $I = (-\infty, \infty)$ we can choose any $a \in \mathbb{R}$ and proceed as with $[a, \infty)$.