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Can the following theorem on intervals be proved by elementary means, without using the outer measure [1, Chap 2] ?

Theorem

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.

$\blacksquare$

I can see how to prove the following special case :

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 the outer measure necessary to solve this problem ?


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

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  • $\begingroup$ What do you mean by $R$ to $L$? $\endgroup$
    – algebroo
    Aug 16, 2023 at 23:19
  • $\begingroup$ @algebroo : Simply right to left order on the real number line, eg. the disjoint intervals $\left(\frac{1}{n+1}, \frac{1}{n}\right]$ for $n \geq 1$ are in R to L order. The disjoint intervals $\left[2 - \frac{1}{n}, 2 - \frac{1}{n + 1}\right)$ for $n \geq 1$ are in L to R order. $\endgroup$ Aug 17, 2023 at 11:39

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