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I have proved the following statement and I would like to know if I have made any mistakes:

$I_1,I_2,\dots$ disjoint sequence of open intervals $\Rightarrow |\bigcup_{k=1}^{\infty} I_k|=\sum_{k=1}^{\infty}\ell(I_k)$

($|\cdot|$ denotes outer measure)

My proof:

By definition of outer measure we know that $\sum_{n=1}^{\infty}\ell(I_k)\geq |\bigcup_{k=1}^{\infty}I_k|$. Now, since $\bigcup_{k=1}^{N}I_k\subset \bigcup_{k=1}^{\infty}I_k$ for $N\geq 1$, from the fact that outer measure preserves order we have that $|\bigcup_{k=1}^{N}I_k|\leq |\bigcup_{k=1}^{\infty}I_k|$ for $N\geq 1$ which implies that the following chain of inequalities holds: $$\sum_{n=1}^{\infty}\ell(I_k)\geq |\bigcup_{k=1}^{\infty}I_k|\geq |\bigcup_{k=1}^{N}I_k|$$ So, if we prove that $|\bigcup_{k=1}^{N} I_k|=\sum_{k=1}^{N} \ell(I_k)$, by taking the limit we would have the desired claim.

We proceed by induction on $N$.

The base case $N=1$ follows from the fact that $|I|=\ell(I)$ if $I=(a,b)$, $a,b\in\mathbb{R}, a<b$.

Suppose now that the claim is valid for $N\geq 1$: we prove it for $N+1$.

First, $|\bigcup_{k=1}^{N+1}I_k|=|\bigcup_{k=1}^{N}I_k \cup I_{N+1}|\leq |\bigcup_{k=1}^{N}I_k|+|I_{N+1}|\overset{\text{ind.hyp.+def.length open interval}}{=}\sum_{k=1}^{N}\ell(I_k)+\ell(I_{N+1})=\sum_{k=1}^{N+1}\ell(I_k)$.

Now, it remains to prove that $\sum_{k=1}^{N+1}\ell(I_k)\leq |\bigcup_{k=1}^{N+1}I_k|$. So, let $U_1,U_2,\dots$ be a sequence of open intervals whose union contains $\bigcup_{k=1}^{N+1} I_k$ and for $n\geq 1$ let $J_n:=U_n\cap (-\infty,a_{N+1}),\ K_n:=U_n\cap (a_{N+1},b_{N+1}),\ L_n:=U_n\cap (b_{N+1},\infty)$. $K_1, K_2,\dots$ is a sequence of open intervals whose union contains $I_{N+1}=(a_{N+1},b_{N+1})$ and $J_1,L_1,J_2,L_2,\dots$ is a sequence of open intervals whose union contains $\bigcup_{k=1}^{N} I_k$. Thus $$\sum_{n=1}^{\infty}\ell(U_n)=\sum_{n=1}^{\infty}(\ell(J_n)+\ell(L_n))+\sum_{n=1}^{\infty}\ell(K_n)\geq |\bigcup_{k=1}^{N}I_k|+|I_{N+1}|\overset{\text{ind.hyp.}}{=}\sum_{k=1}^{N}\ell(I_k)+\ell(I_{N+1})=\sum_{k=1}^{N+1}\ell(I_k)$$ and taking the $\inf$ on both sides we have $|\bigcup_{k=1}^{N+1} I_k|\geq\sum_{k=1}^{N+1}\ell(I_k)$, as desired.

Thank you.

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    $\begingroup$ I think length is a red herring. Any outer measure is countably additive over disjoint intervals. (I think ?) $\endgroup$ Commented Mar 16, 2021 at 18:40
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    $\begingroup$ en.wikipedia.org/wiki/… $\endgroup$ Commented Mar 16, 2021 at 18:44
  • $\begingroup$ @TomCollinge thank you for your interest in my question; I have now managed to give a full proof of the statement: could you please check it? Thank you. $\endgroup$
    – lorenzo
    Commented Mar 16, 2021 at 23:15
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    $\begingroup$ @lorenzo The fact that $|\bigcup_{k = 1}^{\infty}I_k| \leq \sum_{k = 1}^{\infty}l(I_k)$ follows directly from the definition of outer measure; you don't need to use the rather nontrivial fact that $|I| = l(I)$ to prove this. You don't need to prove the base case for $N = 2$. What you do need to prove is the base case for $N = 1$, i.e. that $|I| = l(I)$. If you already have proven that, then your proof looks good to me. $\endgroup$
    – Mason
    Commented Mar 17, 2021 at 0:17
  • $\begingroup$ @Mason thank you for the useful comments, I have corrected the proof accordingly. If you put these observations in your answer below, I would gladly accept it as an answer. $\endgroup$
    – lorenzo
    Commented Mar 17, 2021 at 9:37

2 Answers 2

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One way to proceed is to note that the outer measure of a compact set $K$ is equal to the Darboux upper integral of $\chi_K$, $U(\chi_K)$, aka the Jordan upper content of $K$, due to the finite subcover property for compact sets. Let $\varepsilon > 0$. Choose $N$ large enough and closed intervals $J_k \subset I_k$, $1 \leq k \leq N$, such that $\sum_{k = 1}^{N}l(J_k) \geq \sum_{k = 1}^{\infty}l(I_k) - \varepsilon$. Then appeal to the result that the Jordan content of a disjoint union of a finite number closed intervals is the sum of their lengths, which is simple to prove using theory of the Darboux integral. This proof has the advantage that it does not assume beforehand that $|I| = l(I)$.

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  • $\begingroup$ thank you for your interest in my question; I have now managed to give a full proof of the statement: could you please check it? Thank you. $\endgroup$
    – lorenzo
    Commented Mar 16, 2021 at 23:15
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I am reading "Measure, Integration & Real Analysis" by Sheldon Axler.
Your problem is Exercise 11 on p.24 in Exercises 2A in the above book.

I solved Exercise 11 as follows:

My lemma 1:
Let $a_1,b_1,a_2,b_2,\dots,a_N,b_N$ be any real numbers such that $a_1\leq b_1\leq a_2\leq b_2\leq\dots\leq a_N\leq b_N$.
Then, $$(b_1-a_1)+(b_2-a_2)+\dots+(b_N-a_N)\\=|(a_1,b_1)\cup (a_2,b_2)\cup\dots\cup (a_N,b_N)|\\=|[a_1,b_1]\cup [a_2,b_2]\cup\dots\cup [a_N,b_N]|$$ holds.


My proof of my lemma 1:

We consider the case in which $N=1$.
Let $a_1\leq b_1$.
By 2.14 on p.20 in the book, $|[a_1,b_1]|=b_1-a_1$.
By Exercise 6 on p.23 in Exercises 2A in the book, $|(a_1,b_1)|=b_1-a_1$.
So, $b_1-a_1=|(a_1,b_1)|=|[a_1,b_1]|$ holds.


We assume $$(b_1-a_1)+(b_2-a_2)+\dots+(b_k-a_k)\\=|(a_1,b_1)\cup (a_2,b_2)\cup\dots\cup (a_k,b_k)|\\=|[a_1,b_1]\cup [a_2,b_2]\cup\dots\cup [a_k,b_k]|$$ holds for any real numbers $a_1,b_1,a_2,b_2,\dots,a_k,b_k$ such that $a_1\leq b_1\leq a_2\leq b_2\leq\dots\leq a_k\leq b_k$.


We now prove $$(b_1-a_1)+(b_2-a_2)+\dots+(b_{k+1}-a_{k+1})\\=|(a_1,b_1)\cup (a_2,b_2)\cup\dots\cup (a_{k+1},b_{k+1})|\\=|[a_1,b_1]\cup [a_2,b_2]\cup\dots\cup [a_{k+1},b_{k+1}]|$$ holds for any real numbers $a_1,b_1,a_2,b_2,\dots,a_{k+1},b_{k+1}$ such that $a_1\leq b_1\leq a_2\leq b_2\leq\dots\leq a_{k+1}\leq b_{k+1}$.
Note that $(a_1,b_1)\cup\dots\cup (a_{k+1},b_{k+1})=(a_1,b_{k+1})\setminus\left([b_1,a_2]\cup [b_2,a_3]\cup\dots\cup [b_k,a_{k+1}]\right)$ holds.
By Exercise 3 on p.23 in Exercises 2A in the book and by our induction hypothesis, $$|(a_1,b_1)\cup\dots\cup (a_{k+1},b_{k+1})|\geq |(a_1,b_{k+1})|-|\left([b_1,a_2]\cup [b_2,a_3]\cup\dots\cup [b_k,a_{k+1}]\right)|\\=(b_{k+1}-a_1)-\left((a_2-b_1)+(a_3-b_2)+\dots+(a_{k+1}-b_k)\right)\\=(b_1-a_1)+\dots+(b_{k+1}-a_{k+1})$$ holds.
Of course $$(b_1-a_1)+\dots+(b_{k+1}-a_{k+1})\geq |[a_1,b_1]\cup\dots\cup [a_{k+1},b_{k+1}]|\geq |(a_1,b_1)\cup\dots\cup (a_{k+1},b_{k+1})|$$ holds.
Therefore $$(b_1-a_1)+(b_2-a_2)+\dots+(b_{k+1}-a_{k+1})\\=|(a_1,b_1)\cup (a_2,b_2)\cup\dots\cup (a_{k+1},b_{k+1})|\\=|[a_1,b_1]\cup [a_2,b_2]\cup\dots\cup [a_{k+1},b_{k+1}]|$$ holds for any real numbers $a_1,b_1,a_2,b_2,\dots,a_{k+1},b_{k+1}$ such that $a_1\leq b_1\leq a_2\leq b_2\leq\dots\leq a_{k+1}\leq b_{k+1}$.


We now prove Exercise 11 as follows:
If there exists an open interval $I_k$ such that $I_k=(-\infty,a)$ for some $a\in\mathbb{R}$ or $I_k=(a,\infty)$ for some $a\in\mathbb{R}$ or $I_k=(-\infty,\infty)$, then it is obvious that $$\left|\bigcup_{k=1}^{\infty} I_k\right|=\sum_{k=1}^{\infty} \mathcal{l}(I_k)=\infty$$ holds.
So, we assume that there doesn't exist such an open interval in $\{I_1,I_2,\dots\}$.
By my lemma 1, $$\sum_{k=1}^{N} \mathcal{l}(I_k)=\left|\bigcup_{k=1}^{N} I_k\right|\leq\left|\bigcup_{k=1}^{\infty} I_k\right|\leq\sum_{k=1}^{\infty} \left|I_k\right|=\sum_{k=1}^{\infty} \mathcal{l}(I_k)$$ holds.
So, $$\left|\bigcup_{k=1}^{\infty} I_k\right|=\sum_{k=1}^{\infty} \mathcal{l}(I_k)$$ holds.

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    $\begingroup$ Thanks a lot. :-) $\endgroup$
    – Koro
    Commented May 17, 2023 at 18:43
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    $\begingroup$ @Koro Thank you very much for your edit. $\endgroup$
    – tchappy ha
    Commented May 17, 2023 at 23:56
  • $\begingroup$ This just means $\left|\bigcup_{k=1}^{\infty} I_k\right| \leq \sum_{k=1}^{\infty} \mathcal{l}(I_k)$, which is just the countable subadditivity. It does not show $\left|\bigcup_{k=1}^{\infty} I_k\right|=\sum_{k=1}^{\infty} \mathcal{l}(I_k).$ How to do the other direction inequality ? $\endgroup$ Commented Sep 8, 2023 at 5:30

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