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

Question

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.

Answer 1

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.

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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.

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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 $a_1\leq b_1\leq a_2\leq b_2\leq\dots\leq a_k\leq b_k$.

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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 $a_1\leq b_1\leq a_2\leq b_2\leq\dots\leq a_{k+1}\leq b_{k+1}$.

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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.