Length

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When $a\le b$ the interval $(a,b)$ has a length of $b-a$. In this section we investigate the extent to which we can assign a length to other subsets of $\mathbb{R}$.

Outer measure

If a set $B\subset \mathbb{R}$ is contained within a union of intervals, it seems entirely reasonable that any notion of length we might come up with would say the following: the length of $B$ is not larger than the sum of the lengths of the intervals. That is, if $$B\subset (a_1,b_1)\cup (a_2,b_2)\cup \cdots \cup (a_N,b_N)$$ then $$\mathsf{\text{Length}}(B)\le \sum _{n=1}^N{b}_n-a_n$$ holds.

Our first major idea is that, if we choose the intervals $(a_1,b_1),\dots ,(a_N,b_N)$ wisely, then there may be very little discrepancy between the set $B$ and the union of the intervals; in that case the sum $$\sum _{n=1}^N{b}_n-a_n$$ could be considered very close to the length of $B$.

The above idea is formalized mathematically as follows. We define the outer measure of a set $B\subset \mathbb{R}$ by finding the most efficient way of covering it by a union of intervals. Thus we define $$\mathrm{\Lambda }(B)=inf\{\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n:B\subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)\}$$ for all sets $B\subset \mathbb{R}$. It is the infimum that picks out the most efficient way of covering $B$ by intervals. Notice that, in the definition, we allow not only unions of a finite number of intervals, but unions of an infinite number as well.

The function $\mathrm{\Lambda }$ is defined on $\mathcal{P}(\mathbb{R})$. It assigns a value in $[0,\mathrm{\infty }]$ to every subset of $\mathbb{R}$. It has the following properties, which are very reasonable when thinking of $\mathrm{\Lambda }$ as assigning length to subsets of $\mathbb{R}$.

Theorem

The map $\mathrm{\Lambda }:\mathcal{P}(\mathbb{R})\to [0,\mathrm{\infty }]$ has the following properties.

1. $\mathrm{\Lambda }(\varnothing )=0$.
2. $\mathrm{\Lambda }(A)\le \mathrm{\Lambda }(B)$ whenever $A\subset B$.
3. $\mathrm{\Lambda }(A_1\cup A_2\cup \cdots )\le \mathrm{\Lambda }(A_1)+\mathrm{\Lambda }(A_2)+\mathrm{\Lambda }(A_3)+\cdots$ for all $A_1,A_2,A_3,\dots \subset \mathbb{R}$.
4. $\mathrm{\Lambda }(A-t)=\mathrm{\Lambda }(A)$ for all $A\subset \mathbb{R}$.

Proof:

First we prove (1). We must show that $$\{\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n:\varnothing \subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)\}$$ has an infimum of zero. Since every member of the set is non-negative we can do this by producing a value in the set smaller than any given $ϵ>0$. Since any sequence of intervals covers the empty set, we have complete freedom to choose the endpoints, and can for example take $a_n=0$ for all $n\in \mathbb{N}$ and $b_n=ϵ/2^n$ for all $n\in \mathbb{N}$. For this choice we have $$\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n=\sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{ϵ}{2^n}}=ϵ$$ and therefore $ϵ$ belongs to the above set. Since every member of the set is non-negative and $ϵ>0$ was arbitrary, the infimum must be zero.

For (2) fix $A\subset \mathbb{R}$ and $B\subset \mathbb{R}$ with $A\subset B$. To prove that one infimum is smaller than another, it suffices to prove that one set is larger than the other. Thus, if we can show that $$\begin{array}{rl}& \{\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n:A\subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)\}\\ & \supset \{\sum _{n=1}^{\mathrm{\infty }}{b}_n-a_n:B\subset \bigcup _{n=1}^{\mathrm{\infty }}(a_n,b_n)\}\end{array}$$ then we will immediately have $\mathrm{\Lambda }(A)\le \mathrm{\Lambda }(B)$. But this containment is immediate as any sequence of intervals that covers $B$ must cover the smaller set $A$.

Next we prove (3). Fix $ϵ>0$. For each $i\in \mathbb{N}$ let $$(a_1^i,b_1^i),(a_1^i,b_1^i),\dots ,(a_1^i,b_1^i),\dots$$ be a cover of $A_i$ with $$\mathrm{\Lambda }(A_i)+{\displaystyle \frac{ϵ}{2^i}}\ge \sum _{n=1}^{\mathrm{\infty }}{b}_n^i-a_n^i$$ and let $(c_n,d_n)$ be a sequence of intervals that enumerates all the intervals $(a_n^i,b_n^i)$ for all $i,n\in \mathbb{N}$. Certainly they cover the union of the $A_i$ and therefore $$\begin{array}{rl}\mathrm{\Lambda }\left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)& \le \sum _{n=1}^{\mathrm{\infty }}{d}_n-c_n\\ & =\sum _{i=1}^{\mathrm{\infty }}\sum _{n=1}^{\mathrm{\infty }}{b}_n^i-a_n^i\le \sum _{i=1}^{\mathrm{\infty }}\lambda (A_i)+{\displaystyle \frac{ϵ}{2^i}}\end{array}$$ which gives what we want as $ϵ>0$ was arbitrary.

Lastly, for (4) note that the translate of a cover of $A$ is a cover of $A-t$ and vice versa.▮

Countable additivity

The properties we have proved for $\mathrm{\Lambda }$ are reasonable if we think of $\mathrm{\Lambda }$ as assigning a length. Here is a further reasonable property that we should expect from a device that assigns length.

Countably additive Say that $\mathrm{\Xi }:\mathcal{P}(\mathbb{R})\to [0,\mathrm{\infty }]$ is countably additive if one has $$\mathrm{\Xi }\left(\bigcup _{n=1}^{\mathrm{\infty }}{A}_n\right)=\sum _{n=1}^{\mathrm{\infty }}\mathrm{\Xi }(A_n)$$ whenever $n\mapsto A_n$ is a sequence of subsets of $\mathbb{R}$ that is pairwise disjoint.

Recall that a sequence $n\mapsto A_n$ of sets is pairwise disjoint if $A_i\cap A_j=\varnothing$ for all $i\ne j$. Informally, a mapping $\mathrm{\Xi }$ from $\mathcal{P}(\mathbb{R})$ to $[0,\mathrm{\infty }]$ is countably additive if we can calculate $\mathrm{\Xi }(A)$ by breaking $A$ up into countably many pieces $A_1,A_2,\dots$ and summing up the values $\mathrm{\Xi }(A_1),\mathrm{\Xi }(A_2),\dots$

We are now faced with a central question: is $\mathrm{\Lambda }$ is countably additive?

Example (Vitali set)

Define an equivalence relation $\sim$ on $[0,1]$ by $x\sim y$ if and only if $x-y\in \mathbb{Q}$. Let $V\subset [0,1]$ be a set that contains exactly one member of every equivalence class. Any such set is called a Vitali set.

We can use Vitali sets to prove that $\mathrm{\Lambda }$ is not countably additive!

Theorem

The map $\mathrm{\Lambda }:\mathcal{P}(\mathbb{R})\to [0,\mathrm{\infty }]$ is not countably additive.

Proof:

Fix a Vitali set $V$. First we show that $V-q$ and $V$ are disjoint for every $q\in \mathbb{Q}$. Indeed, suppose that $t\in V$ and $t\in V-q$. Then $t\in V$ and $t+q\in V$. But $t+q\sim t$ so we have contradicted the defining property of $V$.

Now suppose that $\mathrm{\Lambda }$ is countably additive. We want to reach a contradiction. First we verify that $$[0,1]\subset \bigcup _{q\in \mathbb{Q}\cap [-1,1]}V-q\subset [-1,2]$$ holds. For the second containment, since $V\subset [0,1]$ we certainly have $V-q\subset [-1,2]$ for every $q\in [-1,1]$. For the first containment fix $t\in [0,1]$. There is by definition of $V$ some $v\in V$ with $v-t\in \mathbb{Q}$. Thus $v-t=q$ for some $q\in \mathbb{Q}\cap [-1,1]$. We therefore have $t\in V-q$ as desired.

Applying $\mathrm{\Lambda }$ to the above gives $$1\le \mathrm{\Lambda }(\bigcup _{q\in \mathbb{Q}\cap [-1,1]}V-q)\le 3$$ and since $\mathrm{\Lambda }$ is assumed countably additive we have $$1\le \sum _{q\in \mathbb{Q}\cap [-1,1]}\mathrm{\Lambda }(V-q)\le 3$$ after applying $\mathrm{\Lambda }$ to each side. Finally $$1\le \sum _{q\in \mathbb{Q}\cap [-1,1]}\mathrm{\Lambda }(V)\le 3$$ because $\mathrm{\Lambda }(V)=\mathrm{\Lambda }(V-q)$ for all $q$. But the above is impossible no matter the purported value of $\mathrm{\Lambda }(V)$.▮

Next steps

We have seen, from Vitali sets and the above theorems, that there is no mapping $$\mathrm{\Xi }:\mathcal{P}(\mathbb{R})\to [0,\mathrm{\infty }]$$ that has all three of the following properties.

1. Countably additive
2. Assigns lengths to intervals
3. Translation invariant

In order to proceed with a theory of length, we have to give something up. We are going to insist on translation invariance and countable additivity by abandoning the requirement that $\mathrm{\Lambda }$ is defined on $\mathcal{P}(\mathbb{R})$. Our next step, therefore, is to try to identify a rich collection $\mathcal{B}$ of subsets of $\mathbb{R}$ so that $$\mathrm{\Lambda }:\mathcal{B}\to [0,\mathrm{\infty }]$$ is countably additive and translation invariant.