- 1-dimensional volume is the length of a line segment,
- 2-dimensional volume is the area of a surface,
- 3-dimensional volume is the volume that we know of
- and so on.
The basics
- Lebesgue measure of a set of real numbers between $[a, b]$ is $b-a$. This is consistent with our intuitive understanding of the length of a line segment.
- Lebesgue measures are finitely additive. So the measure of a union of the set of points $[a, b]$ and $[c, d]$ is $(b-a) + (d-c)$. I.e. $\text{measure}([a,b]\bigcup[c,d])=(b-a)+(d-c)$. This is again consistent with our intuitive understanding of the sum of two line segments.
- In fact \[\text{measure}(\bigcup_{i=1}^{\infty} A_i)=\sum_{i=1}^{\infty}\text{measure}(A_i)\] The above applies to any $n$-dimensional measure (length, area, volume, etc.) This is consistent with our intuitive understanding of the sum of multiple line segments, areas, volumes, etc.
- countable sets (set of natural numbers, for example),
- uncountable sets (set of real numbers) and
- some seeming bizarre sets for which a Lebesgue measure cannot be defined.
Lebesgue measure for countable sets
- Lebesgue measure of integer points on the real line is $0$. This is because points are just locations on a real line and intuitively, integers are few and far between on a real line.
- Lebesgue measure of a countable set is $0$. This is a stronger statement than the above. It is interesting, since it implies that a set containing countably infinite number of points is also $0$. This is not surprising though, since between any two distinct integer points on the real line lie uncountably many real points. So intuitively, a countable set of points is just not large enough to have a non-zero Lebesgue measure!
- One implication of the above is that rational numbers have a measure of $0$. This is in spite of the fact that rationals are dense - i.e. between any 2 rational numbers, there are countably infinite number of rationals! So, countable infinity is no match for Lebesgue measure.
Lebesgue measure for uncountable sets
- As hinted earlier, Lebesgue measure for an uncountable number of contiguous points is non-zero. So for the set of points $[a, b]$, this is $b-a$.
- Removing a countable number of points from an uncountable set does not change its measure. E.g the measure of the half open set $(a, b]$ (which excludes just the point $a$) or the full open set $(a, b)$, which excludes both points $a$ and $b$ is $b-a$.
- One implication of the above is that if we were to remove, say, all rational numbers from the set $[a, b]$, its measure is still $b-a$. So we removed all countably infinite number of rationals, and yet the measure doesn't budge. Cool!
- Lebesgue measure of an uncountable set need not always be non-zero. Georg Cantor gave the construction of an uncountable set, called the Cantor set, which has uncountably infinite points, and yet has a measure of $0$, as shown in the next section.
Cantor Set - an uncountable set with 0 measure
- Start with the set of points $\mathbb{R}[0, 1]$.
- Divide it into 3 parts - $[0, 1/3]$, $(1/3, 2/3)$, $[2/3, 1]$ and remove the middle part $(1/3, 2/3)$ from the set, as depicted in the diagram below. (Note: the first line in the diagram represents $\mathbb{R}[0,1]$ and those below it represent the construction of the Cantor set as detailed in this and subsequent steps.)
- Of the remaining parts, divide each into 3 parts, as shown in the diagram above. E.g.
- Divide $[0, 1/3]$ into $[0, 1/9]$, $(1/9, 2/9)$, $[2/9, 1/3]$ and again remove the middle part $(1/9, 2/9)$ from the set.
- Similarly, divide $[2/3, 1]$ into $[2/3, 7/9]$, $(7/9, 8/9)$ and $[8/9, 1]$ and remove the middle part, $(7/9, 8/9)$ from the set.
- Keep repeating the above step, as shown in the diagram above.
- The segments that we end up removing have a combined measure of \[\frac{1}{3}+\frac{2}{9}+\frac{4}{27}+\frac{8}{81}\cdots=\frac{1}{3}\sum_{i=0}^{\infty}\bigg(\frac{2}{3}\bigg)^i=1\]
- This means that the number of points left has a Lebesgue measure of 0, even though the number of points left is uncountable!
- This is bizarre indeed. However, there is something even more bizarre. There are some sets that are not Lebesgue measurable! I.e. we cannot define a measure for them. A Vitali set is one such non-measurable set.
Vitali Set - a non-measurable set
Construction of this set relies on the Axiom of Choice and is slightly involved.
Construction of this set relies on the Axiom of Choice and is slightly involved.
- Definition: We define 2 points, $a$ and $b$, as equivalent if $a-b$ is a rational number. We denote this as $a\sim b$. Note that $a$ and $b$ themselves, don't have to be rationals for them to be equivalent. Examples of equivalence are:
- $\frac{1}{2}\sim \frac{1}{3}$, since $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$, a rational number.
- $\pi\sim\pi+\frac{1}{2}$, since $\pi-(\pi+\frac{1}{2})=-\frac{1}{2}$, a rational number.
- We partition the set of real numbers $\mathbb{R}[0, 1]$ into equivalence classes using the above definition. All points within an equivalence class are equivalent to each other, of course.
- Each equivalence class has countably many points, one corresponding to each rational number.
- There are uncountably many such equivalence classes, one corresponding to each irrational number (that is not separated from another irrational number by a rational distance).
- By axiom of choice, we can construct a set containing one point from each of these equivalence classes, which we call $V$, the Vitali set.
- Note that since elements of $V$ are drawn from $\mathbb{R}[0,1]$, $\text{measure}(V)\leq 1$.
- Definition: We define a translation of $V=\{v_0,v_1,v_2,\cdots\}$ called $V_q=\{v_0+q,v_1+q,v_2+q,\cdots\}$ where $q\in\mathbb{Q}[0,1]$, the set of rationals between $[0,1]$. So for each rational number $q$, we have a corresponding uncountable set $V_q$ as shown in the diagram above.
- Note that $\text{measure}(V)=\text{measure}(V_q)\quad\forall q\in\mathbb{Q}[0,1]$, since each $V_q$ is merely a translation of $V$.
- Also note that \[\bigcup_{q\in\mathbb{Q}[0,1]}V_q=\mathbb{R}[0,2]\] i.e. we just partitioned $\mathbb{R}[0,2]$ into countably many partitions, $V_q, q\in\mathbb{Q}[0,1]$, with each partition containing an uncountable number of points.
- With points (8) and (9) in mind, what is the Lebesgue measure of $V$?
- If it is $0$, from (8), it means that each of $V_q$ has a measure of $0$. It implies, from (9), that $\text{measure}(\mathbb{R}[0,2])=0$ which is false, since $\text{measure}(\mathbb{R}[0,2])=2$.
- If it is greater than $0$, say $a$, then from (8), each of $V_q$ has a measure of $a$. Hence, from (9), since there are infinitely many $q$s, $\text{measure}(\mathbb{R}[0,2])=\infty$ which is false, since $\text{measure}(\mathbb{R}[0,2])=2$.
- Hence, $V$ is not Lebesgue measurable!
In closing
We saw that there are some sets with Lebesgue measure $0$ and some with non-zero. We also saw that there are some uncountable sets with measure $0$ and some sets with an undefined measure.
References
We saw that there are some sets with Lebesgue measure $0$ and some with non-zero. We also saw that there are some uncountable sets with measure $0$ and some sets with an undefined measure.
References
- Wapner, Leonard, The Pea and the Sun, 2005.
- The Axiom of Choice (Wikipedia)
- Cantor Set (Wikipedia)
- Vitali Set (Wikipedia)
