About Me

My photo
Ravi is an armchair futurist and an aspiring mad scientist. His mission is to create simplicity out of complexity and order out of chaos.

Sunday, July 10, 2011

Lebesgue Measure

Lebesgue measure is a measure of the length of a set of points, also called the $n$-dimensional "volume" of the set. So, intuitively:
  • 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.
These measures are defined over the space of real numbers $\mathbb{R}$, as opposed to, say, the space of natural numbers $\mathbb{N}$.

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.
Those were the basics. So far so good. We now look at Lebesgue measure for:
  • 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
  1. Start with the set of points $\mathbb{R}[0, 1]$.
  2. 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.)
  1. Of the remaining parts, divide each into 3 parts, as shown in the diagram above. E.g.
    1. 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.
    2. 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.
  2. 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.
  1. 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:
    1. $\frac{1}{2}\sim \frac{1}{3}$, since $\frac{1}{2}-\frac{1}{3}=\frac{1}{6}$, a rational number.
    2. $\pi\sim\pi+\frac{1}{2}$, since $\pi-(\pi+\frac{1}{2})=-\frac{1}{2}$, a rational number.
  2. 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.
  1. Each equivalence class has countably many points, one corresponding to each rational number.
  2. 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).
  3. 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.
  4. Note that since elements of $V$ are drawn from $\mathbb{R}[0,1]$, $\text{measure}(V)\leq 1$.
  5. 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.
  6. 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 $V$.
  7. 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.
  8. With points (8) and (9) in mind, what is the Lebesgue measure of $V$?
    1. 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 $\text{measure}(\mathbb{R}[0,2])=2$.
    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 $\text{measure}(\mathbb{R}[0,2])=2$.
  9. 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

    4 comments:

    1. This comment has been removed by a blog administrator.

      ReplyDelete
    2. Awesome graphic, really helps to visualize the proof. Now why is the Latex not displaying properly?

      ReplyDelete
      Replies
      1. I believe I have Latex working again. Google keeps changing their templating system. So until next time, enjoy!

        Delete