Banach-Tarski Paradox II

This is the second post concerning the Banach-Tarski paradox. For the first, see here.

The Banach-Tarski paradox allows one to, through decomposition and reassembly, turn one three-dimensional ball into two without changing the individual pieces, apparently violating the additivity of volume in Euclidean three-dimensional space. In the previous post, a decomposition of the group F₂, roughly the set of finite strings of the symbols "a" and "b", was shown to yield two copies of the same group when the pieces were "translated" in a certain sense.

In carrying over the properties of F₂ into three-dimensional space, one treats the symbols a and b as rotations about axes in Euclidean three-dimensional space. Traditionally, the axes are considered to be the x- and z-axes of Cartesian coordinates. In fact, the necessity of a choice of axes is why a paradoxical decomposition can only occur in dimensions of three and above, and not in two. This is because, in F₂, the strings ab and ba are distinct; following their respective paths on the Cayley graph yields two different points. In two dimensions, any two rotations about the origin are commutative, i.e. can be performed in either order with the same result. Since the noncommutativity of F₂ cannot be carried over into two-dimensional space, the paradox is not possible there.

The rotations that correspond to a and b are taken to move through an angle of the inverse cosine of 1/3, or about 70.5°. This exact angle choice is unnecessary, but the angles chosen for a and b must be irrational multiples of a right angle. This is because no two linear combinations of them can be allowed to yield the same rotation; all linear combinations of the angles must be distinct. The purpose of this condition is to mimic a property of the group F₂, namely that no two distinct simplified strings represent the same element, or, in other words, no two distinct paths (without retracing) lead to the same point on the Cayley graph.

Next, we briefly restrict our attention to the sphere (which, unlike the ball, does not include the interior area; the sphere is as the surface of the earth and the ball like the surface as well as the interior). The set F₂, which we shall now consider a group of rotations, can act on any point p of the sphere. The set of points thus obtained, following any sequence of rotations (each corresponding to an element of F₂) beginning at p, is called the orbit of p.

In this way, the entire surface of the sphere can be partitioned into an infinite set of these orbits, none of which overlap. Since a set of finite sequences is countable, F₂ is as well. Since the number of points on the sphere is uncountably infinite, it follows that there are uncountably many of these orbits. Here the axiom of choice is invoked to select a single point from each orbit, and collect these into another set M.

The details involved in explicitly applying the decomposition of the sphere are too technical to consider here; the above steps were included to illustrate the use of the axiom of choice. The next step essentially brings about the paradoxical decomposition of the sphere by shifting M by the rotations a and b. Two copies of the sphere arise in a manner similar to that of F₂.

Finally, the result is extended to the three-dimensional ball by performing the decomposition on a continuum of spheres of radii 0<r<R, where R is the original radius of the ball being considered. Each point p on the outer sphere can be paired to a point on any of the smaller spheres by projecting inward along the ray from p to the origin (see below). Clearly the union of all these spheres contains all the points of the ball, with the exception of the origin, O.



The final obstacle, therefore, is proving that the ball with its center removed can be decomposed and reassembled to form the entire ball. In fact, there are subtle difficulties in doing this that do not concern us here. Once this is done, the Banach-Tarski theorem is proven.

Following this technical formulation, it is enlightening to step back and consider the implications of the paradox. It is important to see that the decomposition above could not be applied to a physical object. The above procedure depends on the infinite divisibility of the ball, which an object composed of matter does not possess. Additionally, the pieces in the decomposition, though finite in number, are not "chunks" of the ball but infinite collections of points, and so are not physically continuous.

Though inapplicable to the physical world, the Banach-Tarski paradox helps to elucidate the fundamental differences between mathematical and physical space, and the wide-reaching consequences of assuming statements such as the axiom of choice.

In response to this and similar paradoxes that follow from the axiom of choice, there have been attempts to appropriately weaken the axiom of choice to an axiom which, though giving most of the same benefits, eliminates the paradoxes. One of these is called the axiom of countable choice, which limits the applicability of the axiom to countable sets. This avoids the Banach-Tarski paradox, but some set theoretical results are lost. In addition, the rather arbitrary restriction to countable sets seems inelegant, as it complicates the axiom, bringing in more concepts.

Also, some interesting work has been done since the Banach-Tarski paradox was published in 1924 that has extended the result. First, the final step of the proof above, in its original form, involved a total of 24 pieces. Through an alteration of the orbit scheme above, the number of pieces can be reduced to five.

Furthermore, the beginning and ending sets can be more general than simply a ball and two balls. Clearly, by a repetition of the above process, any (finite) number of balls can be produced by decomposition. It has even been shown that, if the original ball can be decomposed into an infinite number of pieces, one can obtain infinitely many copies of the ball, and even uncountably many. By allowing these decompositions, we can simply conjure up as many balls as we want from a single one!

In fact, the statement has been generalized even further to allow any bounded three-dimensional regions which are not "empty" to be broken up into a finite number of pieces and reassembled into any other of these regions.

The Banach-Tarski paradox is central in proving that there is no finitely additive measure in three-dimensional (and higher) spaces which agrees with the basic conception of volume. In one and two dimensions, there still is no countably additive measure that can be universally applied due to the existence of non-measurable sets (see again the Lebesgue measure series for an example of a measure in mathematics). The above are a few of the surprising geometric applications of the axiom of choice, showing how pervasive this assumption is, even beyond its native set theory.

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Sources: http://www.math.hmc.edu/~su/papers.dir/banachtarski.pdf, http://www.bsu.edu/libraries/virtualpress/mathexchange/05-01/Coleman.pdf, Banach-Tarski Paradox at Wikipedia

Banach-Tarski Paradox

The Banach-Tarski paradox is a very counterintuitive theorem in geometry that, in effect, states that a three-dimensional ball can be broken up and reassembled in such a way that two identical balls of the same size as the first are formed. This "doubling" of the ball seems contrary to the usual ideas of Euclidean geometry, and is therefore exemplifies the controversy over its underlying assumption, the axiom of choice.

The axiom of choice is one of the most debated topics in mathematics. It is an axiom of set theory, and the debate is whether it is right to assume it, and in what "strength" or version. The axiom itself, put informally, runs thus:

From any collection of nonempty sets, there exists a way to choose exactly one member from each set.

The ability to make such choice seems intuitive enough, but as seen elsewhere on this blog, the assumption of the axiom of choice implies the existence of sets that have no "volume", i.e. their size cannot be measured in any way. And this, as we shall see, is a crucial concept needed to derive the Banach-Tarski paradox.

The "translation" of the paradox into set theoretical concepts involves treating the three-dimensional ball as a set, call it A. We consider a division of the ball as a partition of the set A into the subsets A₁, A₂,...,A_n, where none of the subsets overlap, or in set notation, where A_i∩A_j = Ø whenever i ≠ j (i and j run from 1 to n). In this decomposition, n is finite, and only a finite number of pieces are needed to transform one ball into two. This in itself makes the paradox even more unbelievable.

After the ball is partitioned, each piece would then undergo a translation (some movement in space) and a rotation, taking each subset A_i to a corresponding B_i. In this translation, the intrinsic properties of the subsets are not altered; they are simply rearranged in space. The resulting sets, B₁,...,B_n, can be split into two groups in such a way that B₁∪B₂∪...∪B_m is a ball congruent to the first, and the union of remaining subsets, B_{m + 1}∪...∪B_n, is another ball also congruent to the original one. The transformation is illustrated below:



Since Euclidean translations and rotations preserve volume, the Banach-Tarski paradox violates the principle of the additivity of volume, i.e., that the sum of the volumes of two disjoint sets is the volume of their unions. The beginning ball and the (union of) two equivalent balls are composed of the same (translated) pieces, but they have different volumes. The reason this principle is violated is that the sets in question are non-measurable, or cannot be assigned a volume by any measure. Therefore, the idea of volume additivity has no meaning for them. Non-measurable sets, in addition, can only be proved to exist with the axiom of choice or an equivalent statement. We shall assume the existence of these non-measurable sets, but a construction of such a set under the Lebesgue measure can be found here.

The proof of the Banach-Tarski paradox at first does not work directly with the ball, but instead deals with a more abstract group of points. As we shall see, the decomposition of this abstract set can be performed analogously on the ball with slight modifications.

The abstract set in question is called the free group of two generators, or F₂, and it refers, in effect, to the set of strings of the symbols a and b, as well as their inverses, a^-1 and b^-1. For example, the strings aba^-1b and bba^-1baa are members of F₂. We also define e, or the empty string, as the string in F₂ with no members. Finally, we define an operation, *, on strings in F₂ that combines two strings into one by placing the second directly after the first. Therefore ab^-1*b^-1aab = ab^-1b^-1aab. Additionally, strings are simplified by two rules, where x stands for any string in F₂:

1. ex = xe = x
2. aa^-1 = a^-1a = bb^-1 = b^-1b = e

As an example of the above rules, consider the product (under the operation *) of the strings a^-1bab and b^-1a^-1bab^-1.

a^-1bab*b^-1a^-1bab^-1 = a^-1babb^-1a^-1bab^-1 = a^-1baea^-1bab^-1 = a^-1baa^-1bab^-1 = a^-1bebab^-1 = a^-1bbab^-1

The group F₂ can be visualized in several ways. One is as an infinite "tree" of elements, where points on the tree represent strings and "branches" describe the construction of these strings. Below is one example of such a visualization.



A tree of this type is called a Cayley graph of the group F₂. Illustrated are the strings of F₂ up to three symbols in length. The segments connecting the strings (black dots) illustrate the addition of a symbol to the right of the existing string. One begins in the center with e, the empty string, and moves right to add an "a" and up to add a "b". Moving oppositely to either of the above directions appends an inverse of the corresponding symbol. Note that tracing a path and then retracing backward produces a symbol adjacent to its inverse, causing cancellation. The construction of the string ba^-1b^-1 is illustrated; staring at e, one first proceeds upward, then left, then down. The reader can verify this construction.

The decomposition of F₂ requires the use of some new notation. By S(a) denote the set of all strings in F₂ beginning with the symbol a. Define the expression similarly in the case that a^-1, b, and b^-1, respectively, are substituted for a. Next, the notation, aS(a^-1), for example, refers to the set of strings produced by joining the string "a", with a member of the set S(a^-1). Thus there is a string in aS(a) for every string in S(a). It must be made clear, however, that the set S(a) and the other related sets contain only simplified strings. Nowhere in any string in S(a) will there appear a term "aa^-1" or "bb^-1", as these will have already been cancelled.

Armed with these notions, we can construct the "paradoxical" decomposition of F₂. The first step is a simple decomposition involving separating the group into each of four disjoint quadrants, the sets S(a), S(a^-1), S(b), and S(b^-1), and the set {e}, consisting only of the empty string.

The next operation required is a "shift" of the set S(a^-1) by the element a into the set aS(a^-1), which, as discussed above, has as members every string obtained by adjoining a, (through the operation *) on the left, to some member of S(a^-1). The significance of this shift in relation to the sphere will be apparent later. This shifted set is interesting because, when a is combined with a string beginning with a^-1, the two elements cancel. Since, in S(a^-1), the second and subsequent symbols of each member are arbitrary, every string beginning with a^-1, b, or b^-1 is a member of aS(a^-1). For example, the string bab^-1 is the same as a combined with a^-1bab^-1, a member of S(a^-1). Also, strings beginning with a^-1 such as a^-1bb can be formed as the combination of a and a suitable string beginning with "a^-1a^-1", in this case a^-1a^-1bb. Finally, since a^-1 is obviously a member of S(a^-1), the string aa^-1 = e is therefore a member of aS(a^-1). In fact, the only strings that are not included in aS(a^-1) (ironically) are those beginning with a! This is because, for this to occur, the second element of a member of S(a^-1) would have to be a, and this cannot occur, as all strings in S(a^-1) are assumed to be simplified.



The above is another view of the Cayley graph, this time illustrating the quadrant S(a^-1) along with the shifted set aS(a^-1). The only strings not members of the latter set belong to S(a). Therefore, the group F₂ is a union of aS(a^-1) and S(a). A similar procedure, substituting b for a, leads to the conclusion that the union of bS(b^-1) and S(b) is again F₂. In summary,

F₂ = {e}∪S(a)∪S(a^-1)∪ S(b)∪S(b^-1) = aS(a^-1)∪S(a) = bS(b^-1)∪S(b).

The sets S(a^-1) and S(b^-1) from the original have been shifted by a and b, respectively, and been incorporated into two other decompositions of the same group F₂. One copy of F₂ has been made into two. (One might well wonder "what happened to {e}?" It turns out that this piece is simply discarded in the above procedure, but the scheme is modified slightly in application to the ball to correct this subtlety).

The relation between F₂ and the three-dimensional ball, as well as other extensions of the paradox, are found in another post.

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Sources: http://www.kuro5hin.org/story/2003/5/23/134430/275, Banach-Tarski Paradox on Wikipedia

Lebesgue Measure III

For an introduction to the Lebesgue measure and various applications to sets in Rⁿ, see the previous two posts, beginning here.

The utility of the Lebesgue Measure has been described in the previous two posts. To briefly summarize, the Lebesgue measure provides a more general notion of the "size" of a set in space that matches up with the intuitive notions of length, area and volume in 1, 2, and 3 dimensions, respectively, as well as preserving natural properties such as the volume of a union of disjoint sets being the sum of volumes of the individual sets. It even goes beyond these notions to measure countable and uncountable sets, manifolds, and even fractals of various sorts. However, under certain assumptions, there exist sets which cannot be consistently assigned a Lebesgue Measure. Such sets and the validity of the underlying assumption necessary for their existence are discussed in this post.

For the construction of these Lebesgue non-measurable sets or Vitali Sets, one must assume one of the most dubious and controversial statements in mathematics: the axiom of choice.

The axiom of choice is not by any means specific to measure theory, but rather is an axiom of set theory, and therefore lies at the foundation of mathematics. The axiom, put informally, is the seemingly innocent statement below:

For any family of sets, there exists a way of choosing (hence the name "axiom of choice") one member from each set in the family.

Slightly more formally, it states that for any set X containing only nonempty sets as members, there exists a choice function f which selects exactly one member from each member of X. Of course, there may be many choice functions for a collection X of sets; the axiom simply guarantees that at least one exists. For example, set
X = {x₁,x₂,x₃} = {{4,5,6},{1,4,7},{2,7,9}}. A choice function on this collection of sets maps each subset of X to one of its members. For example, one could have
f(x₁) = f({4,5,6}) = 4, and the function would be similarly defined on x₂ and x₃.

The act of selecting one member from each of a class of sets seems completely natural and perhaps even fundamental. It seems impossible to imagine a collection of sets where such a choice function would not be possible. Yet the axiom of choice implies the existence of a set that is not measurable by the Lebesgue Measure, and therefore has no definable volume. That such a set exists and can be embedded in a "well-behaved" space like Rⁿ seems surprising, almost contradictory. Without further ado, let us construct a Vitali Set.

For simplicity, the Vitali Set considered will be in R¹. The first step in constructing a Vitali Set relies on a concept called a quotient group, specifically the quotient of the real numbers and the rational numbers, denoted R/Q.

R/Q contains a number of classes. Each class is the set of rational numbers Q shifted by a real number r. More specifically, each class, denoted Q + r, contains every number that is formed by adding a rational number to r. For example, if r is π, a real number, then Q + π contains members such as 3/4 + π and -9 + π, but not multiples of π such as π/2, nor rational numbers themselves (e.g. 1/2). Conversely, a real number x is a member of Q + r if and only if x - r is rational. Finally, to form the quotient group R/Q, we discard any "repeats", so that each pair of classes Q + r and Q + s in R/Q is disjoint, i.e. the two have no common elements.

Since the real numbers form an uncountable set, but each class Q + r has as many members as the rational numbers and is countable, the quotient group R/Q is an uncountable set, containing countable sets as members. It is also a partition of R, as every real number is contained within one of the classes, but only one, as they are disjoint.

The next step in construction makes use of the axiom of choice. Clearly, each class
Q + r contains members in the interval [0,1] (as each is dense in the real numbers). Therefore, for each class Q + r, the set (Q + r)∩([0,1]) (∩ denotes the intersection of two sets) is nonempty. The axiom of choice then allows one to select exactly one member from each (Q + r)∩([0,1]), and combine them into their own set. The resulting set has one member for each class of R/Q, all within the interval [0,1].

Call the set just formed V. Note that V was not determined in any precise sense, and rather could be any one of an infinity of possible sets satisfying the same conditions, but for a different selection of choice functions. Furthermore, V is uncountable, and for every real number r, there is exactly one v∈V such that v - r is rational. Also, taking r = 0, there is exactly one rational number in V.

V is the Vitali set. Now it must be proven that it does not have a Lebesgue measure. The final step involves translations of V itself, formed in a manner analogous to those of the classes of R/Q. This time, however, the shifts will be of certain rational numbers.

Since the rational numbers are countable, we can make use of a "catalog" of them, indexed by the positive integers. In other words, we can list the rational numbers, and assign every one a number which will serve as its index. This is called an enumeration of the rational numbers. The enumeration, in this case, will be confined to the rational numbers in the interval [-1,1]. As with the selection of choice functions, the enumeration of these numbers chosen is arbitrary, as differences between specific enumerations will not affect the final result. Denote each rational number in the list q_i, for a positive integer i. For example, one could have q₁ = 0, q₂ = 1/2, q₃ = -2/5, etc..

Now we form, for each q_i, the set V_i = V + q_i, or the set of all numbers created by adding q_i to a member of V. It is very important to note that the resulting sets are pairwise disjoint; no two of the translations contain a common member. To see this, remember that each member v of V is representative of a class of the form Q + r. There is one, and only one member of V for every class of the above form in R/Q. The corresponding member in each V_i will also be a member of the class Q + r, as it differs from v only by a rational number. Therefore, it cannot be the same as any member of any V_i formed from another class Q + s. Finally, since the q_i are distinct, no two translations of the member v of the original V yield the same number.

After this fact is accepted, the non-measurability of V can be proven. The set that must next be considered is actually the union of of all the V_i. Since there are countably many V_i, and they are pairwise disjoint, the Lebesgue measure of the union is the sum of the Lebesgue measures of each V_i. Hence,



What is the value of this sum? We can place some limitations on it by an examination of the union of the V_i. Note first that every real number on the interval [0,1] is contained within the union of the V_i, since the original V contains representative members of each class in a partition of R/Q. Since these representative elements are within [0,1], it is only necessary to shift them by a number such that |q| ≤ 1, because a distance of 1 is the "farthest away" two objects in [0,1] can be from one another. This reveals the purpose of the enumeration only being of the rational numbers between -1 and 1. Shifting a single element v of V that represented the class Q + r by all of the q_i (which lie in the interval [-1,1]) will recreate all of Q + r in [0,1], as well as more outside the interval.

The logical basis for the second limitation is even simpler. Since each v in V must be within [0,1], and the maximum shift induced by the q_i to form the V_i has a magnitude of 1, all members of the union of the V_i must be within [-1,2]. More precisely, the smallest possible member of V is 0, causing the smallest possible member of the V_i to be -1, and the largest possible member of V is 1, causing the largest possible member of the V_i to be 2. The above limitations are illustrated in equation below:



We can now use the simple property of the Lebesgue measure that if a set A is contained within another set B, then λ(A) ≤ λ(B). Substituting inequality for inclusion, we can then replace the sets in the above equation with their measures:



Note, finally, that a translation of a set has the same measure as the original set. This is a basic property of the Lebesgue measure, and is reflected in one's intuitive notion of Euclidean geometry; moving figures around in the plane, for example, does not change their area. In light of this property, since each V_i is a translation of V, the last inequality becomes



However, this is a contradiction. Since the expression in the infinite sum is independent of i, it is simply a constant added to itself an infinite number of times. However, the Lebesgue measure of a set can only be a positive real number or zero, and neither of these, when added to itself an infinite number of times, can produce a number between 1 and 3. The assumption that V is measurable is therefore false.

V is not unique of course, due to the different choice function possibilities. In fact, uncountably many Vitali sets arise from the above construction. The idea that so many sets on such a simple structure as the real number line could be non-measurable is remarkable. This brings us back to the main question: is the axiom of choice valid?

There is little consensus on this topic among mathematicians. The axiom itself (in its set-theoretical form) seems intuitive but there are many equivalent or implied statements that seem counterintuitive, such as the result above. Non-measurable sets have been applied to formulate a statement that seems so blatantly contradictory that it is called the Banach-Tarski paradox. It states that, given the axiom of choice, one can subdivide a three-dimensional ball into discrete pieces, and, without altering them, rearrange them in such a way that two three-dimensional balls are formed of identical size to the first. Such an act of "cloning" in reality is contrary to common physical principles. Yet mathematically, many important and seemingly valid theorems depend on the axiom. Therefore, there is evidence for both arguments, and the matter is not likely to be conclusively resolved in the near future.

Returning briefly to the Lebesgue measure, the effectiveness and generality of this instrument, despite the above defect, has led to growth in many related areas of mathematics. Lebesgue himself used the measure and related concepts to more rigorously define properties of functions on Rⁿ, including those which, in earlier periods, were often disregarded. Since the theory of measure is related to that of manifolds, the above developments had great significance in the area of topology as well.

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Sources: Vitali Sets, Axiom of Choice on Wikipedia, http://homepage.univie.ac.at/erhard.reschenhofer/pdf/probstat/P_A.pdf

Lebesgue Measure II

For the definition of the Lebesgue Measure and some of its simple applications to sets, see the previous post.

So far, many of the sets discussed have been products of non-degenerate intervals (or sets approximated by these products of intervals), and these have had positive Lebesgue Measures. Since an interval in R¹ is an uncountable set, so are the products of intervals in Rⁿ. The other sets that have been considered have been either finite or countable, and all of these have had Lebesgue Measure 0. Therefore, thus far, negligibility has coincided with countability. However, there are also uncountable sets which are negligible.

The key concept is that a subset A of Rⁿ is negligible if it has a dimension less than n. Before exploring the implications of this statement, consider a simple example:

Let the set A be interval [0,1] embedded in R² (illustrated below). This, being an interval, is an uncountable set, but it can be shown by a succession of approximations by squares in R² that this set has Lebesgue Measure 0.



A portrayal of the set A in R². The standard Cartesian coordinates and axes are superimposed for clarity. The first approximation (top), uses the single square (2-prism) [0,1]x[0,1], which clearly contains A, to approximate the volume of A. Having side length 1, the volume is 1. In the second approximation (middle), the union of two squares, J₂¹ and J₂², each with side length 1/2, clearly contains A, the total volume of the approximation (denoted by J₂) being (1/2)² + (1/2)² = 1/2. The third approximation (bottom), follows a similar approach, with four squares of side length 1/4 containing A and yielding the approximation J₃ = 1/4 to A. Clearly, if the side length of the approximating squares is halved in each successive approximation, the resulting estimated volume is halved each time. Since this can be done indefinitely, λ(A) = 0.

The above example shows how squares, the most basic units of area in R², can be used to calculate the volumes of other sets, including negligible ones. This example also motivates a useful condition for negligibility: A set S is negligible in Rⁿ if, for every ε > 0, no matter how small, there exists a (finite) collection of rectangular n-prisms J_i, such that S is contained within the union of the J_i and the Lebesgue measure of the union of the J_i is less than or equal to ε.

Since the set A in the example is one-dimensional (in our as yet intuitive sense), its negligibility is consistent with the previous statement that a set of some dimension embedded in a real space of higher dimension is negligible, as 1 < 2. Note that, if the same set A were being considered as a subset of R¹, it would not be negligible, but rather λ(A) would equal 1. This reveals the importance of specifying the dimension of the ambient space Rⁿ. More generally, if A is a "well-behaved" manifold of dimension m embedded within Rⁿ, and m < n, then A is negligible under the Lebesgue Measure.

The purpose of the qualifier "well-behaved" is to distinguish between two types of dimension. The common idea of dimension, the number of perpendicular directions one can "contain" in space, is what is called the topological dimension. However, the Lebesgue Measure also depends on something called the Hausdorff dimension, named after the German mathematician Felix Hausdorff. This alternate measure of dimension applies to a larger class of sets, notably fractals, and does not assign only integer values; many sets have Hausdorff dimension between 1 and 2, for example.

For any space, the Hausdorff dimension of the space is always greater than or equal to the topological dimension (if applicable) of the same space. Thus, even if a set topologically has a smaller dimension than the real space around it, it may still have nonzero Lebesgue Measure.

Take, for example, the Cantor Set. It is produced by removing the middle third of the interval [0,1], and, on every subsequent step, removing the middle third from each remaining line segment. After an infinite number of steps, the set remains. The properties of this set are proved elsewhere. The Cantor set is a fractal, an uncountable set, but not a dense set. Also, if its points are expressed in ternary (base 3) notation, the numerical expansion of each member of the set contains only 0's and 2's (in base 3, the possible digits are 0, 1, and 2). The Hausdorff dimension of the Cantor set is related to its property of containing points which "use" only 2 out of the 3 digits of the ternary system; the dimension is ln(2)/ln(3) ≈ .6309. This figure gives some idea to "how big" the set is. Since this is less than 1, the dimension of the real space in which it is embedded, the Cantor Set has Lebesgue Measure 0.

In the above case, the topological dimension can be considered to be 0, as the set contains only "isolated" points. However, the Hausdorff dimension, though greater, was still less than that of the ambient space, giving the same result: the Cantor set is negligible. In other cases, the distinctness of the Hausdorff dimension does affect the Lebesgue Measure. Consider the figure below.



The above illustrates three stages of the construction of a fractal known as a space-filling curve, or Peano curve. It is a fractal because it is self-similar (structures are duplicated on smaller levels in each step), and after infinite steps, it fills every point in the space it occupies, which, for simplicity, we will assume to be the unit square [0,1]x[0,1] in R². Locally, each part of the space-filling curve looks like a line, giving the set topological dimension 1. However, it fills every point in a two-dimensional region, and has Hausdorff dimension 2. With respect to the Lebesgue Measure on R², the space-filling curve is not a negligible set, but rather has a positive Lebesgue Measure, equal in magnitude to the volume of the unit square, that is, 1.

This post has illustrated the application of the Lebesgue Measure to an even wider variety of sets. However, there are some even more exceptional sets to which the Lebesgue Measure does not assign a value at all! (see the next post)

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Sources: Advanced Calculus of Several Variables by C.H. Edwards Jr., Vitali Sets on Wikipedia