Footnotes:
Lebesgue Measure
Page last updated 18 Sep 2024
Lebesgue measure is a theory that arose from the concept of a “real number line”. Mathematicians began to contemplate what it meant to refer to distances between points on such a line in case of sets of points that had rather involved definitions, and came up with the concept of a “real number line”.
The “Real Number Line”
Mathematicians noticed that if you thought of a line of real numbers like a physical line stretched between two points, then given any real number, you could have a corresponding point on your “real number line”. And, being Platonists, they assumed that such a “real number line” actually exists as a mathematical object, and is composed of an accumulation of points.
This was a fundamental error. The reality is that the notion of a real number line is a notion that is inherently a fractal, where no matter how close one zooms in, the line always looks the same. It may be a simple onedimensional fractal, but a fractal it is, and that means that there never is a situation where the fractality ends and  behold  you then have a solid line where you cannot fit in any more points.
Because of this, there cannot be an actual sequence of all the real numbers between any two values (such as 0 and 1) where every number is set in order according to its value, since for any real number, there is no ‘next’ number. Similarly, there cannot be an actual sequence of points that somehow make an actual line. Moreover, by the very definition of a point, a point has no length or width, so that it is impossible for a collection of points to constitute a line.
But if you recognize that when you define a line where one end corresponds to 0 and the other end corresponds to 1, you are only defining a concept, not describing any actual thing, and that, since there is no limit to how many real numbers you can have between 0 and 1, then similarly, there is no limit to the number of points you can define on this line. But you never actually reach the state where the line is ‘filled’ with points. (Footnote: As Wittgenstein remarked: “But the curve is not composed of points, it is a law that points obey, or again, a law according to which points can be constructed.” as in Philosophical Grammar, University of California Press, 2005.)
This is in direct opposition to the Platonist stance which insists that all the points on the line ‘exist’ simultaneously, thus constituting an entire continuous “real number line”. For more on Platonism see Platonism, The Myths of Platonism, Platonism’s Logical Blunder, Numbers, chairs and unicorns and the posts Moderate Platonism and Descartes’ Platonism.
An example of Lebesgue measure theory
First, a couple of definitions:
An open interval is an interval that does not include the endpoints that define that interval (for example the open interval whose endpoints are ^{1}⁄_{3} and ^{1}⁄_{2} is the set of all points between ^{1}⁄_{3} and ^{1}⁄_{2} but not including the points ^{1}⁄_{3} and ^{1}⁄_{2}).
A closed interval is an interval whose endpoints are included in the interval.
Now, let’s consider a definition of a set A of ever decreasing intervals that is defined like this:
We start with the closed interval between 0 and 1. Now take a suitable listing (Footnote: See Onetoone correspondences and Listing the rationals.) of the rational numbers between 0 and 1 (for details see below A specific listing of rational numbers). Then, going through this list of rational numbers, for the first rational we define an associated open interval ^{1}⁄_{10} wide with that rational at the midpoint of the interval; our set now includes all the numbers in that interval (not including the endpoints). For the next number, define an associated open interval ^{1}⁄_{100} wide with that rational at the midpoint of the interval; we add those numbers to our set. For the next number, define an associated open interval ^{1}⁄_{1000} wide with that rational at the midpoint of the interval; we add those numbers to our set. And so on, with each subsequent open interval being ^{1}⁄_{10} of the length of the previous interval. (Footnote: Note that an iterative process is not in fact required: we can define the set A without any reference to iterations, see Formal set definitions.)
Given this definition, there are only two possibilities. Either:
 The entire closed interval between 0 and 1 is covered by the intervals, or
 There are some irrational points between 0 to 1 that are not covered by any interval (obviously, by the definition, there cannot be any rational points that are not covered by some interval).
Now, according to conventional mathematics using the axioms of Lebesgue measure theory, the total length of the intervals of the set A must be less than ^{1}⁄_{9}. This number is calculated by using a calculation that gives the limiting value of the sum of ^{1}⁄_{10} + ^{1}⁄_{100} + ^{1}⁄_{1000} + ^{1}⁄_{10000} + … which gives the value of ^{1}⁄_{9}. (Footnote: See Geometric Series for how this is calculated.) Note that this is a maximum value, since if some intervals overlap, the limiting sum will be less than ^{1}⁄_{9}.
Which means that since the length of the original interval (from 0 to 1) is 1, then the remaining length, according to the axioms of Lebesgue measure theory, must be at least ^{8}⁄_{9}. And so, according to this theory, there must be sufficiently many points remaining that can account for this value ^{8}⁄_{9}.
Analyzing the Lebesgue result
Given that there are points not in the set A, then it is clear that any such points cannot be rational, and so they must be irrational; we will call this set of such irrationals the set B.
Complete Interval: Note that in the following we call an interval of A a “complete interval” if it is not a subinterval of any interval of A (apart from itself).
Now, if there are any points in B, then, according to the definition, there must be complete intervals of A that have left and right endpoints (the points that are the lower and upper bounds of the complete interval ). These endpoints cannot be rational, since every rational is the midpoint of some enumerated interval, hence any such endpoints must be irrationals. (Footnote: This is an indication that the notion that such a set which contains irrationals is welldefined by the strict application of the above definition of open intervals cannot be correct.) Furthermore, from the definition of the set A it follows that there cannot be any degenerate interval (Footnote: A degenerate interval is a single point.) in the set A, and that every interval in the set B is degenerate. (Footnote: Clearly, there could not be any intervals in the set B that contain more than a single point, since for any two real numbers, there are infinitely many rationals between them.)
It is easy to show that the irrationals in the set B cannot possibly give rise to a total width greater than the total width of the set A, as follows:
Every rational has an associated n by the enumeration listing of the rationals. And every rational is in some complete interval of A. For any given rational, either it is the rational with the lowest associated n of all the rationals in that complete interval or it is not. If it is, then there is a unique association of that natural number n with the left irrational endpoint of that complete interval of A. Since every rational is enumerated, every complete interval of A is included by such an association, and hence every left endpoint of a complete interval is accounted for by such an association. (Footnote: In the same way, there is also a unique correspondence of the lowest associated n of all the rationals in that complete interval and the right endpoint of that interval.) Hence one can say that the endpoints of A are denumerable (since they are a subset of the rationals) where to be denumerable means that there must be an enumeration function that defines that enumeration, so that for a given natural number n, the function uniquely gives the appropriate endpoint. In this case however, while there can be a definable function that uniquely defines the association of every complete interval of A with every left endpoint, if it is impossible to define the function from the set description, then while there can be such a function, there can be no way, for any given enumeration function, to confirm that it is the required function. (Footnote: While there may be no theoretical reason why any such particular enumeration function might not be definable in principle, since there are infinitely many possible enumerations of rational numbers, there are, of course, real world limitations to the actual creation of definitions of enumeration functions by some sort of physical means.)
Clearly, the endpoints of the complete intervals of A, all of which are of zero size, cannot constitute a greater measure than those intervals for which they are endpoints, since all those intervals are of nonzero size. But according to the axioms of Lebesgue measure theory, the total measure of all the complete intervals of A must be less than ^{1}⁄_{9} , and that the remaining measure of at least ^{8}⁄_{9} is accounted for by points that are not in the set A.
Some people, when faced with this unpalatable contradiction, make the bizarre attempt to get around the contradiction by claiming that there must be other points in the set B as well as those single points that are between two intervals of A. I have asked these people to prove me wrong by providing a logical argument to back up their claims but they have been unable to do so.
Contradiction
This brings us to the crux of the matter. There is a contradiction inherent in Lebesgue measure theory, since it asserts that the endpoints of the intervals of A, where each such endpoint has zero measure, has a total measure greater than that of the intervals between those endpoints  and infinitely many of these intervals have nonzero measure. That is an absurdity.
Yes, according to the axioms of Lebesgue measure theory, that set of single irrational points (each of which has precisely zero length) has a total length greater than ^{8}⁄_{9} , while the intervals of set A, which includes every rational (and also many irrationals) must have a total length less than ^{1}⁄_{9} .
Conventional mathematics claims that although there are infinitely many intervals between these irrational points, these irrational points constitute a “bigger” infinity than that of the intervals  that there are somehow ‘more’ of these points than the intervals between them ! And that somehow (although exactly how is never divulged) because there is a ‘bigger’ infinity of these single points, they have a total measure of at least ^{8}⁄_{9} even though each such single point has a measure of precisely zero. But, as we have seen above, the infinity of these irrational points cannot possibly be of a “bigger” infinity than that of the rational numbers.
Further absurdity
While the case above is over a unit interval, it is a simple matter to scale everything by any given multiple n as a natural number. We let the new overall interval be n times the previous case, so it is now [0, n]. In the new sequence of intervals, each interval is now n times the previous size, hence the sequence is now ^{n}⁄_{10}, ^{n}⁄_{100}, ^{n}⁄_{1000} …. For the new enumeration, each number in the enumeration is obtained by multiplying the corresponding number of the previous enumeration by n.
Hence everything is simply scaled by n. Hence, according to Lebesgue measure, the measure of the points in A cannot be more than ^{n}⁄_{9}, and the measure of all points not in A must be at least ^{8n}⁄_{9}, in other words, the measure of all points not in A has been multiplied by n. But the scaling simply means that any single irrational points that were not in the original set A have only been repositioned to a new position (the new position is given by multiplying the previous position by n). Yet the claim is that, according to Lebesgue measure, these points now somehow constitute a measure that is n times the size of the original measure of at least ^{8}⁄_{9} (but less than 1).
For example if n =1,000,000, these points, simply because they have moved to a different location, according to Lebesgue measure now constitute a measure that is at least ^{8,000,000}⁄_{9} instead of the previous value of between ^{8}⁄_{9} and 1. So where does the extra measure of 888,888 units come from? Do the points somehow magically become a million times wider? And simply by choosing a different n, we can change the Lebesgue “measure” of this set of single isolated points by as much as we want.
This nonsensical result is a stark demonstration of the inherent absurdity of Lebesgue measure.
Welcome to a fantasy land where hocuspocus takes precedence over logic.
Attempts to evade the contradictions
Some people appear to have some difficulty accepting that Lebesgue theory results in a contradiction. As shown above, the only possibility is that the set A contains only nondegenerate intervals, and the other set B contains only single point degenerate intervals, and there are no points common to both.
Yet there are people (for example, see the comments) who claim that there are points in B that are not single points between two open nondegenerate intervals of A. But there is no other possibility  they cannot be points in nondegenerate intervals of B because B cannot have any nondegenerate intervals. I deal with this in more detail in an appendix below The converging sequence plea.
In several of the comments to this page, the commenters devise an argument that gives the result that there are points in B other than single points between intervals of A, and they proudly proclaim that their argument proves me wrong. They don’t seem to understand that generating a contradiction doesn’t somehow mean that they have discovered that there are magic points in B that can somehow be impossible and possible at the same time.
No  if you get a contradiction, that simply means that there is something wrong with your argument. That’s a basic principle of mathematics. It’s ironic that some commenters keep trying to find contradictions in my arguments, and when they fail, they believe that they can prove me wrong by coming up with their own contradictory argument. The capacity of such people for selfdeception is astonishing. (Footnote: It seems to be a characteristic of the Platonist mindset to refuse to let contradictions get in the way of their favored notions. For an example of how Platonists manage to pretend that there isn’t a contradiction involved, see the webpage An apparent inconsistency of Lebesgue measure. As Wilfrid Hodges (PDF An Editor Recalls Some Hopeless Papers, The Bulletin of Symbolic Logic, Vol 4, No. 1, March 1998) has remarked, with reference to flawed attempts to attack the diagonal argument: ‘to attack an argument, you must find something wrong in it. Several authors believed that you can avoid [that] by simply doing something else.’ But that is precisely what the protagonists on the webpage do  they attempt to avoid the contradiction by doing something else other than finding something wrong with the contradictory statement. They appear to be unable to see that the presence of a contradiction is telling them that there is something fundamentally wrong with their mathematical foundations, but they cannot contemplate the possibility that it might be an inconsistent theory. See also Counterargument against Lebesgue Measure? for another example of the irony of such people telling me that my argument is “wishywashy” while they provide no rational argument whatsoever.)
Misunderstandings of this topic almost invariably involve an avoidance of any consideration of the application of limit states to summations involving infinitely many items. The page Understanding sets of decreasing intervals explains why such definitions of everdecreasing intervals are inherently contradictory because by such definitions, without any limit included in the definition, the endpoints of any union of overlapping intervals of the set A cannot be defined as either rational or irrational, since for any endpoint of an interval given by the definition of A there is always another interval that extends that interval past that rational endpoint. The fallacies involved in attempts to circumvent these contradictions are analyzed in detail on the page Understanding Limits and Infinity and also in the paper PDF On SmithVolterraCantor sets and their measure which explains how all contradictions can be avoided by a logical analysis of the definitions involved.
Now for a few details of Lebesgue’s theory of measure.
The rules of Lebesgue’s theory
Lebesgue’s theory of measure is a theory that has to be bolted on to conventional number theory, by inventing axioms that are not inherent in fundamental number theory. (Footnote: Note that Lebesgue measure theory has never had any confirmation of any efficacy in relation to any real world application  unlike the conventional usage of numbers, which have been used time and time again in real world applications.) The reason for this necessity for bolting on is that in conventional number theory, for any two different numbers, there is a numerical value that is simply the difference between those two numbers, while the difference between a number and itself is precisely zero. But when you have the concept of a “real number line”, the notion of an interval now corresponds to the notion of the difference between two numbers. And what people refer to as a single point on the real number line corresponds to a single number; this isn’t really an interval, but sometimes it is referred to as a degenerate interval  in which case the measure of such a degenerate interval is precisely zero (the difference between a number and itself).
A measure, in its very simplest form, is simply the difference between two real numbers. And one expects that more complex measures would be dependent on multiples of such basic measures. But Lebesgue measure manages to assume that a collection of single zeros (each consisting of the difference between a number and itself) can somehow constitute a measure that is greater than zero. Yes, really !
The key assertions in Lebesgue theory are essentially: (Footnote: These are, of course, somewhat simplified here, but the essential facets of the theory are given by this.)
 For any set of single degenerate points that is denumerable, the Lebesgue measure of that set is zero.
 For a set of nonoverlapping intervals, but only provided the intervals are denumerable, the Lebesgue measure is the sum of the lengths of all of the intervals. (Footnote: It also assumes that there is always a simple summation of the lengths of infinitely many everdecreasing intervals, which is incorrect, see below Different orders of summation.)
 For a set of numbers between two numbers a and b that is not made up of either of the two above types, the Lebesgue measure cannot be deduced directly, but is given by subtracting the total of Lebesgue measures of the sets of type (i) and (ii) from the overall length between a and b. (Footnote: Note: In an attempt to avoid contradictions when Lebesgue theory is used along with set theory and the axiom of choice, then it has to be claimed that there must be sets of points that don’t have any measure  not a zero measure, nor some finite measure, nor an infinite measure  just no measure at all. Which means that, when using set theory and the Axiom of Choice, there are sets of points for which the axioms of the Lebesgue theory of measure cannot give a measure for those sets. See also The Axiom of Choice.)
The axioms of Lebesgue theory of measure are based around the requirement that if, for a given interval, any set A is defined for that interval, then the sum of the Lebesgue measures of the that set and its complement (the set of elements not in A, see complement) must always sum up to the total length of the interval. Now, while it might be nice to have that requirement satisfied, the Lebesgue method of doing so comes at a high price. The downsides are many. One major downside is that it is never explained how a collection of infinitely many zeros (the measures of single points separated by nondegenerate intervals) can be a finite nonzero value.
But the principal downside is that it leads to a direct contradiction  as in the case described above of ever decreasing intervals.
Limits and Limitlessness
The problems arise because of a failure to acknowledge that some definitions involve limitlessness, such as the recursive algorithm defined above that never terminates. Now, although a definition involves limitlessness, what you can do is to applying a limiting condition. But you must be careful. If there is a choice of limiting conditions that can be applied, then you must be sure to choose the limiting condition that corresponds to whatever aspect of the limitlessness that you are attempting to calculate a limiting value for.
In the case of the ever decreasing intervals as described above, some people appear to fail to take into account the fact that, whatever finite number n you choose, and then consider that n rationals have been assigned as midpoints of intervals, there are, not a billion times as many rationals not so assigned, nor a billion to the power of a billion times as many rationals not so assigned, nor a googleplex times as many rationals not so assigned, nor Graham’s number times as many rationals not so assigned, there will be infinitely many times that quantity of rationals still to be assigned an interval with its midpoint as that rational. Regardless of however large your n is, there will always be infinitely many times that quantity of rationals still to be assigned an interval with its midpoint as that rational. In other words, you never “run out” of rationals onto which a nondegenerate interval can be attached. That is why there can be two different limits that can be associated with the definition  you can either:
 calculate a limiting condition for the total length of intervals of decreasing size, without including any consideration of the relationships between the endpoints of the intervals
or  calculate a limiting condition for the totality of points that are in the set of points given by all defined intervals, without including any consideration of the limit of the summation of lengths of intervals. This is done by applying a limit to every union of overlapping intervals of A, which results in irrational values for the endpoints of the complete intervals of A.
In case (a), you get a value of number theory: a numerical value of ^{1}⁄_{9}.
In case (b), you get a value of set theory: a set of points between 0 and 1.
These are two completely different types of values. To assume that the value (a) must imply the other case (b) is absurd, and indicates a complete failure to understand limitlessness. Also note that the measure of the remaining intervals are decreasing, and there is no finite width at which this stops, so that the limiting value is zero width.
You can also see a formal paper on how to eliminate contradictions in the calculation of the total measure of sets that are defined in terms of limitlessness, see PDF On SmithVolterraCantor sets and their measure. For more on set theory, see the pages that give an overview of set theory, starting at Overview of set theory: Part 1: Different types of set theories.
An Alternative basis for Calculating Measure
If we reject the notion that a set of points can have a measure even if both that set and its complement do not define any nondegenerate intervals, then we can have a valid basis for the determination of the measure of a set, by specifying which sets have properties that contribute to a valid determination of measure, as follows:
A definition of a set of points A is valid for the calculation of a measure if and only if every point of the set A is either a point of a nondegenerate interval of that set or else it is the endpoint of the open end of an interval of the complement of the set A.
Note that a point that is an endpoint that is the open endpoint of an interval necessarily implies that that interval is nondegenerate, since:
A degenerate hypothetical interval (x, x) that is open at both ends is contradictory, since x is defined to be both the left endpoint of the interval and not in the interval, and also the right endpoint of the interval and not in the interval. Hence there can be no points in this hypothetical interval, and such an interval does not exist.
Similarly, the hypothetical degenerate semiopen (Footnote: An interval that has one endpoint that is in the interval and one endpoint that is not in the interval; also can be called semiclosed.) interval (x, x] is contradictory, since on the one hand x is defined to not be within the interval (since it must be the left endpoint of an interval where the left endpoint is not within that interval), while on the other hand it is defined to be within the interval (since it must be the right endpoint of an interval where the right endpoint is within that interval). Hence such an interval does not exist. The hypothetical degenerate semiopen interval [x, x) is similarly contradictory, and such an interval does not exist.
Note also that a sequence of nested intervals that are all semiopen do not have a valid limit state since, as shown above, the hypothetical interval (x, x] or [x, x) is contradictory, and such an interval does not exist.
If we take as an example, the set of all rationals, or the set of all irrationals, then since every point is a degenerate closed interval, no point can be the open endpoint of an interval of the complementary set, since all the points of the complementary set are all also degenerate closed intervals.
Different orders of summation
In the assertion that the set A has a measure of ^{1}⁄_{9} it is assumed that it is a very simple matter  that one simply adds up an interval of length ^{1}⁄_{10} then add another interval of length ^{1}⁄_{100} and so on  and that you can extrapolate that to infinity. But the definition of each of those lengths is dependent on the interval it is associated with  each length is defined by the left endpoint and the right endpoint for each case. So while it is simply asserted that the total length obtained by adding up infinitely many decreasing fractions, this conceals the fact that the calculation that is actually being defined is:
R_{1} − L_{1} + R_{2} − L_{2} + R_{3} − L_{3} + …
where L_{1} and R_{1} are the left and right endpoints of the first interval, L_{2} and R_{2} are the left and right endpoints of the second interval, L_{3} and R_{3} are the left and right endpoints of the third interval, and so on.
For a special case where each subsequent interval is added so as to ‘touch’ the previous one (is adjacent to) the previous one  the endpoints coincide  then we can have:
^{1}⁄_{10}  +  ^{1}⁄_{100}  +  ^{1}⁄_{1000}  +  ^{1}⁄_{10000} 
0.10 − 0.00  +  0.11 − 0.10  +  0.111 − 0.110  +  0.1111 − 0.1110 
For a finite sum, the left endpoint of one interval coincides with the right endpoint of the previous interval, and so the corresponding endpoint numbers cancel out  in the above 0.10, 0.11, and 0.111 cancel out, leaving 0.1111 as the correct summation. If the process continues infinitely, the limiting value is 0.111… which is equal to ^{1}⁄_{9} . We can also note that if we start at zero, the sequence of midpoints is ^{1}⁄_{20}, ^{21}⁄_{200}, ^{221}⁄_{2000} … and the limiting value of these midpoint values is also ^{1}⁄_{9}. (Footnote: Note that for a finite sum of intervals, the two endpoints of any given interval do not need to appear consecutively in any such ordering. For example, if L_{1}, R_{1}, L_{2}, R_{2}, L_{3}, R_{3}, L_{4}, R_{4} are in ascending order, then the total length of a finite number of such intervals (L_{1}, R_{1}), (L_{2}, R_{2}), (L_{3}, R_{3}), (L_{4}, R_{4}) can be given by R_{4} − L_{1} + R_{3} − L_{2} + R_{1} − L _{2} + R_{2} − L_{3} + R_{3} − L_{4}.)
But for the case of intervals that are not adjacent, and where the process continues infinitely, there is not necessarily any such simple limiting summation value. Unlike the case above where the midpoints of the sequence converge neatly towards a single point, any enumeration of the rationals over a given interval must always zigzag across that interval, and can never converge to a single point. The failure to take this into account may be that part of the reason that some people have difficulty with this matter  since besides the fact that for any finite n in the recursive process there always remain infinitely many rationals that have not been covered by any interval, it is also the fact that those remaining rationals never fall into a tidy sequence that converges towards a single point.
It is well known that for an infinite series that has both positive and negative terms, the limiting sum is dependent both on the values and the order in which they appear in the series (see Sums of infinitely many fractions: 1). The Lebesgue measure value of ^{1}⁄_{9} arises from the naive assumption that you can take a specific case such as the limit of the adding the interval 0 to ^{1}⁄_{10} , then the interval ^{1}⁄_{10} to ^{11}⁄_{100} , then ^{11}⁄_{100} to ^{111}⁄_{1000} , and so on, where the intervals follow on directly adjacent to each other and converge to one single limiting point, and apply that same assumption to a completely different scenario. The history of mathematics is not short of errors that resulted from the assumption that a general case follows from a specific case. And, in terms of set membership, as previously noted, we can define the set A in a way that does not specify any order of addition of intervals at all, see Formal set definitions.
The simplistic summation of infinitely many interval lengths overlooks the crucially important fact that, in the application of a limit, there is no particular order of incrementation of intervals. That is, for any point that is not in A, that point must be obtained by a limit, and for multiple such points, it is not the case that a limit is “reached” for one point, and then, subsequently, a limit is “reached” for another point, and then another. No, in the application of a limit state, there is no particular order of “reaching” the limit points, and hence there is no particular order of summation and subtraction of the endpoints, and one cannot assume that one can apply the simplistic case to this scenario. The naive assumption that one can always calculate the size of such a set by simply adding the lengths has no logical basis. Since there are infinitely many different possible orderings, ignoring the fact that different orderings can result in different limiting values, and that a limit state does not specify a particular limiting order, is absurd. The assertion that the total length of the set A can be no more than ^{1}⁄_{9} is an absurdity that should be obliterated from mathematics.
One correct calculation of measure?
If Platonism is correct, then the measure of any set of points must be an intrinsic property of the set  rather than being merely a human invention that is used for certain purposes. And so, if Platonism is correct, then there can only be one correct calculation of the measure of any set of points. Clearly, Lebesgue measure cannot be the correct Platonist theory of measure, since it leads directly to a blatant contradiction. There is no logical reason to suppose that Lebesgue theory is a theory that reflects some Platonist measure that exists independently of the human mind. It follows that there is no reason to promote Lebesgue measure theory as the ‘correct’ theory of measure. (Footnote: A reader suggested that there is more than one way of examining measure in order to see the full picture, using the analogy of blindfolded men examining an elephant. But the measure of a simple interval between two points is, very simply, the difference of their values. There’s no nuance involved. And if Platonism were correct then, any set of points must either have a total measure that “exists” independently of any method of calculation, and which corresponds to such a simple interval, or else there is no measure of that set that is an intrinsic property of measure for the set and which “exists” independently of any method of calculation.)
Also see Understanding Limits and Infinity which demonstrates the logical fallacies in some arguments regarding the set A.
Note: Mention in a sci.math forum
I came across a sci.math forum A Counterargument against Lebesgue Measure Theory? that makes some erroneous claims about my site. In case anyone might think that there is some substance in what is discussed there, I thought I should point out a few facts about what is written there. Below relates to what was there on 01 Oct 2021.
Two of the posters claim that I assert that the points of the complement of A constitute a denumerable set. The set of rational points, as in section Analyzing the Lebesgue result above that constitute the lowest enumeration numbers of the complete intervals of A is a denumerable set, and hence the associated irrational endpoints of the complete intervals also constitute a denumerable set. I have pointed out that there is association between the points of the complement of A and a set of certain rationals of A, as detailed in section Analyzing the Lebesgue result above (the lowest enumeration numbers of the complete intervals of A). That set of rationals is a denumerable set, and that implies that the associated irrational endpoints of the complete intervals also constitute a denumerable set. However, the indication is that it may be impossible to define an enumeration function that gives the set of all those associations from the definition of the set A. But as I indicate above, that does not imply that a definition of that enumeration function is impossible. I have since updated this page to deal with that misconception.
The posters there also glibly assert that all intervals of the set A must have rational endpoints  and simply ignore the consequent contradiction. As I point out on this site, for any set whose intervals are open and nondegenerate, an endpoint of any interval of it is, by definition, not a point of that set, but a point of its complement. And for the case in question this generates a contradiction, since all endpoints of the intervals of the set A are (unless the definition includes a limit) defined to be rational, while all points of the complement of A must also, by the same definition, be irrational. And it cannot be the case that the points of the complement of A are not the endpoints of the intervals of A, since that would imply that, since the points are different, then there must be a nondegenerate interval between them, and that nondegenerate interval would then have to be in the complement of A, which is impossible.
One poster claims that while he could demonstrate by a counterexample that my argument is wrong, he declines to do so, so we only have his word for it  and yet this same person complains that I engage in “wishywashy wordy arguments and unsupported claims”, apparently completely unaware of the irony. In addition, he also concludes that I am a crank, not because he has demonstrated any error in my writing, but because he thinks that noncrank mathematicians do not use certain terms of the English language. Logic is clearly not his strong point. I suspect that his claimed counterexample might be the same as that claimed in a comment to this site, see my response to this claim, where I point out the fallacy therein.
When these people observe that I get a result that differs to their results, they simply assume that that must mean that their result is the correct one and that I am wrong. It never occurs to them that perhaps it might indicate yet another contradiction/paradox to add to the list of all the contradictions/
The converging sequences plea
This is the plea that refers to sets where a point of a closed interval of one set is such that an open interval of the complementary set cannot be defined with that point as it’s endpoint, and an example is given, such as this:
Given the interval 0 to 1, where n is a nonzero natural number, we define:
S is the set that is the limit of the union of all open intervals (^{1}⁄_{n + 1} , ^{1}⁄_{n} )  (hence does not include 0, which is the limit case is (0, 0), which is not a valid interval, since the endpoints of an open interval are not in the interval).
T is the complement of S, that is, the set that is the limit of the union of all points ^{1}⁄_{n} (hence this also includes 0).
By this definition, 0 is in the set T and is a degenerate single point interval, but it is impossible to define an interval of S whose left or right endpoint is 0. Various people have asserted that this has great significance in the question of the consistency of Lebesgue measure. They claim that it proves that there are points in T that are not endpoints of complete intervals of S, and that this rescues Lebesgue measure theory from contradiction.
But all that is necessarily claimed for the case discussed on this page is that for any two complementary subsets A and B of a set, in terms of intervals we can state the following:
 No two complete intervals of the same subset can be immediately adjacent (otherwise the two intervals would constitute one complete interval of that set). (Footnote: By immediately adjacent we mean that there are no points between the left interval and the right interval.)
 A leftside open interval cannot be immediately adjacent to a rightside open interval  that is, for any right endpoint of a rightside open interval that is numerically less than the left endpoint of a leftside open interval, there are always points between those two points.
 A leftside closed interval cannot be immediately adjacent to a rightside open interval  that is, for any right endpoint of a rightside closed interval that is numerically less than the left endpoint of a leftside closed interval, there are always points between those two points.
 Given the above, we can say that there can be no open intervals that are immediately adjacent, and no closed intervals that are immediately adjacent. It follows that if one set A consists only of open intervals, there can be no such intervals immediately adjacent to each other, and similarity, if one set B consists only of closed intervals, there can be no such intervals immediately adjacent to each other.
 This of itself does not imply that for a given endpoint of an interval it is always possible to define an interval of the complementary set with that endpoint as one of its endpoints. So the converging sequences plea isn’t introducing anything that conflicts with the analysis on this page.
But in any case, the converging sequences plea doesn’t provide any sort of rebuttal, as we show in the following. In the converging sequence example above, there are two converging sequences that both converge (from either side) to the value 0. For this to apply in the case analyzed on this page, the same would have to be the case  that is, for every such purported additional point, there would have to be two converging sequence of irrational points of B that are separated by complete intervals of A, and which both converge to the same point, one from the left side, the other from the right side. (Footnote: There cannot be a valid degenerate interval that is halfopen, such as (a, a], since the point a is defined as not being in the interval, by the left side “(a ”, and at the same time it is defined as being in the interval, by the right side “a]”. ) And we can furthermore observe that for every one of those irrational points that are the points of such a pair of converging sequences of B (points that are separated by complete intervals of A), there already is a corresponding converging sequence of overlapping intervals of A  in other words we can say that, for every such additional purported point, there would have to be:
a pair of converging sequences of infinitely many converging sequences of infinitely many overlapping intervals of A, where each pair converge to the same point,
and which gives us gives us one point not already in the set B. But that only accounts for one point  for the plea to account for a nondenumerable infinity of points, there would have to be:
an infinite set of such points,
that is, there would have to be:
an infinite set of paired converging sequences of infinitely many converging sequences of infinitely many overlapping intervals of A, where each pair converges to the same point.
Furthermore, that infinite set would have to be nondenumerable  we will consider that shortly, but before addressing that difficulty, in order to rescue Lebesgue measure theory, one would also have to assume that:

for every possible real number interval, and

for every possible enumeration of the rationals, and

for every possible initial fraction less than ^{1}⁄_{2} rather than the ^{1}⁄_{10} in the definition above,
there exists a nondenumerable infinity of paired converging sequences of infinitely many converging sequences of infinitely many overlapping intervals of A.
But there is no evidence nor logical reason to support any notion that a converging sequence can somehow exist independently of any definition  a Platonist claim that such sequences might exist independently of any definition is faith based wishful thinking and has no place within serious mathematics. And no logical argument has been proffered to prove why there would have to be such an infinite set of such converging sequences for every possible case, never mind a nondenumerable set. The existence of only one case without such a nondenumerable infinite set invalidates the plea. And we can note that, for the definition of A, as on this page, the endpoints of the complete intervals of A are scattered over the interval 0 to 1, and no logical argument has been proffered to prove why there would have to be such paired infinite sequences among the points of B. (Footnote: Each complete interval of A is itself composed of infinitely many overlapping intervals, and no two intervals of the enumeration (as in the definition of the set A) are the same size.)
But there’s more  even making the above assumption, that in every possible case there are infinitely many converging sequences of infinitely many converging sequences of infinitely many overlapping intervals of A, it still would not account for any nonzero measure for the set B, because:
For each such point postulated, that point would require a pair of converging sequences of irrational endpoints of complete intervals of A. And since there is a denumerable set of the irrational endpoints of complete intervals of A (Footnote: Each is associated uniquely to a complete interval of A by being its left endpoint.) there could only be a denumerable set of such paired sequences of the endpoints of the complete intervals of A. But it is fundamental to Lebesgue measure theory that a denumerable set of degenerate points has zero measure. Hence the plea, even if it were applicable, could not possibly, by the very rules of Lebesgue measure theory itself, generate any nonzero measure at all. (Footnote: Similarly, for the everdecreasing sequence example above with the intervals (^{1}⁄_{n + 1} , ^{1}⁄_{n} ), between 0 and any given interval (0 , ^{1}⁄_{n} ) there is zero measure. And within any such interval (0 , ^{1}⁄_{n} ) every point of T is flanked by nondegenerate intervals of S, so that within (0 , ^{1}⁄_{n} ) the points of T cannot possibly be of greater measure than their associated nondegenerate intervals of S.)
The fact that the everdecreasing sequence plea is brought up at all as a defense seems to indicate a degree of desperation, since as noted above, it cannot in any case generate any measure not already accounted for.
A specific listing of rational numbers
An example of an enumeration of the rationals can be given by using the pattern of rationals:
^{1}⁄_{2}  ^{1}⁄_{3}  ^{1}⁄_{4}  ^{1}⁄_{5}  ^{1}⁄_{6}  … 
^{2}⁄_{3}  ^{2}⁄_{4}  ^{2}⁄_{5}  ^{2}⁄_{6}  …  
^{3}⁄_{4}  ^{3}⁄_{5}  ^{3}⁄_{6}  …  
^{4}⁄_{5}  ^{4}⁄_{6}  …  
^{5}⁄_{6}  … 
This gives a list that begins as ^{1}⁄_{2,}^{1}⁄_{3,} ^{2}⁄_{3,} ^{1}⁄_{4,} ^{3}⁄_{4,} ^{2}⁄_{4,} ^{1}⁄_{5,} ^{2}⁄_{5,} ^{3}⁄_{5,} ^{4}⁄_{5,}^{1}⁄_{6,} ^{2}⁄_{6,} ^{3}⁄_{6,} ^{4}⁄_{6,} ^{5}⁄_{6,}… Note that this gives duplicates like ^{2}⁄_{4,} ^{3}⁄_{21}, but the important fact is that the enumeration includes every rational between 0 and 1 (not including 0 or 1). The definition of the list can be made so as to remove the duplicates if required.
Clearly, there can be infinitely many similar enumerations based on similar definitions. An example is given below  note that this enumeration follows a pattern that for each subsequent denominator, the values run from the lowest to the highest value of the numerator. For every subsequent denominator, this gives a pattern of rationals across the interval 0 to 1. This patterning continues infinitely as the terms progress. The enumeration can be represented by an algorithm as follows, where the calculation of the n^{th} rational does not require the calculation of all of the rationals prior to n in the enumeration:
 Let t = 0
 If √ is a Natural number, then:
 m = (−1 + √ and the )/2n^{th} rational is (m − t)/(m + 1)
 Otherwise let t = t + 1 and repeat from step 2.
This can easily be made into a program as has been done here:
Rationale: Every logical argument must be defined in some language, and every language has limitations. Attempting to construct a logical argument while ignoring how the limitations of language might affect that argument is a bizarre approach. The correct acknowledgment of the interactions of logic and language explains almost all of the paradoxes, and resolves almost all of the contradictions, conundrums, and contentious issues in modern philosophy and mathematics.
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