Measure Theory 4 | Not everything is Lebesgue measurable - Subtítulos bilingües
Hello and welcome back to measure theory.
This is part 4 in our series and today I will finally explain you why we can't define the Lebesgue measure on the whole
power set.
And this should give you at least one motivation why it's so important to study the measure theory.
For the start here, let me state the measure problem again.
We search a measure mu on the power set of r.
So as you can see, this is the one-dimensional case.
So we first deal with the real number line before we go to higher dimensions.
We call this the measure problem because we want now two natural properties that are fulfilled by this measure move.
First, the measure of a normal interval should be just the length of this interval.
Hence, just b minus a, where I say that b is, of course, larger than e.
Of course, this is what we want.
We want to generalize the measuring of lengths.
And now the second natural property is that we have some we call translation invariance.
This means that if we have a set A and shift that by a fixed vector x,
we don't change the length or the measure of this set.
This means that we get our move of A on the right-hand side here.
And there you have it, that is what we call our measure problem.
and such a measure mu would truly generalize the length measure of intervals.
However, for more than 100 years, it is known that this measure problem is not
solvable, so we can't define such a measure mu on the whole power set.
Therefore, it makes sense to build the whole measure theory with sigma algebra.
In the end, we will see that we can choose a suitable sigma algebra where we indeed can solve this measure problem.
However, in this video, I start by proving that the measure problem on the power set does not have a solution.
The claim I will prove now is the following.
Let me be a measure on the power set with the property 1 and 2, but I will weaken the property 1 a little bit.
I just want that the measure of the unit interval is finite, so not infinity.
This is of course fulfilled when property 1 is fulfilled, and why I excluded a 0 and this interval here you will see later.
In addition, MOU should now also fulfill property 2, which means it is translation invariant.
The result is now that there is only one measure that satisfies this.
And this is the zero measure.
And of course, the trivial measure is not what we want.
The trivial measure does not fulfill property one in our measure problem, so this explains this implication here.
The whole video is now about proving this.
and indeed this is a typical exercise you find in measure theory lectures.
Therefore I strongly advise you to try for yourself and then maybe fill in the details by using this And of course,
if you're not interested in technical details, you can happily skip this video and go to the next one in the series.
For everyone else now comes the proof.
I will structure it a little bit so I start with A by giving some definitions.
We will study the interval where we know that the measure is finite, so this is what I call capital I.
which means the unit interval where I exclude the 0 and include the 1.
And on i, I now define an equivalence relation.
And this equivalence relation should ignore rational numbers inside the interval, which means I define x and y as equivalent.
if and only if the following holds.
The difference between both of them is only a rational number.
So x minus y is in q.
This means that we don't want to distinguish numbers that only differ by a rational number.
We want to deal with these numbers as the same.
Therefore, the equivalence class of such an x can be written as x plus all rational numbers r.
So we put all these numbers in one box and call it the x with the brackets.
However, keep in mind, we defined the relation on our set i, so now we just live
on the unit interval, therefore these numbers should also live on the unit interval.
Therefore I have to add here that also x plus r lies still in the unit interval, so I can't.
big question on numbers because then I would leave the unit them all.
Otherwise, now we hit all the numbers we want.
There we have the well-defined equivalence relation with the well-defined equivalence classes.
To give some visualization I made throughout these boxes for real.
They are sets, sets with elements, so I could call this x1 and have one box here.
Then I have another one, so this would be x2, x3 and x4, and of course this goes on forever.
We don't know if we have countable many boxes as the picture suggests here,
so please be careful, but we know that we have disjoint the composition of the unit in wall.
This is what we always have if we define an equivalence relation.
Now we already reach the essential part of the proof, because we define a set A in i that describes all these boxes.
That means each element A describes where we present exactly one of these boxes.
In this picture I would look like,
okay, I choose one element here, maybe I call this A1, then I go to next box, I pick an element A2, here I pick out an element A3, here and so on,
and then I put all these elements.
And this is what I call the set A.
This looks nice now,
but keep in mind if we have uncountable many boxes,
this picture is not the correct way to represent this and therefore we need another definition for us.
But of course with exactly the same idea behind The first property is therefore for each box.
So for each equivalence class x I find such an a in a This just means picking out an element a out of the
And then the second property should tell us that this element is unique, so I only choose one element of each box.
So for all a,
b, and a, we have the property that a and b come out of the same box, x, we can imply that a is equal to b.
Hence, the definition tells us that the set A has exactly one representative out of each box.
The equivalence classes are therefore exactly represented by the set A.
However, it's not clear how to do this representation.
You don't find and construction how we can find the set A.
It looks very nice in the picture and there you might see we have a lot of possibilities to define such a set A.
But keep in mind if we are in the uncountable picture we might not know if such a set A could exist.
And indeed, the justification is very strong here.
What we need here is indeed the axiom of choice that is given in the set theory.
Therefore, it's an axiom that guarantees the existence of such a set A with exactly these two properties.
Okay, that was a lot, but we are still not finished with all the definitions I want to give.
Now we have fixed the set A, and now I want to shift it a little bit.
So I translate it by a rational number.
So in this word I define as An.
So this would be Rn plus the set A.
And Rn is a rational number,
more concretely, I want a sequence Rn, so this goes over the natural numbers, that enumerate the whole rational numbers.
That's not completely correct,
I just want an enumeration of the rational numbers into Yeah,
real interval minus one till one and of course why I want that you will see later However,
you should see that we can use that the rational numbers are countable when we want to apply the sigma adaptivity later and
Now we can finally go over to part B of my First, we show here that the sets I here defined are indeed disjoint.
So we have a n intersected with a m and this gives you the empty set if n is not equal to m.
The proof works easily by contraposition.
This means you could read this one here as an implication, so if n is unequal m, this applies that this sets are disjoint.
Now contraposition now means, okay, they are not disjoint and this applies n equals m.
So this is logically equal to m.
However, not being disjoint means there is an element we can choose out of this intersection.
So this is what we do.
And this implies immediately two properties.
Being an means, I can write x, s, r, n plus sum, no.
The lower case a in a so I have an a here or being in a m means I can write it
as Rm plus some a But of course these could be two different a's so I also give here an index to say this Okay,
but of course the X is the same on the left, so I can equal them.
This means that I have Rn plus An equals to Rm plus Am.
And then you see what I can do is put all the As on the one side and all the R's on the other side.
And now on the right, we subtract two rational numbers, so you know what comes out is also a rational number.
Now, remind yourself that we defined an equivalence relation exactly when the difference of two numbers is a rational number.
So in other words, a n and a m are now equivalent.
Well, to put this in other words, we also say our an is in the equivalence class represented by am.
If you want, you could also now add am here as well, so of course am is also in the equivalence We're that.
do And then you see we have property two for set A.
And it tells you if two elements come out of the same box or the same equivalence class, they have to be the same.
So the conclusion here is a n is equal to a n.
Okay, so this tells us the left hand side here is zero, but then also the right hand side is zero, and therefore we can again imply that also the rational numbers here are the same,
so Rn is equal to Rn.
but the RMs were chosen as an enumeration of the rational numbers and therefore here also the indices have to coincide.
And this proves now the claim by quantum position.
Very good.
Now I want to go over to the next part.
In part C I want to look at the union of these disjoint sets.
union n over n.
Okay, now keep in mind what the definition of a and wars, that was defined by our set A, that lives in the unit interval, shifted by rational
numbers, and that live in minus one to one.
This means by using the union,
I still should not be able to leave the interval that is given by minus one,
and now I shift one with maximum one, so here I have two.
And on the other side,
you can use how we defined a,
so this one, was representation of all the equivalence classes and where the equivalence classes were defined by the differences with the rational numbers.
And now I add back all the rational numbers.
So I should get at least the unity level out again.
Okay, so this one is to claim we should prove here in part C.
But already told you most of the things you need, therefore I think it's not hard for you to do the proof for yourself.
So the proof is an exercise for you.
Just put all the ideas I gave you now into formulas.
Okay, with all these parts in mind, I can now go to the core of the proof.
We now assume that we have a measure on the whole power set of the real line,
and it also should fulfill the two properties we have given in the claim.
And now we can use all the things above.
For example,
by our translation invariance to we can write that the measure mu of ARN plus
a is the same as the measure mu of a and this holds for all natural numbers n.
And maybe let us,
you see here immediately,
we know that the measure is always monotonic,
which means the measure of this set is less or equal than the measure of this set
and this is less or equal than the measure of this set.
So we have exactly this inequality here.
We need it later again, so I call the inequality here by star.
Before we go further, let us use now our second condition here, that the measure of the unit interval is at least finite.
Let us give this number a name, so I write the measure of zero one equals two number, and I call just capital C.
With this, we can indeed calculate the measure here.
We can use the properties of a measure, namely the sigma additivity.
So I split the set into unit intervals or shifted unit intervals.
here I go to zero and include it and then I have a disjoint union when I exclude the zero here and go to one,
include it here, again one and here two.
Okay, and here we can now use the sigma adaptivity.
Now we can write this as measure of the set plus measure of this set plus measure of this
set and by using the translation invariance into, you know, all this measure have the same value, namely C.
Hence we have C plus C plus C, so we see.
Okay, very good.
Now I want to use this inequality I called star before,
so this one,
what you should see immediately now is that on the left we have see itself, but on the right we now calculate it, please.
So let us write it down.
So I have c less than or equal than.
And now I can use this sigma editivity as always, because this one is disjoint.
Now this was part b from before.
So I have now here the sum or the series from 1 to infinity of mu of a n.
And this is less for equal than 3 times c.
And now we also know that we can get rid of the n here, because here you see it, this is an.
So, the translated version of a.
But by translation invariance, we know the measure is the same, so we can write move of A on this case.
And I want to write that down as important inequality here,
so C less or equal, the series of move A, and less or equal than 3.
So, please look closely at this inequality.
You see a fixed number mu of a in the middle and then this series.
Then you know the series will explode.
It you infinity if mu of a is created in zero.
So, the only possible case when this is finite needs mu of a equals to zero.
And this is the case because we know that c is less than infinity.
So the series has to be convergent.
By this calculation, we now can conclude that the measure of our set A has to be 0.
However, this means now the value of the series is zero, so we have zero in the middle and
left and right we have c and 3c, so there's also no other possible way, other than c to be zero.
However, remind yourself that c was just a short notation for the measure of the unit double.
Hence, this measure is also just zero.
And indeed, this helps us now, calculating the measure of the whole real line, just because I have translation invariance and sigma adaptivity.
So we split the real line into unit intervals and shift them.
So what we can do is just use an interval, starting with an integer m, and then go to m plus 1.
And if I use the union here,
which is then that is joined union,
where m goes over all integers,
I have what I want,
now I use sigma adaptivity, and then translation invariance, and then I get all all sorts out, adding zeroes stays at zero.
This now means that the volume or length of the whole read line measured in mu is just zero.
So we are dealing with the trivial measure, the zero measure.
And in fact, this is what we wanted to prove.
And there we have it, the full proof how to see that the measure problem is not solvable.
I know it was a long proof with a lot of technical details but I hope you learned something here.
And maybe...
this proof with some interpretation of the whole thing.
You saw that it is possible to construct such a set A, where we can't have a reasonable length or measure.
We would get contradictions if we are not dealing with the trivial zero measure.
So the only possibility to deal with such sets that behave so strangely that we can measure them is to exclude them from the beginning.
We won't measure all possible subsets.
We just deal with sets we then call the measurable sets.
Now, these are the sets that behave nicely enough such that we can solve the measure problem in this so-called sigma algebra then.
In fact, the bare sigma algebra you learned in part 2 of the series about is a correct choice to solve the measure problem.
We will go into details about this in the future,
but first in the next videos, I want to talk about maps that preserve our measurable structure here, so these are so-called measurable maps.
Okay, then.
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