Okay, so to finish this material on measure theory,
we're going to look at how you show that there is
a non-Lebesgue-measurable set,
which will also be a non-Borel set as well.
Because every Borel set is Lebesgue measurable.
Now you don't have to do this
to get a non-Borel set.
In fact, you can prove that
the number, cardinality-wise, of Borel sets is,
the same as the number of real numbers. And there are
an awful lot more subsets of the real numbers than that!
So there ... so you can get
immediately, just by a cardinality argument, that
very few subsets of the real, you know ... cardinality-wise ...
a negligible number of subsets of the real line are Borel sets.
Because the number of Borel sets is just the number of points in
the real line, and the number of
subsets of the real line much bigger than that.
The fact that any set you ever write down turns out
to be a Borel set has got nothing to do with it!
Right? Most sets are not Borel sets,
but you'll just never meet one!
And Lebesgue measurable is even worse!
Lebesgue measurable ... unless you use,
the axiom of choice somewhere,
everything could be Lebesgue measurable.
But if we use the axiom of choice, then we will get
some sets that are not Lebesgue measurable.
Oh, and there are some explicitly defined non-Borel sets
around in the literature, but we won't worry about those.
Now, according to Wikipedia,
Vitali did this in 1905.
So ... over a hundred years ago now! And it looks very innocuous.
We're going to define
an equivalence relation on the unit interval
by saying that two elements of the unit interval are related
if their difference is rational.
Now it's really easy to check that that's an equivalence relation.
Obviously x minus x is zero,
so x is related to x, because zero is rational.
x - y is rational if and only if y - x is rational.
And transitivity ... oops, lost my pen!
Transitivity is because the sum of two rationals is rational.
So it's an equivalence relation,
It partitions the closed unit interval into equivalence classes,
because that's what equivalence relations do.
The equivalence classes themselves are
basically translates of the rationals, intersected with [0,1].
So they're ... each equivalence class is countable, in itself.
So if you look at any one of these equivalence classes it's
only got countably many elements.
What does that tell you about how many
equivalence classes there must be ...
in the closed unit interval?
If each of them is countable ...
if each of the equivalence classes is countable,
how many equivalence classes are there going to have to be ...
in order to get ... roughly speaking?
Student: uncountably many. Uncountably many, that's right!
If you had only countably many equivalence classes,
then you'd take their union,
would be a countable union of countable sets, would be countable.
But the closed unit interval isn't countable ...
so ... as we know by Cantor's argument,
so there must be uncountably many equivalence classes.
So there's the first thing.
We don't seem to have done anything much yet,
but there are uncountably many equivalence classes in here.
And what we're going to do is take one point from each.
Each of these equivalence classes has got infinitely many points in.
So there's plenty of points to choose from!
All you have to do is take one point from each.
That sounds harmless enough!
But in doing so, you'd use the axiom of choice.
All you have to do is believe that if you've got infinitely many sets,
each of which has got infinitely many elements,
that you can find some set ... well, ok ...
and they're disjoint as well, remember. The equivalence classes
don't overlap each other.
All you have to do is choose one point from each.
Seems reasonable that
you should be able to find a set like that.
But you will never meet that set in the real world!
Otherwise you will have a ...
well ... prove something almost as good as the axiom of choice.
You will never actually come across this set in the real world.
This set is really nasty! Right ... so ...
There's your set E: exactly one point from each equivalence class.
There's uncountably many of them, so this is an uncountable set.
But it is a weird uncountable set!
Its translates by rational numbers don't meet each other,
that's the first thing, no matter how you translate it around.
Well they can't, because,
if two of these translates met each other,
that would result in there being two different points in your
set E which are equivalent to each other,
which would mean their difference would be rational ...
or rather the other way round, there would be two different points
in E whose difference is rational,
and then they'd be equivalent to each other.
But E has only got one point from each equivalence class.
So none of these translates overlap.
I'll leave it to you to check the details of that.
Also, any y you take in the closed unit interval
is equivalent to something in E. I'll do this one!
... y in [0,1] ...
and E has one point from every equivalence class.
So there is an x in E which is equivalent to y.
But then y - x is rational,
and ... call it q.
Of course x is in E ... y is equal to q + x,
which is one of the elements of q + E.
But there's a little bit extra I've said up there,
just a tiny bit extra. What did I say?
I said that q was rational and was
between -1 and 1 inclusive.
You might be able to do slightly better than that.
Why is q between -1 and 1?
Well, q is y minus x, x is in E,
E is a subset of [0,1],
y is in [0,1]. So, if you take the difference of two numbers in [0,1],
you get something between -1 and 1, right?
Okay, so that did that bit.
I left one little bit in there as an exercise,
the fact that translates were disjoint.
But that's again ...
that's just for the reason that there's only one ...
that there's only one point from each equivalence class in E.
Let's call the intersection of Q with [-1,1]
S, just to save a bit of writing.
That's just a nice countable set.
And they're the translates we're going to be using.
And ... oh, you ... use a sequence to use them up,
just so that we know that that we're taking a union of a sequence.
But if you want to ... you know ... if you're happy working with just
countable unions anyway that we didn't have to do this.
But just so that I can put it into
the obvious sequence form, so that we can use our
measure and outer measure stuff in the usual way,
let's form a sequence which, which has ...
which goes through the whole of S, every point mentioned
exactly once in the sequence, since it's countable.
And now we look at all
these translates, the ones we mentioned a moment ago,
all the q + E's where q is in S.
Which you could also write as the union of the q_k + E's.
And, as we said a moment ago,
anything, any y in [0,1], is in one of these translates.
So what that tells you is that the closed interval [0,1]
is a subset of this union.
That's just from above.
What about this one?
That A is a subset of [-1,2]?
Well, E's a subset of [0,1]. And now you're going to look at
translating it by numbers between -1 and 1.
So you can't escape from the closed interval [-1,2].
So that's what's going on there.
We've got: A is sandwiched between two sets.
So we know what lambda star does, roughly speaking,
lambda star is monotone. Whatever lambda star of A is,
it's going to be between 1 and 3.
Because the length of this interval is one,
and the length of this interval is three,
and lambda star is monotone,
so lambda star of A has to be between 1 and 3.
Now we're going to see, well,
so far, everything is just true.
What we're going to do now is to get a contradiction by
assuming that it's measurable.
So this is what I've just said.
One has to be less than or equal lambda star of A has to be
less than or equal to three, by monotonicity.
Also, those translates have all got the same lambda star,
Lebesgue outer measure, as the original set. Translates have got ...
We know that [taking] translates doesn't change length or
even this generalized notion of length that this lambda star assigns.
Now we're finally going to make
an assumption to get a contradiction from.
Let's suppose that E is Lebesgue measurable.
Then the translates are Lebesgue measurable too,
because that's one of the properties we said about
Lebesgue measurable sets.
They're Lebesgue measurable and they're pairwise disjoint,
because they're translates by rational numbers ...
We know that no two rational numbers
will give you an overlapping translates of ...
of this weird set E.
That's how E was chosen!
Exactly one point from each equivalence class.
So translating by different rational numbers can't lead to overlaps.
And that means you can add up!
Adding up is because lambda is a measure.
And we're talking Lebesgue measurable sets now.
In fact, if you go back and look at this afterwards,
the conclusion is that the lambda stars definitely don't
add up for these correctly.
That's what we're going to prove in the end,
because we're about to get a contradiction ...
The lambda stars do not add up correctly for these.
If you assume they're Lebesgue measurable,
then they should: lambda and lambda star would agree,
and the Lebesgue measures should add up correctly.
But there is a problem!
It's supposed to be coming out as something between minus one ...
between one and three.
But we're adding the same number together infinitely often.
Now, think of that ... imagine that number is 1.
Then what you're doing is 1+1+1 ...
+1+1+1+1 ... and you should get infinity.
On the other hand, if the number is 0,
then you're doing 0+0+0+0+0... and
you get 0. And you think about it just a little bit ... [it] doesn't
take long to realize that you've either got to get 0 or infinity,
if you add the same number to itself infinitely often.
This is a problem because it's supposed to be between 1 and 3!
Ok?
So if it's either 0 or infinity,
it can't be between 1 and 3,
and that's your contradiction. Conclusion?
Conclusion 1 is that E is not Lebesgue measurable.
Conclusion 2: even without ...
even without that, that set E,
the lambda stars definitely don't add up correctly.
And that's even ... that's out ... that lives out ... that lives on
outside the contradiction argument.
So lambda star goes wrong when you add them up there.
You can also prove that lambda star must be positive
for that set, because in fact
if lambda star of a set is 0,
then it's countably subadditive,
and you'd have got ... everything would have to be 0 again.
So you can prove that it must have positive lambda star,
disjoint translates which have also each got positive lambda star,
but the lambda star of the union is no more than three.
So, very much, the lambda stars don't add up.
So you can't ... you can't come up with
any good way to define a total length of this set,
that works with translation and adds up
correctly on these translates, and so on.
If you've got the kind of properties we're after, it's hopeless!
So E is a really horrible set,
much worse than anything we've met before.
And the existence of such a horrible set,
requires the axiom of choice. So it's guaranteed to be a
really horrible set that you'll never meet in the real world.
Then if you want to work with area, you can just take E cross [0,1].
You could probably use E cross E as well, but
E cross [0,1] is a bit more obvious, which gives you the sort of ...
a vertical line at every point of E.
And then the translates will cover the unit square,
and you get exactly the same problem working with area.
So area has problems too,
just E cross [0,1], and then you can go for volume as well, and so on.
And finally, then, some optional reading.
We've just done Measure Theory and
we didn't do integration theory.
You ... integration theory is where the measure theory ...
measure and integration gets really good!
But it's another huge chapter which we don't have time for,
because we want to do some functional analysis in this module!
We could throw out the functional analysis and do
integration theory instead, but that's not the plan!
So if you want find out about integration theory,
there are lots of books you can look in:
Rudin, Real and Complex Analysis,
or there's Chapter 4 of my Measure and Integration stuff,
which is available online, with audio.
So there's lots of different ways
you can find out about integration theory.
And there you can find out about
the Monotone Convergence Theorem, and
Fatou's Lemma, and the Dominated Convergence Theorem,
and great stuff like that.
And finally find out how to integrate
the characteristic function of the rationals, and get 0,
which the Riemann integral couldn't do.
The Lebesgue integral says,
well, the Lebesgue measure of the rationals is 0,
so the integral of the characteristic function of the rationals
is zero too. And that's it! In fact, line 1
of the definition of the Lebesgue ... essentially the first ...
the very first bit of
defining the Lebesgue integral, immediately gives you that
the characteristic function of the rationals integrates to 0.
And you've finished.
And that was the one that the Riemann integral couldn't do.
So that's got to be a good sign, when the theory of
Lebesgue integration gets you to
solve the characteristic function of the rationals instantly!
So I highly recommend that you find out about the
Lebesgue integral, as a very interesting study.
But we've reached the end
of this topic, as far as this module's concerned,
and so we'll stop there for today.