The unmeasurable constructed in the post is called Vitali unmeasurable set,in fact,from its construction,Vitali set is highly undetermined,because it uses Axiom of Choice. The outer measure of the Vitali set ranges over ,in fact,I think from its construction,we can that the outer measure of a Vitali set can be any real number in .

It need not to be dense(But it could be dense),but it is must be uncountable,because any countable set is measurable.

Continuously differentiable reminds of the motion of a mass point in physics. is continuously differentiable.

Maybe with this condition,we can compute the length of the curve?Does a curve which has length in higher dimensional space has outer measure zero?What does a curve which has length mean?
———–

The above is what I thought when I didn’t know how to solve this problem.Now I figure out it,the condition “continuously differentiable” enable us to approximate this curve by a polygonal line very well!Note that “continuous” is also needed,otherwise,the curve may change so rapidly in an arbitrarily small scale that we can not approximate it well by using polygonal line.

Unfortunately(Or fortunately?),there is real number system,so we need Lebesgue measure.

Jordan measure is suitable for ,Lebesgue measure is suitable for .

So if we only use Jordan measure in ,that will be not enough.

Jordan measure is dense,but Lebesgue measure is complete!

Complete stuff is real infinity,and dense stuff is not real infinity.Like a natural number,a natural number is finite,but the set of all the natural number is infinite!

In my comment http://terrytao.wordpress.com/2010/09/04/245a-prologue-the-problem-of-measure/#comment-226452,I have proved this result in the case of Jordan measure.

Now I begin to prove this result in the case of Lebesgue measure.Of course,I will try to base my argument on my proof in the case of Jordan measure.

So the proof becomes simple,based on two facts:

1.A Lebesgue measurable set can be arbitrarily approximated by a Jordan measurable set.

2.A linear transformation from to itself will not transform a set whose outer measure is very small to a set whose outer measure is far larger.

Fact 1 can be easily proved by the definition of Lebesgue measure.

Fact 2 can be easily proved by the change of variables formula in Jordan measure.

Inner regularity is just a counterpart to outer regularity.Just consider the complement of the measurable set will be enough.

But now I still have something to think.In the case of outer regularity(lemma 9),why has to be open?

Open set.I frequently see open set in real analysis.Why open set?I think one reason mathematician like open set,is that,open set has a good property:

Every open set in can be expressed uniquely into at most countable union of open boxes.

This property makes us define the outer measure more conveniently.Because it is easy to define the volume of at most countable union of open boxes.

So if we don’t use open set in outer measure,to reach the same effect,we have to use at most countable union of elementary boxes instead,this becomes more inconvenient in language.

This exercise is stated as follows:

“Show that if the are all Lebesgue measurable, and converge pointwise to , then is Lebesgue measurable also.”

I love this exercise,because this exercise makes me think again about the distinction between the uniform convergence and pointwise convergence.

I think uniform convergence correspond to the finite intersection of sets,and pointwise convergence correspond to the infinite(countable or uncountable) intersection of sets.

In this particular exercise,pointwise convergence correspond to the countable intersection of sets(Because is countable).

It seems that the condition here is stronger than in Mr.Tao’s book
“Analysis”,exercise 18.2.3.

In that book,the Downward monotone convergence theorem is stated as
follows:

Show that if is a decreasing
sequence of measurable sets,and ,then we have
.

At that time,my proof to that exercise is stated as follows:
——————————–

Firstly,by the algebra property,we know that
is measurable.

Then,it is easy to verify that .So .

According to the finite additivity of disjoint measurable sets,we have
.
( is finite,so the operation of minus makes sense).

According to Exercise 11,(1)(Upward monotone convergence),we have
,so .

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But,surprizingly,the condition in Exercise 11,(2) is a bit
stronger than in Mr.Tao’s book, does not need to be
finite,in the case of when is not finite, is
finite,the theorem also holds!

So I think I need to change my proof a bit to prove this stronger
theorem.Now I figure out it,I even do not need to change my
proof,because ”the stronger theorem” only make use of the fact that
the first finite elements of a sequence do not affect its limit!HaHaHaHa…

This lemma tells us the simple fact that disjoint Lebesgue measurable sets has the property of countable additivity,the proof,however,is rather tough,and long.

When faced with such a proof,I usually do not like to read it,because even though I read it,the proof still belongs to the one who find out it.Mathematics,do not like other subjects such as language——Even if you see a proof many times , it still do not belongs to you.In order to truely understand a proof,you have to find it out yourself,or,if the proof is so hard that it does not worth to invest too much time in it, at least you should try to simplify it or rewrite it in your own way,or decompose it in simpler steps.

So I find a proof myself,I think my proof is a bit simpler than Mr.Tao’s proof.Now let me start the proof.

As a preliminary,I have already proved that a set is L-measurable iff it satisfy the Carathéodory criterion.

Let ,We first prove that is L-measurable,that is ,according to Caratheodory criterion,,.

According to the sub-additivity property of outer measure,we know that ,so we only need to prove that .

According to the Morgan law,we just need to prove that

.

According to the outer regularity(Lemma 9,see also my comment http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/#comment-227019 ),we just need to prove that

.

According to the Monotonicity property of outer measure,we know that .

So we just need to prove that

………….(1)

According to the L-measurability of (together with the mathematical induction),we know that

,

so (1) holds.So is L-measurable.Now we prove that

.

According to the outer regularity(Lemma 9,see also my comment http://terrytao.wordpress.com/2010/09/09/245a-notes-1-lebesgue-measure/#comment-227019 ),we have

.

Then according to the finite additivity of disjoint L-measurable sets(The finite additivity of disjoint L-measurable sets can be easily proved by using Caratheodory criterion together with the mathematical induction.),we have .

Done.

Outer regularity can also be stated as this way,which is also the way I prefer:

“Let be an open set. Then can be expressed as the countable union of almost disjoint boxes (and, in fact, as the countable union of almost disjoint closed cubes).”

I think that the following is also true:

Let be an open set. Then can be expressed UNIQUELY as the countable union of disjoint open boxes.