A generalization of this theorem is stated as follows,but it is not true:

(A false generalization of Egorov’s theorem) Let be a sequence of measurable functions that converge pointwise almost everywhere to another function , and let . Then there exists a Lebesgue measurable set of measure at most , such that converges locally uniformly to outside of ,and can not reach outside of .

Because if is from to ,then in the process of the proof, will not hold.

]]>I prove Lusin’s theorem myself,because I don’t like to see proofs provided by other people.

Theorem 18 (Lusin’s theorem) Let be measurable,and finite almost

everywhere, and let . Then there exists a Lebesgue

measurable set of measure at most such that the restriction of to the complementary set is continuous on that set.

Proof:For sake of convenience we ignore the measure zero set,let we

assume that is finite everywhere and measurable.Suppose that is pointwise approximated by a sequence of simple functions

.According to Egorov’s theorem,for any

given ,there exists a measurable such that ,and such that can be locally uniformly approximated by .

There is little difference between a simple function and a finite

linear combination of indicator functions on boxes.In fact,

, where a vanishing function means a function whose

support set has measure zero.

So it in fact does no harm to assume that is a finite linear combination of indicator

functions on at most countable numbers of elementary box(Why it does

no harm?).

Suppose that the support set of is ,where ,

is an elementary box who has positive measure.And , and are almost disjoint.

Now ,we take (or

),we cut a very small fraction of out,which is denoted by ,whose measure is ,after cut , is still an elementary box,and such that .

By using the trick "give yourself an epsilon of room",we can let

as small as possible.

So after this operation,the support set of becomes

,and we let be restricted on .After is restricted,it can be easily be verified that is continuous.

It is easy to verify that a if a sequence of continuous function convergent locally uniformly to a function,then that function is also continuous.

So Lusin's theorem is proved!

]]>My above comment prove Egolov’s theorem,however,my writing is bad,and

lose several key point.

In my opinion,the key words of proving Egolov’s theorem is “localize

and minimize”.We localize the region on which tends to infinity so as to make use of the downward monotone convergence theorem(In order to manage it successfully,we must make use of the regularity of ,because can be pointwise approximated by a sequence measurable functions,so itself must be measurable.Measurable function is lovely,it is very regular,i.e,it can be pointwise approximated by simple functions),after that,we use the trick “give yourself an epsilon of room”,to minimize the region on which tends to infinity.

We also localize the region on which is bounded,to avoid such case as “escape to horizontal infinity” happen.

]]>Now I understand that In exercise 4, is the uniform limit of bounded simple function means there EXISTS a sequence of bounded simple functions which approximated uniformly to …So the counterexample moving bump is no longer a counterexample,because can be approximated uniformly by .

]]>I think Exercise 4 is wrong,it says:

Exercise 4 Let . Show that is a bounded unsigned measurable function if and only if is the uniform limit of bounded simple functions.

A counterexample is the moving bump in Remark 7…

In order to repair,I think should be bounded in two directions,both horizontal and vertical.

]]>Now I disprove this question after it is fixed.The counterexample is in Remark 7—— moving bump.I think moving bump is an interesting example,it is as matter of fact a relationship between the reality and potential. is approximated by ,this is the reality.But can be approxiamted by ,this is the potential.

So if I change my question in this way,then it will be true:

Let be Lebesgue measurable,then there

exists a sequence of simple functions

which convergent pointwise to ,such that for any real number and positive real number which makes finite,there exists positive integer k such that .

I proved this theorem myself.Now I provide my general idea for proving this theorem.For sake of convenience,I take as a function from to .

For any point ,and any , If is bounded on ,then let it be.

If is not bounded on ,then let be as small

as possible,if when is small enough so that

becomes bounded on ,then let it be,in such case,we call as “not real infinity point”.If

becomes “arbitrarily” small but still keeps unbounded on ,then just let be small enough,in this case,we call as “real infinity point”.

In doing so,we will separate all those open interval in on which tends to infinity,and because we let be small enough,the region we separated has measure as small as

possible(Why?hint:consider the trick give yourself an epsilon of room

http://terrytao.wordpress.com/2009/02/28/tricks-wiki-give-yourself-an-epsilon-of-room/).

Now we consider those open interval on which is bounded.According to http://terrytao.wordpress.com/2010/09/19/245a-notes-2-the-lebesgue-integral/#comment-228116 It is easy to verify that on those interval, converges uniformly to .

So the Egorov’s theorem is proved!

]]>I enjoy Mr.Tao’s proof of this lemma very much.It is a beautiful,elegant proof!I love this proof very much!

In fact,I prove it in somewhat different way,first,I approximate by simple functions,then I prove the triangle inequality for simple functions.

But Mr.Tao’s proof is really an amazing proof,in fact,Mr.Tao’s proof has a strong geometric meaning,that is,the smallest path between twp point is a line,because any curve (or polyline) through the two points can project on the line,and the function of projection can only make the length shorter !

Especially I enjoy the formula

This formula strongly reminds me of Fourier transform(Which I only learned from Mr.Tao’s Analysis II),so I think this formula combined with my geometric meaning will give me a very vivid geometric meaning about Fourier transform!

]]>Let be Lebesgue measurable,and which is pointwise approximated by simple functions .For any real number and positive real number ,there exists positive integer such that .

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