Dear Shakeel,

You would find it from the fact that the collection of measurable sets is a sigma-algebra. In case you are not aware of what is sigma-algebra, please have a look at G. de Barra's book on measure theory & Integration, page 30, theorem 4 and definition 3. And probably page 15 will give you the answer how to start, where he defined the symmetric difference of two sets.

Hope it will help.

http://books.google.co.in/books/about/Measure_Theory_And_Integration.html?id=yKQOPfcEIFUC&redir_esc=y

If you want to go bare bones and prove from scratch you would probably have to prove that the symmetric difference can be expressed as the union of two measurable sets. I would suggest thinking of ways to re-express the sets into pieces, each of which are measurable.

Let A and B be two \mu-measurable sets, i.e. 0 \leq \mu(A) , \mu(B) < \infty. Let furthermore A and B be elements of a \Sigma-algebra.

The symmetric difference is defined as the disjoint union (A\B) \cup (B\A). Note that by construction both (A\B) \subset A and likewise (B\A) subset B are elements of the \Sigma-algebra and thus each (A\B) and (B\A) are \mu-measurable (e.g. \mu(A\B)

You say that proving measurability of union and of section was easy. In my reference book the proof for the union is 1/3 page. Then the book is clever and turns from the union directly to the complement: the proof takes 1/2 page and looks a bit more involved than the proof for the union. From this it builds section, difference, and symmetric difference by successive application of union and complement, i.e. by operations that were proved to transform measurable sets into measurable sets.

So you see, you need the prove of: The complement of a measurable set is measurable. Unfortunately, there are many ways of defining the concept of measurability. The details of any proof strongly depend on the definition under consideration. Consulting a book or asking your teacher (he knows what definition he was using) is the most efficient way to give the final answer to your question.

Note that many books employ an axiomatic approach (from your words it is clear that this is n o t the situation you have in mind, and so Tapan has to ask himself where his thought were when he was writing his answer) and introduce measurable sets as any sigma-algebra of sets. The axioms for sigma algebras say that the contain unions and complements to all its members, so that the symmetric differences are covered in a trivial manner. The non-trivial part is only deferred, however: when later specific measures and specific systems of measurable sets are to be studied one has to prove that some system of sets is in fact a sigma algebra and encounters just the proof problem for the complement I was speaking about before.

@Markus.

You misunderstand measurability as the property of having finite measure. These properties are completely unrelated however. R is Lebesque-measurable although the Lebesque measure has an infinite value for it. Look into one of the many books and wonder.

@Ulrich.

you are absolutely right. When restricting A and B to \Sigma-algebras the claim is a "Taschenspielertrick" (or triviality) as demonstrated. For the more general case we have to put a bit more effort into the problem. And we also need to specify the measures, put conditions on the sets A and B to prevent any Banach-Tarski paradox. P.S.: I am an old-fashioned constructivist and have never used Zorn's Lemma... Do we really need infinity?

I got my answer.If someone can prove the union and intersection to be measure-able then one can use those results in the definition of symmetry to prove it measure-able.Sir Ulrich Mutze is right..I am thankful to all of you and specially to Sir Mutze.He talked about it in detail.

@Shakeel: saying > makes your statement exactly right. Don't worry!