# Lebesgue measure on borel set

1. Mar 17, 2007

### redrzewski

Can someone show me an example to clarify this statement from Royden's Real Analysis:

The Lebesgue measure restricted to the sigma-algebra of Borel sets is not complete.

Now, from the definition of a complete measure space, if B is an element of space M, and measure(B) = 0, and A subset of B, then A is an element of M.

But my understanding of the Borel sets is that it is the smallest algebra containing all the open and closed sets. Hence A would be in the Borel set, hence A would be in M.

So I'm obviously missing something.

thanks

2. Mar 17, 2007

### matt grime

Why is A in the set? There is nothing in what you wrote that compels a subset of a set of measure 0 to be in the Borel sigma algebra. You say it is the *smallest* sigma algebra, but behave as if it is the *largest*. A would be in the set if it could be obtained from the open (or closed) sets by operations of intersection, union, and complement.

3. Mar 17, 2007

### redrzewski

That makes sense. I guess I misinterpreted the definition.

"The collection B of Borel sets is the smallest sigma-algebra which contains all the open sets."

I read "contains" to mean: every open set A is an element of collection B.

But apparently, what "contains" means here is that every open set A is a subset of some element of B.

thanks for the help.

4. Mar 17, 2007

### matt grime

A is *not* an open set. That's the point.

5. Mar 17, 2007

### redrzewski

Thanks again.

I think I've got it now.

6. Mar 17, 2007

### AKG

The collection B of Borel sets is the smallest sigma-algebra containing every open set. In particular, this means that if A is an open set, then A is an element of B. The Lebesgue measure m restricted to B is incomplete. This means that there is some A in B such that m(A) = 0 which has some set C as a subset of A such that C is not in B. The only open set with measure 0 is the empty set, and every subset of the empty set is contained in B. But there is some other element of B which isn't an open set, it has measure 0, and it also has a subset which isn't an element of B. This set A is probably some whacky set. A sigma-algebra which contains all the open sets doesn't only contain open sets, it also contains all countable intersections of open sets (which aren't necessarily open), and all countable unions of countable intersections of open sets, and all countable intersections of countable unions of countable intersections of open sets, etc.

7. Mar 18, 2007

### matt grime

I seem to remember reading somewhere that to make such an A requires the axiom of choice, but I don't know the proof.