# Lebesgue measure on borel set

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

matt grime
Homework Helper
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.

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.

matt grime
Homework Helper
A is *not* an open set. That's the point.

Thanks again.

I think I've got it now.

AKG