Can Borel sets be incomplete in Lebesgue measure?

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Discussion Overview

The discussion revolves around the completeness of the Lebesgue measure when restricted to the sigma-algebra of Borel sets. Participants explore the implications of the definitions of Borel sets and complete measure spaces, particularly focusing on whether subsets of measure zero sets are included in the Borel sigma-algebra.

Discussion Character

  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant requests clarification on the statement that the Lebesgue measure restricted to Borel sets is not complete, indicating a misunderstanding of the definitions involved.
  • Another participant argues that a subset of a set of measure zero is not necessarily in the Borel sigma-algebra, challenging the initial assumption made by the first participant.
  • A clarification is provided that the Borel sigma-algebra is the smallest sigma-algebra containing all open sets, which leads to a discussion about the implications of this definition.
  • It is noted that the set A in question is not an open set, which is a key point in understanding the completeness issue.
  • One participant explains that there exist Borel sets of measure zero that have subsets not contained in the Borel sigma-algebra, emphasizing the incompleteness of the Lebesgue measure in this context.
  • A later reply mentions the potential requirement of the axiom of choice to construct such a set A, although the proof is not provided.

Areas of Agreement / Disagreement

Participants express differing interpretations of the definitions related to Borel sets and measure completeness. There is no consensus reached on the implications of these definitions, and the discussion remains unresolved regarding the specifics of constructing a set A that illustrates the incompleteness.

Contextual Notes

Participants highlight limitations in their understanding of the definitions of Borel sets and the properties of sigma-algebras, particularly regarding the inclusion of subsets of measure zero sets.

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

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

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