# Tagged Questions

**10**

votes

**2**answers

365 views

### Is sigma-additivity of Lebesgue measure deducible from ZF?

Is sigma-additivity (countable additivity) of Lebesgue measure (say on measurable subsets of the real line) deducible from the Zermelo-Fraenkel set theory (without the axiom of choice)?
Note 1. ...

**4**

votes

**2**answers

590 views

### Are all models of ZF + DC + “All set of reals are lebesgue measurable” also models of CH? [duplicate]

Possible Duplicate:
Lebesgue Measurability and Weak CH
I have studied a little set theory and I found that Solovay constructed a model of ZF+DC+"All set of reals are Lebesgue measurable" and ...

**4**

votes

**3**answers

290 views

### Bounded operators and axiom of choice

In the article below, it is shown that the proposition "Every linear operator defined on a whole Hilbert space is bounded" is consistent with the axioms of ZF + a weakened version of the axiom of ...

**1**

vote

**3**answers

495 views

### Axiom of Choice and Vitali's theorem [duplicate]

Possible Duplicate:
Axiom of choice and non measurable set
I am told somebody has shown the equivalence of the Axiom of Choice with
existence of non-measurable Lebesgue sets on the real line, ...

**18**

votes

**2**answers

2k views

### Can a Vitali set be Lebesgue measurable? (ZF)

Here is the definition of Lebesgue measure.
The standard proof that Vitali sets are not Lebesgue measurable uses countable additivity of Lebesgue measure, which is not a theorem of ZF. (In ...

**6**

votes

**5**answers

873 views

### Is Lebesgue/Borel non-measurability actually caused by non-uniqueness?

In ZFC, every construction of a Lebesgue or Borel non-measurable set uses the axiom of choice. None of them that I've seen use choice to define a unique set, even though it's entirely possible to do ...

**5**

votes

**4**answers

2k views

### Does constructing non-measurable sets require the axiom of choice?

The classic example of a non-measurable set is described by wikipedia. However, this particular construction is reliant on the axiom of choice; in order to choose representatives of $\mathbb{R} ...

**27**

votes

**4**answers

7k views

### Non Borel sets without axiom of choice

This is a simple doubt of mine about the basics of measure theory, which should be easy for the logicians to answer. The example I know of non Borel sets would be a Hamel basis, which needs axiom of ...