# is the caratheodory Criterion both necessary and sufficient?

Let F be a field over probability space ยง..the caratheodory extension criterion only extends this to a sigma algebra generated by F..i know that a larger sigma algebra cannot satisfy the countable additivity of lebesgue outer measure, but is there one which can satisfy the properties of probability function even if not that of lebesgue measure?

• To improve your question, please take out the ALL CAPS phrases, they look like you are shouting at us. Also, please fix the symbols in your question. – Lee Mosher Aug 1 at 14:27
• Sure, will do that..but I don't know how to get the omega symbol.. – A.G Aug 1 at 14:49
• The reason for CAPS was to emphasize the difference between this and a very similar question asked on this site – A.G Aug 1 at 14:57
• To improve your symbols, take a look at the mathjax guide. – Lee Mosher Aug 1 at 15:21

## 1 Answer

Let $$m^*$$ be Lebesgue outer measure on $$[0,1]$$. Let $$A \subseteq [0,1]$$ be a non-measurable set. That is, $$m^*(A) + m^*([0,1]\setminus A) > 1$$. Of course $$A$$ fails the Caratheodory criterion. But there is a countably-additive extension $$\lambda$$ of Lebesgue measure to a sigma-algebra including both the Lebesgue measurable sets and $$A$$.

Is that the question?

• No..my question was over any space and any field over it..not only the real line and borel field.. – A.G Aug 1 at 17:08
• Yes I know it fails the caratheodory criterion..but why is the criterion necessary for the sets to satisfy the properties of the more fundamental probability measure? – A.G Aug 1 at 17:15