Sigma Algebras on [0,1]: Understanding Measures and Sub-Intervals

[Old Monk]

This topic came up while studying measures on sub-intervals of [0,1]. The collection of all intervals in [0,1] is a semi-algebra, say J. Now from finite disjoint union of members of J let's say we form a set A.

I was able to prove that A is an algebra, since for any C,D ε A, C$\cap$D and C$^{c}$ belong to A.

I'm not able to understand why A isn't a σ-algebra. Can anyone please outline a proof or give me a counter-argument.

 

[micromass]

Are singletons in A?? Is $\mathbb{Q}$ in A?

 

[Old Monk]

Yeah singletons are in A. Since singletons of the form (a,a) exist in J, the single union of such elements should exist in A. That I assume implies, all the rationals in [0,1] are also in A.

 

[micromass]

Yeah singletons are in A. Since singletons of the form (a,a) exist in J, the single union of such elements should exist in A.

Isn't $(a,a)=\emptyset$?? Singletons would rather come from $[a,a]=\{a\}$. But ok, let's assume that the singletons are in A.

That I assume implies, all the rationals in [0,1] are also in A.

Why??

 

[Old Monk]

Yeah sorry, that's [a,a].

We have [0,r)$\cup$(r,1] in A, if r is any rational in [0,1]. Therefore the element {r} ε A. So I think we could generalize this to all r ε Q$\cap$[0,1].

 

[micromass]

Yeah sorry, that's [a,a].

We have [0,r)$\cup$(r,1] in A, if r is any rational in [0,1]. Therefore the element {r} ε A. So I think we could generalize this to all r ε Q$\cap$[0,1].

Yeah, I agree that every {r} is in A. But why does that imply that $\mathbb{Q}\in A$. Is it even true??