Measure Theory

[Dhamnekar Win,odd]

The class of finite unions of arbitrary (also unbounded) intervals is an algebra on Ω=R (but is not a σ -algebra). How to prove it? 

I know an algebra is a class of sets A⊂2Ω which holds the following conditions

 

1)Ω∈A

 

2)A is closed under complements.

3)A is closed under unions.

σ -algebra fulfills all these three conditions. But in addition, it is closed under countable unions. Here countable means finite or countably infinite. 

I also know that R=(−∞,+∞) which is uncountably infinite unions of arbitrary (also unbounded) intervals.(is that correct?  🤔) 

Now, with these information available to me,, how can I answer this question? 

Edited March 16 by Dhamnekar Win,odd
I added more text.

[Dhamnekar Win,odd]

Errata: an algebra is a class of sets \(\mathcal{A} \subset 2^{\Omega}\) 

[Dhamnekar Win,odd]

Arbitrary Intervals are those intervals which include both bounded intervals and unbounded intervals. Bounded Intervals are the intervals with finite endpoints. e.g. [a,b], (a,b), (a,b], [a, b). Unbounded intervals are the intervals with at least one infinite endpoint. e.g. (- ∞,a), (a, ∞), (b,∞ ), (-∞, b), (-∞,∞) etc.

Now the class of finite unions of arbitrary including unbounded intervals \(\mathcal{A}\) is an algebra on [math]\Omega =\mathbb{R}[/math] because it is \( \cup \)-closed. e.g. \( \displaystyle\cup_{n=1}^{\infty} [n, n +1) \) is the finite unions of arbitrary bounded intervals belonging to the class \(\mathcal{A}\). 

It is also \( \setminus \)-closed.e.g \( \mathbb{R} \setminus \displaystyle\cup_{n=1}^{\infty} [n, n+1) \in \mathcal{A} \)   

\(\Omega \in \mathcal{A}\) Hence it is a algebra. 

But \(\mathcal{A}\) is not a \(\sigma\)- algebra because it is not closed under countable unions of finite arbitrary bounded intervals.e.g \( \displaystyle\cup_{n=1}^{\infty} [n,n+1 ) \)  is a countably infinite unions of bounded intervals.

Edited March 22 by Dhamnekar Win,odd