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### A Sigma Algebra that Contains Intervals (a,infinity) Contains Every Interval

```Date: 09/21/2010 at 11:03:55
From: saba
Subject: segma- algebra (real analysis)

Prove that if a sigma-algebra of the subset of real numbers contains all
intervals of the form (a,infinity), then it contains all intervals.

I know that the sigma-algebra of a set is a Lebesgue measurable set, which
contains all intervals and all open sets. Now, if a sigma-algebra contains
an interval of the form (a,infinity), then the complement of this interval
also, by definition, is contained by the sigma-algebra. I know that every
interval is measurable. And I know that the collection of measurable sets
is a sigma-algebra.

But if I start with the interval (a,infinity), how do I prove that other
intervals are contained by the sigma-algebra?

```

```
Date: 09/22/2010 at 21:11:23
From: Doctor Jordan
Subject: Re: segma- algebra (real analysis)

Hi Saba,

As I understand it, you want a proof that if F is a sigma-algebra (the
elements of which are subsets of R) that contains all intervals of the
form (a,infinity), then F contains all intervals.

So we want to show for any real numbers a < b that the interval (a,b) is
an element of F.

We are given that (a,infinity) and (b,infinity) are elements of F. The
complement of (b,infinity) is (-infinity,b], and this is an element of F
because F is a sigma-algebra. The intersection of (a,infinity) and
(-infinity,b] is (a,b], and this is an element of F because F is a
sigma-algebra.

However, we want to show that (a,b) is an element of F, not that (a,b] is
an element of F. At this point, I got stuck for a while.

If we knew that [b,infinity) were an element of F, then its complement
(-infinity,b) would be an element of F. The intersection of (a,infinity)
and (-infinity,b) is (a,b), which is what we want to be an element of F.
Therefore, to show that (a,b) is an element of F, it suffices to show that
[b,infinity) is an element of F.

Because F is a sigma-algebra, if A_1,A_2,... are elements of F, then their
intersection is an element of F; namely, a sigma-algebra is closed under
countable intersections.

Let

A_n = (b - 1/n,infinity)    for n = 1,2,....

Each set A_n is an element of F, because it is an interval from some real
number to infinity. Therefore the intersection of the A_n's is an element
of F. The intersection of the A_n's is [b,infinity). Check this by showing
that every element of the intersection of the A_n's is in [b,infinity),
and that every element of [b,infinity) is in the intersection of the A_n's.

Therefore [b,infinity) is an element of F, and hence (a,b) is an element
of F. Thus if every interval of the form (a,infinity) is an element of F,
then every interval is an element of F.

- Doctor Jordan, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 09/22/2010 at 22:17:21
From: saba
Subject: Thank you (segma- algebra (real analysis))

Thank you very very very much.
```
Associated Topics:
College Analysis

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