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This question is based on the following from Elements of Integration and Lebesgue Measure by Bartle. We are given the following example and exercise:

Let X be the set ${\mathbf R}$ of real numbers. The Borel algebra is the $\sigma$-algebra generated by all open intervals $(a,b)$ in ${\mathbf R}$. Observe that the Borel algebra ${\mathbf B}$ is also the $\sigma$-algebra generated by all closed intervals $[a,b]$ in ${\mathbf R}$

Exercise 2.B. Show that the Borel algebra ${\mathbf B}$ is also generated by the collection of all half-open intervals $(a,b]=\{x\in {\mathbf R}: a<x\leqslant b\}$

The definition of sigma algebras we are given is that they are unions and complements of a family of sets on a universal set $X$.

My thought on the way to solve this was to use something like $$ (a,b)=\cup_{n=1}^\infty[a+1/n,b-1/n] $$ and similarly $$ (a,b)=\cup_{n=1}^\infty(a,b-1/n] $$ to show that we can construct $(a,b)$ as a union of either the closed or the half-open intervals.

What I'm wondering about is, first I would like to check that the above is a correct answer of course, and also I wonder whether I've missed some other trick which doesn't involve these constructions.

Suzu Hirose
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2 Answers2

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Yes, you have the right idea. But the solution you gave is not complete.

Let $\mathfrak{B}$ be the Borel set generated by intervals of the form $[a,b]$ and $\mathfrak{B}^*$ be the Borel set generated by elements of the form $(a,b)$

By the relation you gave,

$$(a,b)=\bigcup_{n=1}^\infty [a+1/n,b-1/n]$$

You showed that any open interval $(a,b)$ can be written as a countable union of closed sets.

Hence, $(a,b)\in \mathfrak{B}$

Now since each element of $\mathfrak{B}^*$ is a countable union, intersection and complement of open intervals. Thus you have

$$\mathfrak{B}^*\subseteq\mathfrak{B}$$

Now, you also need to show $\mathfrak{B}\subseteq\mathfrak{B}^*$.

I'm sure, since you have the right idea, that you'll be able to show that too.

Swapnil Tripathi
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  • Thanks for your confirmation. I was hoping there was some kind of clever trick without all these constructions. – Suzu Hirose Nov 29 '14 at 02:05
  • "Now since each element of  is a countable union, intersection and complement of open intervals. " Why would any Borel set in $B^*$ be of that form? – Aman Nov 29 '14 at 05:38
  • @Aman: Let $x\in\mathfrak{B}^*$ then $x$ is a countable union, intersection and complement. Wherever, there is an open interval in the expansion of $x$ you can replace it by the above expression and thus you have. $x\in\mathfrak{B}$ – Swapnil Tripathi Nov 29 '14 at 08:57
  • The point I wanted to make was that not every borel set is a countable union, intersection, complement of open sets, even though borel sigma algebra is generated by open sets. It is just the smallest sigma algebra containing open sets. – Aman Nov 29 '14 at 12:59
  • Ok. So just showing the containment of an open interval suffices? – Swapnil Tripathi Nov 29 '14 at 13:57
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You showed that the $\sigma$-algebra generated by the closed (resp. half-open) intervals is a superset of the Borel algebra. The other inclusion is still missing: Obtain $[a,b]$ using nothing but countable unions and complements of open intervals. A straight-forward possiblity is $$[a,b]=\mathbb R\setminus\left(\bigcup_{n=1}^\infty\left(\mathbb R\setminus(a-1/n,b+1/n)\right)\right)\text.$$

Reusing this representation, we obtain $$(a,b]=(a,(a+b)/2)\cup[(a+b)/2,b]$$ for half-open intervals.