# Borel sets including $\infty$

Can someone give me a hint how to show the equality $\{ B \cup F \mid B\in \sigma(\{(-\infty,a) \mid a\in \mathbb{R}\}),\ F\subseteq\{-\infty,+\infty\} \}=\sigma(\{ [-\infty,a) \mid a \in \mathbb{R} \})$ ?

The set $\sigma(\{(-\infty,a) \mid a\in \mathbb{R}\})$ represents the usual Borel sets on the real line (and $\{(-\infty,a)\mid a\in \mathbb{R}\}$ is its generator).

What puzzles me is the fact that for example $(-\infty,1)\cup \{+\infty \}$ is in the set on the LHS of the equality, but I don't know how to produce this set on the RHS of the equality (making me wonder, if the above equality is true.)

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Do want $F \subseteq \{ - \infty , + \infty \}$ in the definition on the left-hand-side?  (Otherwise $[ - \infty , + \infty ]$ is not in the collection defined on the left-hand-side, but it is in the collection defined on the right-hand-side.) – Arthur Fischer Dec 10 '12 at 18:16
@ArthurFischer Ah, yes. Sorry. – user52145 Dec 10 '12 at 18:18

## 2 Answers

I will denote by $\overline{\mathbb{R}}$ the set $\mathbb{R} \cup \{ - \infty , + \infty \}$.

To produce $( - \infty , 1 ) \cup \{ + \infty \}$ on the right-hand-side, note that

• for all $a \in \mathbb{R}$ we have that $[ a , + \infty ] = \overline{\mathbb{R}} \setminus [ - \infty , a )$, and so these sets belong to the $\sigma$-algebra on the right-hand-side.

Then $$( - \infty , 1 ) \cup \{ + \infty \} = \left( \bigcup_{n=1}^\infty [ - \infty , 1-{\textstyle \frac{1}{n}} ] \cap \bigcup_{n=1}^\infty [ -n , + \infty ] \right) \cup \bigcap_{n=1}^\infty [ n , + \infty ].$$

To show equality, you can proceed as follows:

1. Show that the collection on the LHS is itself a $\sigma$-algebra. (Showing closure under countable unions should be quite easy, and complementation will follow after noting that $\overline{\mathbb{R}} \setminus ( B \cup F ) = ( \mathbb{R} \setminus B ) \cup ( \{ - \infty , + \infty \} \setminus F)$ for $B \subseteq \mathbb{R}$ and $F \subseteq \{ - \infty , + \infty \}$.) (This can be seen as taking the "join" of two $\sigma$-algebras on disjoint sets.)
2. To show RHS $\subseteq$ LHS: Simply note the every set in the generating set for the RHS is an element of the LHS. (Should be very easy.) Since LHS is a $\sigma$-algebra, the set inclusion is done.
3. To show LHS $\subseteq$ RHS: Show, using ideas similar to the example above, that every set of the form $( - \infty , a )$ belongs to the RHS, as well as $\{ - \infty \}$ and $\{ + \infty \}$. (From this you will be able to show that all Borel subsets of $\mathbb{R}$ belong to RHS, as well as all subsets of $\{ - \infty , + \infty \}$.)
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To see that this the RHS contains the LHS:

It suffices to show that this set contains $\infty$, $-\infty$ and the open intervals (since it is by assumption a $\sigma$-algebra)

$-\infty$ :

It contains $[- \infty, -n)$ for all $n$ and is closed under intersection.

$\infty$:

It is closed under compliment so contains $[n, \infty]$ for every $n$ and again is closed under intersection.

$(a,b)$ :

It contains $[-\infty, a]$ by taking the intersection of $[-\infty, a + \frac{1}{n})$ and it contains $[-\infty, b)$. By virtue of being a $\sigma$-algebra it is then closed under the difference of these two sets, so it contains $(a,b)$ .

To see that the LHS contains the RHS:

Since the LHS contains $[-\infty, a)$ for all $a$, you just need to show that the LHS is actually a $\sigma$-algebra.

You should try do this, all you need to do is show that the set is closed under countable unions and compliments. You should do this using the fact that the Borel sets are a $\sigma$-algebra, plus possibly some cases based on if you include $-\infty, +\infty$ or both.

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