To prove facts about Borel sets, use closure properties

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To prove facts about Borel sets, use closure properties

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[QUICK DESCRIPTION]
The Borel sets for a given topology are defined to be the smallest class of sets that contains all open sets and is closed under countable unions and intersections. Therefore, to prove that every Borel set has a certain property, it is sufficient to prove it for open sets (or even for simpler sets such as intervals), and then to prove that countable unions or countable intersections of sets with the property have the property. It is also possible to express such arguments in terms of [[transfinite induction]].

[PREREQUISITES]
Basic real analysis, measure theory.

[EXAMPLE]
The following problem arises in connection with measure differential inclusions, which are problems of the form
[maths] \frac{dx}{dt}\in F(t,x(t)), [/maths]
where $F$ is a closed convex set-valued function with possibly unbounded values.

Given a vector measure $\mu$ with values in $\R^n$ and a non-negative measure $\nu$ (positive for open sets), the problem is to show that if
[maths open|Start] \frac{\mu(U)}{\nu(U)}\in K [/maths]
for any open set $U$ where $K\subseteq \R^n$ is a closed convex set, then
[maths borel|Target]\begin{align}  
\frac{\mu(E)}{\nu(E)}&\in K,\qquad \nu(E)>0,\\
\mu(E)&\in K_\infty,\qquad \nu(E)=0,
\end{align}[/maths]
where
[maths asympt-cone|Asymptotic cone]K_\infty = \{\lim_{i\to\infty} t_i x_i\mid t_i\downarrow 0,\;x_i\in K\}[/maths]
which is known as the ''asymptotic'' or ''recession'' cone of $K$.

Since we are assuming the result for open sets, it is sufficient to prove that the property is closed under countable unions and intersections. We can make this task slightly easier by noting that finite unions and intersections of sets with the desired property also have the desired property; this in turn means that we only need to deal with ''nested'' countable unions and intersections.

Considering nested unions, suppose $E=\bigcup_{i=1}^\infty E_i$ with $E_i\subseteq E_{i+1}$.  Then either $\nu(E_i)=0$ for all $i$, or there is an $i$ where $\nu(E_i)>0$.  In the former case, $\mu(E_i)\in K_\infty$ for all $i$, and so taking limits, $\mu(E)\in K_\infty$ as $K_\infty$ is a closed convex cone.  In the latter case, we have $\mu(E_j)/\nu(E_j)\in K$ for all $j\geq i$, and so taking limits the result again holds.

Considering nested intersections, suppose $E=\bigcap_{i=1}^\infty E_i$ with $E_i\supseteq E_{i+1}$.  Then either $\nu(E)>0$ or $\nu(E)=0$.  In the former case, $\nu(E_i)>0$ for all $i$, and taking limits of $\mu(E_i)/\nu(E_i)\in K$ gives the desired result.  If $\nu(E)=0$ then either $\nu(E_i)=0$ for some $i$ or $\nu(E_i)>0$ for all $i$.  In $\nu(E_i)=0$ then $\nu(E_j)=0$ for all $j\geq i$ and taking limits of $\mu(E_j)\in K_\infty$ again gives the desired result.  Finally, considering $\nu(E_i)>0$ for all $i$ we see that
[maths]\mu(E_j) = \nu(E_j)\,x_j,\qquad x_j \in K.  [/maths]
Taking limits and using the definition of $K_\infty$ gives the desired: $\mu(E)\in K_\infty$.

[GENERAL DISCUSSION]
One can think of the Borel sets as being built up from the open sets by means of successive operations of countable unions and intersections. However, there are transfinitely many stages to this process, so [[transfinite induction]] is needed.

More precisely, we build up the Borel sets for a metric space as follows: for each ordinal $\alpha$ we have a class of sets ${\bold B}_\alpha$, where
* ${\bold B}_0$ is the set of open sets,
* ${\bold B}_{\alpha+1}$ is the set of countable unions or countable intersections of sets in ${\bold B}_\alpha$,
* ${\bold B}_\alpha = \bigcup_{\beta<\alpha}{\bold B}_\beta$ is $\alpha$ is a limit ordinal.
Then ${\bold B}_{\omega_1}$ is the set of Borel sets where $\omega_1$ is the first uncountable ordinal.

Note that ${\bold B}_1$ contains all the closed sets (for a metric space) and so contains all the complements of sets in ${\bold B}_0$.
Thus, the complement of any set in ${\bold B}_\alpha$ can be found in ${\bold B}_{\alpha+1}$ for finite $\alpha$; if $\alpha$ is infinite, then complements of sets in ${\bold B}_\alpha$ are in ${\bold B}_\alpha$.

To prove the above result, we could use a transfinite induction on the levels of the Borel hierarchy. We start with the fact that [eqref borel] holds for for all $E\in{\bold B}_0$.

If [eqref borel] holds for all ${\bold B}_\beta$ for $\beta<\alpha$ with $\alpha$ a limit ordinal, then clearly by definition of ${\bold B}_\alpha$, [eqref borel] also holds for all $E\in{\bold B}_\alpha$.

To deal with successor ordinals, one has to prove that if [eqref borel] holds for sets $E_1,E_2,\dots$ then it holds for $\bigcup_{i=1}^\infty E_i$ and $\bigcap_{i=1}^\infty E_i$. This we showed above.

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