Borel Sigma-Algebra on Euclidean Space by Monotone Class
Theorem
Let $\sqbrk {\R^n, \tau}$ be the $n$-dimensional Euclidean space.
Then:
- $\map \BB {\R^n, \tau} = \map {\mathfrak m} \tau$
where $\BB$ denotes Borel $\sigma$-algebra, and $\mathfrak m$ denotes generated monotone class.
Proof
Let $U \in \tau$ be an open set, and define $C$ by:
- $C := X \setminus U$
hence $C$ is a closed set.
Further, define, for all $n \in \N$:
- $C_n := \ds \bigcup_{c \mathop \in C} \map B {c; \frac 1 n}$
where $B$ denotes open ball.
The $C_n$ are open sets, being the union of open balls.
It is clear that $C \subseteq C_n$ for all $n \in \N$.
Conversely, as $U$ is open, for any $u \in U$ (that is, $u \notin C$), find $n \in \N$ such that:
- $\map B {u; \dfrac 1 n} \subseteq U$
as is possible from the definition of open set in a metric space.
Thus, for all $c \in C = X \setminus U$, this means:
- $\map d {u, c} \ge \dfrac 1 n$
whence $u \notin C_n$.
That is, we have established that:
- $c \in C \iff \forall n \in \N: c \in C_n$
Phrased in terms of intersection, this means:
- $C = \ds \bigcap_{n \mathop \in \N} C_n$
Thus, since $C_n \in \tau \subseteq \map {\mathfrak m} \tau$:
- $C \in \map {\mathfrak m} \tau$
Now define:
- $\relcomp X \tau := \set {X \setminus U: U \in \tau}$
Then we have shown:
- $\map {\mathfrak m} {\tau \cup \relcomp X \tau} \subseteq \map {\mathfrak m} \tau$
and the reverse inclusion (and hence equality) follows from Generated Monotone Class Preserves Subset.
Now applying Generated Sigma-Algebra by Generated Monotone Class: Corollary:
- $\map \sigma \tau = \map {\mathfrak m} {\tau \cup \relcomp X \tau}$
Combining these two equalities gives the result, by definition of Borel $\sigma$-algebra.
$\blacksquare$
Sources
- 2005: René L. Schilling: Measures, Integrals and Martingales ... (previous) ... (next): $\S 3$: Problem $12$