Equivalence of Definitions of Sigma-Algebra
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Theorem
The following definitions of the concept of Sigma-Algebra are equivalent:
Definition 1
Let $X$ be a set.
Let $\Sigma$ be a system of subsets of $X$.
$\Sigma$ is a $\sigma$-algebra over $X$ if and only if $\Sigma$ satisfies the sigma-algebra axioms:
\((\text {SA} 1)\) | $:$ | Unit: | \(\ds X \in \Sigma \) | ||||||
\((\text {SA} 2)\) | $:$ | Closure under Complement: | \(\ds \forall A \in \Sigma:\) | \(\ds \relcomp X A \in \Sigma \) | |||||
\((\text {SA} 3)\) | $:$ | Closure under Countable Unions: | \(\ds \forall A_n \in \Sigma: n = 1, 2, \ldots:\) | \(\ds \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma \) |
Definition 2
Let $X$ be a set.
Let $\Sigma$ be a system of subsets of $X$.
$\Sigma$ is a $\sigma$-algebra over $X$ if and only if $\Sigma$ satisfies the sigma-algebra axioms:
\((\text {SA} 1')\) | $:$ | Unit: | \(\ds X \in \Sigma \) | ||||||
\((\text {SA} 2')\) | $:$ | Closure under Set Difference: | \(\ds \forall A, B \in \Sigma:\) | \(\ds A \setminus B \in \Sigma \) | |||||
\((\text {SA} 3')\) | $:$ | Closure under Countable Disjoint Unions: | \(\ds \forall A_n \in \Sigma: n = 1, 2, \ldots:\) | \(\ds \bigsqcup_{n \mathop = 1}^\infty A_n \in \Sigma \) |
Definition 3
A $\sigma$-algebra $\Sigma$ is a $\sigma$-ring with a unit.
Definition 4
Let $X$ be a set.
A $\sigma$-algebra $\Sigma$ over $X$ is an algebra of sets which is closed under countable unions.
Proof
Definition 1 implies Definition 3
Let $\Sigma$ be a system of sets on a set $X$ such that:
- $(1): \quad X \in \Sigma$
- $(2): \quad \forall A, B \in \Sigma: \relcomp X A \in \Sigma$
- $(3): \quad \ds \forall A_n \in \Sigma: n = 1, 2, \ldots: \bigcup_{n \mathop = 1}^\infty A_n \in \Sigma$
Let $A, B \in \Sigma$.
From the definition:
- $\forall A \in \Sigma: A \subseteq X$.
Hence from Intersection with Subset is Subset:
- $\forall A \in \Sigma: A \cap X = A$
Hence $X$ is the unit of $\Sigma$.
So by definition 2 of $\sigma$-ring it follows that $\Sigma$ is a $\sigma$-ring with a unit.
Thus $\Sigma$ is a $\sigma$-algebra by definition 3.
$\Box$
Definition 3 implies Definition 1
Let $\Sigma$ be a $\sigma$-ring with a unit $X$.
By definition, $X \in \Sigma$.
From definition 2 of $\sigma$-ring, $\Sigma$ is:
- $(1) \quad$ closed under set difference.
- $(2) \quad$ closed under countable union
From Unit of System of Sets is Unique, we have that:
- $\forall A \in \Sigma: A \subseteq X$
from which we have that $X \setminus A = \relcomp X A$.
So $\Sigma$ is a $\sigma$-algebra by definition 1.
$\Box$
Definition 1 implies Definition 2
First, $(\text {SA} 1')$ is the same as $(\text {SA} 1)$.
Secondly, $(\text {SA} 3')$ is a special case of $(\text {SA} 3)$.
It remains to show $(\text {SA} 2')$.
Let $A, B \in \Sigma$ be arbitrary.
Observe:
\(\ds A \setminus B\) | \(=\) | \(\ds A \cap \relcomp X B\) | Set Difference as Intersection with Relative Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \relcomp X {\relcomp X A} \cap \relcomp X B\) | Relative Complement of Relative Complement | |||||||||||
\(\ds \) | \(=\) | \(\ds \relcomp X {\relcomp X A \cup B}\) | De Morgan's Law |
By $(\text {SA} 2)$:
- $\relcomp X A \in \Sigma$
Furthermore by $(\text {SA} 3)$:
- $\relcomp X A \cup B \in \Sigma$
Finally, by $(\text {SA} 2)$:
- $A \setminus B = \relcomp X {\relcomp X A \cup B} \in \Sigma$
Hence $(\text {SA} 2')$ is verified.
$\Box$
Definition 2 implies Definition 1
First, $(\text {SA} 1)$ is the same as $(\text {SA} 1')$.
Secondly, $(\text {SA} 2)$ is a special case of $(\text {SA} 2')$, since for all $A \subseteq X$:
- $\relcomp X A = X \setminus A$
by definition of relative complement.
It remains to show $(\text {SA} 3)$.
Let $A_n \in \Sigma$ be arbitrary for $n = 1, 2, \ldots$.
Let $F_1 := A_1$.
For all $n \ge 2$, let recursively:
- $F_n := A_n \setminus \paren {F_1 \sqcup \cdots \sqcup F_{n - 1} }$
Sublemma
For all $n = 1, 2, \ldots$ we have:
- $F_n \in \Sigma$
and:
- $A_1 \cup \cdots \cup A_n = F_1 \sqcup \cdots \sqcup F_n$
Proof of Sublemma
We shall prove by induction.
For $n = 1$ the claim is trivial, as $F_1 = A_1$.
Suppose that the claim is true for $n = k - 1$.
In particular:
- $F_1 \sqcup \cdots \sqcup F_{k - 1} \in \Sigma$
Therefore by $(\text {SA} 2')$:
- $F_k = A_k \setminus \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \in\Sigma$
Furthermore:
\(\ds A_1 \cup \cdots \cup A_k\) | \(=\) | \(\ds \paren {A_1 \cup \cdots \cup A_{k - 1} } \cup A_k\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \cup A_k\) | induction assumption | |||||||||||
\(\ds \) | \(=\) | \(\ds \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \sqcup \paren {A_k \setminus \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } }\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {F_1 \sqcup \cdots \sqcup F_{k - 1} } \sqcup F_k\) |
$\Box$
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Therefore:
- $\ds \bigcup_{n \mathop = 1}^\infty A_n = \bigsqcup_{n \mathop = 1}^\infty F_n$
where the right hand side belongs to $\Sigma$ by $(\text {SA} 3')$.
Hence $(\text {SA} 3)$ is verified.
$\Box$
Definition 1 implies Definition 4
Immediate from the definition of algebra along with the added condition of closure under countable unions.
$\Box$
Definition 4 implies Definition 1
By definition $1$ of algebra of sets, an algebra has the properties:
\((\text {AS} 1)\) | $:$ | Unit: | \(\ds X \in \Sigma \) | ||||||
\((\text {AS} 2)\) | $:$ | Closure under Union: | \(\ds \forall A, B \in \Sigma:\) | \(\ds A \cup B \in \Sigma \) | |||||
\((\text {AS} 3)\) | $:$ | Closure under Complement Relative to $X$: | \(\ds \forall A \in \Sigma:\) | \(\ds \relcomp X A \in \Sigma \) |
Replacing $(\text {AS} 2)$ with closure under countable unions immediately yields the first definition.
$\blacksquare$
Sources
- 1970: Avner Friedman: Foundations of Modern Analysis ... (previous) ... (next): $\S 1.1$: Rings and Algebras
- 1970: Avner Friedman: Foundations of Modern Analysis ... (previous) ... (next): $\S 1.1$: Rings and Algebras: Theorem $1.1.1$
- 1984: Gerald B. Folland: Real Analysis: Modern Techniques and their Applications : $\S 1.2$