Theorem: Measurable Sets form Algebra of Sets

Let $\mu^*$ be an outer measure on a set $X$.
Then the set of $\mu^*$-measurable sets is an algebra of sets.

Proof:
For a subset $S \subseteq X$, let $\relcomp X S$ denote the relative complement of $S$ in $X$.
We first prove the second property of an algebra of sets, as described on that page.
Let $S$ be $\mu^*$-measurable. For any subset $A \subseteq X$:
{{begin-eqn}}
{{eqn | l = \map {\mu^*} A
      | r = \map {\mu^*} {A \cap S} + \map {\mu^*} {A \cap \relcomp X S}
}}
{{eqn | r = \map {\mu^*} {A \cap \relcomp X {\relcomp X S} } + \map {\mu^*} {A \cap \relcomp X S}
      | c = Complement of Complement
}}
{{end-eqn}}
as desired.
Now we prove the first property.
Suppose that $S_1$ and $S_2$ are $\mu^*$-measurable sets.
Let $A$ be any subset of $X$.  
Since:
{{begin-eqn}}
{{eqn | l = A \cap \paren {S_1 \cup S_2}
      | r = \paren {\paren {A \cap \paren {S_1 \cup S_2} } \cap S_1} \cup \paren {\paren {A \cap \paren {S_1 \cup S_2} } \setminus S_1}
      | c = Set Difference Union Intersection and Union is Commutative
}}
{{eqn | r = \paren {A \cap \paren {\paren {S_1 \cup S_2} \cap S_1} } \cup \paren {\paren {\paren {S_1 \cup S_2} \cap A} \setminus S_1}
      | c = Intersection is Associative and Intersection is Commutative
}}
{{eqn | r = \paren {A \cap \paren {\paren {S_1 \cup S_2} \cap S_1} } \cup \paren {\paren {\paren {S_1 \cup S_2} \setminus S_1} \cap A}
      | c = Intersection with Set Difference is Set Difference with Intersection
}}
{{eqn | r = \paren {A \cap S_1} \cup \paren {\paren {S_2 \setminus S_1} \cap A}
      | c = Intersection Absorbs Union and Set Difference with Union is Set Difference
}}
{{eqn | r = \paren {A \cap S_1} \cup \paren {\paren {S_2 \cap A} \setminus S_1}
      | c = Intersection with Set Difference is Set Difference with Intersection
}}
{{eqn | r = \paren {A \cap S_1} \cup \paren {\paren {A \cap S_2} \setminus S_1}
      | c = Intersection is Commutative
}}
{{eqn | r = \paren {A \cap S_1} \cup \paren {\paren {A \setminus S_1} \cap S_2}
      | c = Intersection with Set Difference is Set Difference with Intersection
}}
{{end-eqn}}
{{wtd|I'm pretty sure the above, or at least significant parts of the above, have/has already been posted up as proofs in their own right. If I remember, or rather, when I'm in the mood, I may go through and find them.}}
By subadditivity of an outer measure:
:$\map {\mu^*} {A \cap \paren {S_1 \cup S_2} } \le \map {\mu^*} {A \cap S_1} + \map {\mu^*} {\paren {A \setminus S_1} \cap S_2}$
Thus:
{{begin-eqn}}
{{eqn | r = \map {\mu^*} {A \cap \paren {S_1 \cup S_2} } + \map {\mu^*} {A \setminus \paren {S_1 \cup S_2} }
}}
{{eqn | r = \map {\mu^*} {A \cap \paren {S_1 \cup S_2} } + \map {\mu^*} {\paren {A \setminus S_1} \setminus S_2}
      | c = Set Difference with Union
}}
{{eqn | o = \le
      | r = \map {\mu^*} {A \cap S_1} + \map {\mu^*} {\paren {A \setminus S_1} \cap S_2} + \map {\mu^*} {\paren {A \setminus S_1} \setminus S_2}
      | c = by the above argument
}}
{{eqn | r = \map {\mu^*} {A \cap S_1} + \map {\mu^*} {A \setminus S_1}
      | c = {{Defof|Measurable Set#Measurable Sets of an Arbitrary Outer Measure|Measurability}} of $S_2$
}}
{{eqn | r = \map {\mu^*} A
      | c = {{Defof|Measurable Set#Measurable Sets of an Arbitrary Outer Measure|Measurability}} of $S_1$
}}
{{end-eqn}}
The result follows by the subadditivity of an outer measure.
{{refactor|Create separate proof|level = medium}}
Alternatively, one could use the equality
{{begin-eqn}}
{{eqn | l = \map {\mu^*} {A \cap \paren {S_1 \cup S_2} }
      | r = \map {\mu^*} {A \cap \paren {S_1 \cup S_2} \cap S_1} + \map {\mu^*} {A \cap \paren {S_1 \cup S_2} \setminus S_1}
}}
{{eqn | r = \map {\mu^*} {A \cap S_1} + \map {\mu^*} {A \cap S_2 \setminus S_1}
}}
{{end-eqn}}
to prove the result directly without the use of subadditivity.
{{qed}}
Category:Measurable Sets
Category:Algebras of Sets
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