Measurable Sets form Algebra of Sets

Theorem
The set $$\mathfrak {M} \ $$ of measurable sets, a subset of the power set $$\mathcal{P} \left({\R}\right) \ $$ of the reals, is an algebra of sets.

Proof
There are two properties a collection of sets must fulfill to be an algebra of sets, laid out on that definition page.

The second of these properties, as decribed on that page, follows trivially from taking $$A \ $$, as defined there, to be the universal set.

Thus we must prove only the first property.

We use $$m, m^* \ $$ as defined on the page Definition:Lebesgue Measure and take $$\mathbb{U} \ $$ as the universal set.

Suppose $$E_1, E_2 \ $$ are measurable sets. Let $$A \ $$ be any set.

Since $$E_2 \ $$ is measurable, we have


 * $$m^*(A \cap \mathcal{C}_\mathbb{U} \left({E_1}\right) ) = m^*(A \cap \mathcal{C}_\mathbb{U} \left({E_1}\right) \cap E_2) + m^*(A \cap \mathcal{C}_\mathbb{U} (E_1) \cap  \mathcal{C}_\mathbb{U} (E_2) ) $$

Since:
 * $$A \cap (E_1 \cup E_2) = (A \cap E_1) \cup (A \cap E_2 \cap \mathcal{C}_\mathbb{U} (E_1) ) $$

by De Morgan's Laws, we have


 * $$m^*(A \cap (E_1 \cup E_2)) \leq m^*(A \cap E_1) + m^*(A \cap E_2 \cap \mathcal{C}_\mathbb{U} (E_2) )$$

Thus:

$$ $$ $$ $$

again by De Morgan's Laws and the measurability of $$E_1 \ $$.

Since:
 * $$\mathcal{C}_\mathbb{U} (E_1 \cup E_2) = \mathcal{C}_\mathbb{U} (E_1) \cap \mathcal{C}_\mathbb{U} (E_2)$$

this shows $$E_1 \cup E_2 $$ is measurable.