User:Caliburn/s/mt/CohnRoad/ex

Theorem
Let $X$ be a set.

Let $\mathcal S \subseteq \map {\mathcal P} X$ be the set of singletons of $X$.

Then $\map \sigma {\mathcal S}$, the $\sigma$-algebra generated by $\mathcal S$, consists precisely of the co-countable subsets of $X$.

Theorem
Let $\map {\mathcal P_K} \R$ be the set of compact subsets of $\R$.

Then $\map \sigma {\map {\mathcal P_K} \R}$, the $\sigma$-algebra generated by the compact subsets of $\R$, consists precisely of the Borel $\sigma$-algebra $\map {\mathcal B} \R$.

Theorem
Let $A \subseteq \R$.

Let $\lambda$ be the Lebesgue measure and $\lambda^\ast$ be the Lebesgue pre-measure on $\R$.

Then there exists a Borel subset $B \subseteq A$ such that $\map \lambda B = \map {\lambda^\ast} A$.

Theorem
Let $B \subseteq \R$.

Let $\lambda^\ast$ be the Lebesgue pre-measure on $\R$.

Then $B$ is $\lambda^\ast$-measurable :


 * for each open interval $I \subseteq \R$ we have:


 * $\map {\lambda^\ast} I = \map {\lambda^\ast} {I \cap B} + \map {\lambda^\ast} {I \cap B^c}$

Theorem
Let $I \subseteq \R$ be a bounded interval.

Let $\lambda^\ast$ be the Lebesgue pre-measure on $\R$.

Let $B \subseteq I$

Then $B$ is $\lambda^\ast$-measurable :


 * $\map {\lambda^\ast} I = \map {\lambda^\ast} B + \map {\lambda^\ast} {I \cap B^c}$