for the following task i need some help. Maybe someone can explain it to me.
thank you in advance for your efforts:).
(i)Show that, all $\lambda_*$ - Null sets $N \subseteq \mathbb{R}^n$ are Carathéodory measurable, i.e. that for all $A \subseteq \mathbb{R}^n$ $$\lambda_*(A)=\lambda_*(A \cap N) + \lambda_*(A \cap (\mathbb{R}/N))$$ is true.
(ii)Show that for $\Omega = \mathbb{R}^n$, $\mu_* = \lambda_*$ with $\mu_C$ from theorem $10.11$ holds:
$A \subseteq \mathbb{R}^n$ is Carathéodory-measurable if and only if for all $\epsilon > 0$ there exist an open set $\mathcal{O} \subseteq \mathbb{R}^n$ with $A \subseteq \mathcal{O}$ such that $$\lambda_*(\mathcal{O}/A) \le \epsilon.$$
- theorem 10.11: The sets of $\mu_*$-measurable sets forms a $\sigma$-algebra $\mathcal{A}_C$ and $\mu_C$=$\mu_*|_{\mathcal{A}_C}$ is a measure on $\mathcal{A}_C$
What I know:
the definition of a measure $\mu$:
A measure on a measurable room $(\Omega,\mathcal{A})$ is a map $\mu:\mathcal{A} \to [0,1]$ with the following properties:
(i) $\mu(\emptyset)=0$
(ii) Let $(A_i) \subset \mathcal{A}$ be a countable family of pairwise disjunct sets, then $$\mu(\bigcup_i A_i)=\sum \limits_{i}\mu(A_i)$$
the definition of an outer-measure $\mu_*$:
An outer-measure on $\Omega$ is a map $\mu_*:\mathcal{P}(\Omega) \to [0,1]$ with the following properties:
(i)$\mu_*(\emptyset)=0$
(ii)(monotony) $A \subseteq B$ implies $\mu_*(A) \le \mu_*(B)$
(iii)($\sigma-subadditivity$) for countable families $(A_i)$ of subsets from $\Omega$ aplies that $$\mu_*(\bigcup_i A_i) \le \sum \limits_{i}\mu_*(A_i)$$
the definition of $\lambda_*$:
- $\lambda_*(A)$ = inf {$\sum_{j \in \mathbb{N} } \vert Q_j \vert$ | $A \subseteq \bigcup_{j\in\mathbb{N}} Q_j $}
for (i) the following could be helpful:
- Let be $N \subseteq \mathbb{R}^n$ a Null set. Then for all $$\lambda_*(A \cup N)= \lambda_*(A)$$ $\underline{proof}:$
It is valid with the monotonicity of the outer Lebesgue measure $$\lambda_*(A) \le \lambda_*(A \cup N) $$ since $A \subseteq A \cup N.$ Further applies because of the $\sigma$-subadditivity $$\lambda_*(A \cup N) \le \lambda_*(A) + \lambda_*(N) = \lambda_*(A)$$
for (ii) the following could be helpful:
- the identity A = B\(B\A)
- contruct a measurable B with the property $\lambda_*$(B\A)=0
My Problems:
(i)
- i dont know how I show the equation for all $\lambda_*$-Null sets
(ii)
- This task is difficult for me because I do not understand the equivalence described. I would appreciate an explanation of the equivalence and when this could be a useful tool