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# 数学代写|测度与积分代写Measure And Integration代考|MATH30360 Outer Measures

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## 数学代写|测度与积分代写Measure And Integration代考|Outer Measures

Definition 6.1. A function $\mu^: \mathcal{P}(X) \rightarrow[0, \infty]$ is an outer measure if (1) $\mu^(\emptyset)=0$
(2) $\mu^\left(\cup A_i\right) \leq \sum \mu^\left(A_i\right)$
(3) $\mu^(A) \leq \mu^(B)$ if $A \subset B$.
Proposition $6.2$ (Example of an outer measure.). Let $\mathcal{E} \subseteq \mathcal{P}(X)$ be arbitrary collection of subsets of $X$ which contains both $\emptyset, X \in \mathcal{E}$. Let $\rho: \mathcal{E} \rightarrow[0, \infty]$ be a function such that $\rho(\emptyset)=0$. For any $A \subseteq X$, define
$$\mu^(A)=\inf \left{\sum_i \rho\left(E_i\right): A \subseteq \cup E_i, E_i \in \mathcal{E}\right} .$$ Then $\mu^$ is an outer measure.
Proof. It is clear that $\mu^(\emptyset)=0$ and $\mu^(A) \leq \mu^(B)$ for $A \subseteq B$. Suppose that $A_i \in \mathcal{P}(X)$ and $\mu^\left(A_i\right)<\infty$ for all $i$, otherwise there will be nothing to prove. Let $\epsilon>0$ and choose $E_{i j} \in \mathcal{E}$ such that $A_i \subseteq \bigcup_{j=1}^{\infty} E_{i j}$ and
$$\mu^\left(A_i\right) \geq \sum_{j=1}^{\infty} \rho\left(E_{i j}\right)-2^{-i} \epsilon .$$ Then \begin{aligned} \mu^\left(\cup A_i\right) & \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \rho\left(E_{i j}\right) \ & \leq \sum_{i=1}^{\infty}\left(\mu^\left(A_i\right)+2^{-i} \epsilon\right)=\sum_{i=1}^{\infty} \mu^\left(A_i\right)+\epsilon \end{aligned}

## 数学代写|测度与积分代写Measure And Integration代考|Carathéodory’s Construction Theorem

Theorem $6.5$ (Carathéodory’s Construction Theorem). Let $\mu^$ be an outer measure on $X$. Let $\mathcal{M}=\mathcal{M}\left(\mu^\right)$ then $\mathcal{M}$ is a $\sigma$-algebra and $\left.\mu \equiv \mu^*\right|_{\mathcal{M}}$ is a complete measure.

Proof. Clearly $\emptyset, X \in \mathcal{M}$ and if $A \in \mathcal{M}$ then $A^c \in \mathcal{M}$. So to show that $\mathcal{M}$ is an algebra we must show that $\mathcal{M}$ is closed under finite unions, i.e if $A, B \in \mathcal{M}$ and $E \in \mathcal{P}(X)$ then
$$\mu^(E) \geq \mu^(E \cap(A \cup B))+\mu(E \backslash(A \cup B)) .$$
Using the definition of $\mathcal{M}$ we have
\begin{aligned} \mu^(E) & =\mu^(E \cap A)+\mu^(E \backslash A) \ (6.5) & =\mu^(E \cap A \cap B)+\mu^(E \cap A \backslash B)+\mu^((E \backslash A) \cap B)+\mu^(E \backslash(A \cup B)) . \end{aligned} We will now make use of the identity: $$E \cap(A \cup B)=(E \cap A) \cup(E \cap B)=[(E \cap A) \backslash B] \cup[(E \backslash A) \cap B] \cup(E \cap A \cap B)$$ which in words states that $E \cap A \cap B \cup(E \cap A \backslash B) \cup((E \backslash A) \cap B)$ is equal to the points common to $E, A$ and $B$ union the points in $(E$ and $A$ but not $B)$ union the points in $(E$ and $B$ but not $A)$ which is equal to those points in $E$ which are also in either $A$ or $B$. Using this identity in Eq. (6.5) along with the subadditivity of $\mu^$ shows
$$\mu^(E) \geq \mu^(E \cap(A \cup B))+\mu^(E \backslash(A \cup B))$$ which implies that $A \cup B \in \mathcal{M}$ and therefore shows that $\mathcal{M}$ is an algebra. Now suppose $A, B \in \mathcal{M}$ are disjoint, then taking $E=A \cup B$ in Eq. (6.4) implies $$\mu^(A \cup B)=\mu^(A)+\mu^(B)$$

## 数学代写|测度与积分代写MEASURE AND INTEGRATION代考|OUTER MEASURES

$2 \mu^{\left(\cup A_i\right)} \leq \sum \mu^{\left(A_i\right)}$
$\left.\left.3 \mu^{\prime} A\right) \leq \mu^{\prime} B\right)$ 如果 $A \subset B$.

$$\mu^{(A,)} \geq \sum_{j=1}^{\infty} \rho\left(E_{i j}\right)-2^{-i} \epsilon$$

$$\mu^{\left(\cup A_i\right)} \leq \sum_{i=1}^{\infty} \sum_{j=1}^{\infty} \rho\left(E_{i j}\right) \quad \leq \sum_{i=1}^{\infty}\left(\mu^{\left(A_i\right)}+2^{-i} \epsilon\right)=\sum_{i=1}^{\infty} \mu^{\left(A_i\right)}+\epsilon$$

## 数学代写测度与积分代写MEASURE AND INTEGRATION代考|CARATHÉODORY’S CONSTRUCTION THEOREM

$$\left.\left.\mu^{(} E\right) \geq \mu^{(} E \cap(A \cup B)\right)+\mu(E \backslash(A \cup B)) .$$

$$\left.\left.\left.\left.\left.\left.\left.\mu^{(} E\right)=\mu^{(} E \cap A\right)+\mu^{(} E \backslash A\right)(6.5) \quad=\mu^{(} E \cap A \cap B\right)+\mu^{(} E \cap A \backslash B\right)+\mu^{(}(E \backslash A) \cap B\right)+\mu^{(} E \backslash(A \cup B)\right)$$

$$E \cap(A \cup B)=(E \cap A) \cup(E \cap B)=[(E \cap A) \backslash B] \cup[(E \backslash A) \cap B] \cup(E \cap A \cap B)$$

$$\left.\left.\left.\mu^{(} E\right) \geq \mu^{(} E \cap(A \cup B)\right)+\mu^{(} E \backslash(A \cup B)\right)$$

$$\left.\left.\left.\mu^{(} A \cup B\right)=\mu^{(} A\right)+\mu^{(} B\right)$$

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