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# 数学代写|实分析代写Real Analysis代考|Math444

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## 数学代写|实分析代写Real Analysis代考|Completion of a Measure Space

If $(X, \mathcal{A}, \mu)$ is a measure space, we define the completion of this space to be the measure space $(X, \overline{\mathcal{A}}, \bar{\mu})$ defined by
\begin{aligned} \overline{\mathcal{A}} & =\left{\begin{array}{l|l} E \Delta Z & \begin{array}{l} E \text { is in } \mathcal{A} \text { and } Z \subseteq Z^{\prime} \text { for } \ \text { some } Z^{\prime} \in \mathcal{A} \text { with } \mu\left(Z^{\prime}\right)=0 \end{array} \end{array},\right. \ \bar{\mu}(E \Delta Z) & =\mu(E) . \end{aligned}
It is necessary to verify that the result is in fact a measure space, and we shall carry out this step in the proposition below. In the case of Lebesgue measure $m$ on the line, when initially defined on the $\sigma$-algebra $\mathcal{A}$ of Borel sets, the sets in $\sigma$-algebra $\overline{\mathcal{A}}$ are said to be Lebesgue measurable.

Proposition 5.38. If $(X, \mathcal{A}, \mu)$ is a measure space, then the completion $(X, \overline{\mathcal{A}}, \bar{\mu})$ is a measure space. Specifically
(a) $\overline{\mathcal{A}}$ is a $\sigma$-algebra containing $\mathcal{A}$,
(b) the set function $\bar{\mu}$ is unambiguously defined on $\overline{\mathcal{A}}$, i.e., if $E_1 \Delta Z_1=$ $E_2 \Delta Z_2$ as above, then $\mu\left(E_1\right)=\mu\left(E_2\right)$,
(c) $\bar{\mu}$ is a measure on $\overline{\mathcal{A}}$, and $\bar{\mu}(E)=\mu(E)$ for all sets $E$ in $\mathcal{A}$.

(d) if $\tilde{\mu}$ is any measure on $\overline{\mathcal{A}}$ such that $\tilde{\mu}(E)=\mu(E)$ for all $E$ in $\mathcal{A}$, then $\tilde{\mu}=\bar{\mu}$ on $\overline{\mathcal{A}}$
(e) if $\mu(X)<+\infty$ and if for $E \subseteq X, \mu_(E)$ and $\mu^(E)$ are defined by $$\mu_(E)=\sup {A \subseteq E, A \in \mathcal{A}} \mu(A) \quad \text { and } \quad \mu^(E)=\inf {A \supseteq E, A \in \mathcal{A}} \mu(A) \text {, }$$
then $E$ is in $\overline{\mathcal{A}}$ if and only if $\mu_(E)=\mu^(E)$.
Proof. For (a), certainly $\mathcal{A} \subseteq \overline{\mathcal{A}}$ because we can use $Z=Z^{\prime}=\varnothing$ in the definition of $\overline{\mathcal{A}}$. Since $(E \Delta Z)^c=(E \Delta Z) \Delta X=(E \Delta X) \Delta Z=E^c \Delta Z, \overline{\mathcal{A}}$ is closed under complements.
To prove closure under countable unions, let us first prove that
$$\overline{\mathcal{A}}=\left{\begin{array}{l|l} E \cup Z & \begin{array}{l} E \text { is in } \mathcal{A} \text { and } Z \subseteq Z^{\prime} \text { for } \ \text { some } Z^{\prime} \in \mathcal{A} \text { with } \mu\left(Z^{\prime}\right)=0 \end{array} \end{array}\right} .$$

## 数学代写|实分析代写Real Analysis代考|Fubini’s Theorem for the Lebesgue Integral

Fubini’s Theorem for the Lebesgue integral concerns the interchange of order of integration of functions of two variables, just as with the Riemann integral in Section III.9. In the case of Euclidean space $\mathbb{R}^n$, we could have constructed Lebesgue measure in each dimension by a procedure similar to the one we used for $\mathbb{R}^1$. Then Fubini’s Theorem relates integration of a function of $m+n$ variables over a set by either integrating in all variables at once or integrating in the first $m$ variables first or integrating in the last $n$ variables first. In the context of more general measure spaces, we need to develop the notion of the product of two measure spaces. This corresponds to knowing $\mathbb{R}^m$ and $\mathbb{R}^n$ with their Lebesgue measures and to constructing $\mathbb{R}^{m+n}$ with its Lebesgue measure.

In the theorem as we shall state it, we are given two measures spaces $(X, \mathcal{A}, \mu)$ and $(Y, \mathcal{B}, v)$, and we assume that both $\mu$ and $v$ are $\sigma$-finite. We shall construct a product measure space $(X \times Y, \mathcal{A} \times \mathcal{B}, \mu \times v)$, and the formula in question will be
\begin{aligned} \int_{X \times Y} f d(\mu \times v) & \stackrel{?}{=} \int_X\left[\int_Y f(x, y) d v(y)\right] d \mu(x) \ & \stackrel{?}{=} \int_Y\left[\int_X f(x, y) d \mu(x)\right] d v(y) . \end{aligned}
This formula will be valid for $f \geq 0$ measurable with respect to $\mathcal{A} \times \mathcal{B}$.
The technique of proof will be the standard one indicated in connection with proving Corollary 5.28. We start with indicator functions, extend the result to simple functions by linearity, and pass to the limit by the Monotone Convergence Theorem (Theorem 5.25). It is then apparent that the difficult step is the case that $f$ is an indicator function. In fact, it is not even clear in this special case that the inside integral $\int_Y I_E(x, y) d v(y)$ is a measurable function of $X$, and this is the step that requires some work.

## 数学代写|实分析代写Real Analysis代考|Completion of a Measure Space

\begin{aligned} \overline{\mathcal{A}} & =\left{\begin{array}{l|l} E \Delta Z & \begin{array}{l} E \text { is in } \mathcal{A} \text { and } Z \subseteq Z^{\prime} \text { for } \ \text { some } Z^{\prime} \in \mathcal{A} \text { with } \mu\left(Z^{\prime}\right)=0 \end{array} \end{array},\right. \ \bar{\mu}(E \Delta Z) & =\mu(E) . \end{aligned}

(a) $\overline{\mathcal{A}}$是包含$\mathcal{A}$的$\sigma$ -代数，

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