## 数学代写|概率论代写Probability theory代考|Abundance of Integrable Sets

2023年4月7日

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## 数学代写|概率论代写Probability theory代考|Abundance of Integrable Sets

In this section, let $(\Omega, L, I)$ be a complete integration space.
Let $X$ be any function defined on a subset of $\Omega$ and let $t$ be a real number. Recall from the notations and conventions described in the Introduction that we use the abbreviation $(t \leq X)$ for the subset ${\omega \in \operatorname{domain}(X): t \leq X(\omega)}$. Similar notations are used for $(X<t),(X \leq t)$, and $(X<t)$. We will also write ( $t<$ $X \leq u)$ and similar for the intersection $(t<X)(X \leq u)$ and similar. Recall also the definition of the metric complement $J_c$ of a subset $J$ of a metric space.

In the remainder of this section, let $X$ be an arbitrary but fixed integrable function. We will show that the sets $(t \leq X)$ and $(t<X)$ are integrable sets for each positive $t$ in the metric complement of some countable subset of $R$. In other words, $(t \leq X)$ and $(t<X)$ are integrable sets for all but countably many $t \in(0, \infty)$

First define some continuous functions that will serve as surrogates for step functions on $R$. Specifically, for real numbers $s, t$ with $0<s<t$, define $g_{s, t}(x) \equiv$ $(t-s)^{-1}(x \wedge t-x \wedge s)$. Then, by Assertion 1 of Proposition 4.3 .4 and by linearity, the function $g_{s, t}(X) \equiv(t-s)^{-1}(X \wedge t-X \wedge s)$ is integrable for each $s, t \in R$ with $0<s<t$. Moreover, $1 \geq g_{t^{\prime}, t} \geq g_{s, s^{\prime}} \geq 0$ for each $t^{\prime}, t, s, s^{\prime} \in R$ with $t^{\prime}<t \leq s<s^{\prime}$. If we can prove that $\lim {s \uparrow t} I g{s, t}(X)$ exists, then we can use the Monotone Convergence Theorem to show that the limit function $\lim {s \uparrow t} g{s, t}(X)$ is integrable and is an indicator of $(t \leq X)$, proving that the latter set is integrable. Classically the existence of $\lim {s \uparrow t} I g{s, t}(X)$ is trivial since, for fixed $t$, the integral $I g_{s, t}(X)$ is nonincreasing in $s$ and bounded from below by 0 . A nontrivial constructive proof that the limit exists for all but countably many $t$ ‘s is given by [Bishop and Bridges 1985], who devised a theory of profiles for that purpose. In the following, we give a succinct presentation of this theory.

Definition 4.6.1. Profile system. Let $K$ be a nonempty open interval in $R$. Let $G$ be a family of continuous functions on $R$ with values in [0,1]. Let $t \in K$ and $g \in G$ be arbitrary. If $g=0$ on $(-\infty, t] \cap K$, then we say that $t$ precedes $g$ and write $t \diamond g$. If $g=1$ on $[t, \infty) \cap K$, then we say that $g$ precedes $t$ and write $g \diamond t$.
Let $t, s \in K$ and $g \in G$ be arbitrary with $t<s$. If both $t \diamond g$ and $g \diamond s$, then we say that the function $g$ separates the points $t$ and $s$, and write $t \diamond g \diamond s$. If for each $t, s \in K$ with $t<s$, there exists $g \in G$ such that $t \diamond g \diamond s$, then we say that the family $G$ separates points in $K$.

## 数学代写|概率论代写Probability theory代考|Uniform Integrability

In this section, let $(\Omega, L, I)$ be a complete integration space.
Proposition 4.7.1. Moduli of integrability for integrable functions. Let $X \in L$ be arbitrary. Recall Definition 4.6.11. Then the following conditions hold:

1. Let $A$ be an arbitrary integrable set. Then $X 1_A \in L$.
2. $I\left(|X| 1_A\right) \rightarrow 0$ as $\mu(A) \rightarrow 0$, where $A$ is an arbitrary integrable set. More precisely, there exists an operation $\delta:(0, \infty) \rightarrow(0, \infty)$ such that $I\left(|X| 1_A\right) \leq \varepsilon$ for each integrable set $A$ with $\mu(A)<\delta(\varepsilon)$, for each $\varepsilon>0$.
3. $I\left(|X| 1_{(|X|>a)}\right) \rightarrow 0$ as $a \rightarrow \infty$. More precisely, suppose $I|X| \leq b$ for some $b>0$, and let operation $\delta$ be as described in Assertion 2. For each $\varepsilon>0$, if we define $\eta(\varepsilon) \equiv b / \delta(\varepsilon)$, then $I\left(|X| 1_{(|X|>a)}\right) \leq \varepsilon$ for each $a>\eta(\varepsilon)$.
4. Suppose an operation $\eta>0$ is such that $I\left(|X| 1_{(|X|>a)}\right) \leq \varepsilon$ for each $a>$ $\eta(\varepsilon)$, for each $\varepsilon>0$. Then the operation $\delta$ defined by $\delta(\varepsilon) \equiv \frac{\varepsilon}{2} / \eta\left(\frac{\varepsilon}{2}\right)$ satisfies the conditions in Assertion 2.

Proof. 1. Let $n \geq 1$ be arbitrary. Then $n 1_A$ is integrable. Hence by Assertion 1 of Proposition 4.3.4, the function $|X| \wedge\left(n 1_A\right)$ is integrable. Now consider each $\omega \in \operatorname{domain}(X) \cap \operatorname{domain}\left(1_A\right)$. Suppose
$$\left|X(\omega) 1_A(\omega)\right| \wedge n \neq|X(\omega)| \wedge\left(n 1_A(\omega)\right) .$$
Then $1_A(\omega) \neq 0$. Hence $1_A(\omega)=1$. It follows that $\left.|X(\omega)| \wedge n \neq|X(\omega)| \wedge n\right)$, which is a contradiction. Thus
$$\left|X(\omega) 1_A(\omega)\right| \wedge n=|X(\omega)| \wedge\left(n 1_A(\omega)\right)$$
for each $\omega \in \operatorname{domain}(X) \cap \operatorname{domain}\left(1_A\right)$. In other words, $\left|X 1_A\right| \wedge n=|X| \wedge$ $\left(n 1_A\right)$. We saw earlier that $|X| \wedge\left(n 1_A\right)$ is integrable. Hence $\left|X 1_A\right| \wedge n$ is integrable.

# 概率论代考

## 数学代写|概率论代写Probability theory代考|Uniform Integrability

1. 让 $A$ 是一个任意的可积集。然后 $X 1_A \in L$.
2. $I\left(|X| 1_A\right) \rightarrow 0$ 作为 $\mu(A) \rightarrow 0$ ， 在哪里 $A$ 是 任意可积集。更准确地说，存在一个操作 $\delta:(0, \infty) \rightarrow(0, \infty)$ 这样 $I\left(|X| 1_A\right) \leq \varepsilon$ 对于 每个可积集 $A$ 和 $\mu(A)<\delta(\varepsilon)$, 对于每个 $\varepsilon>0$.
3. $I\left(|X| 1_{(|X|>a)}\right) \rightarrow 0$ 作为 $a \rightarrow \infty$. 更准确地 说，假设 $I|X| \leq b$ 对于一些 $b>0$, 让操作 $\delta$ 如断 言 2 中所述。对于每个 $\varepsilon>0$, 如果我们定义 $\eta(\varepsilon) \equiv b / \delta(\varepsilon)$ ，然后 $I\left(|X| 1_{(|X|>a)}\right) \leq \varepsilon$ 每 个 $a>\eta(\varepsilon)$.
4. 假设一个操作 $\eta>0$ 是这样的 $I\left(|X| 1_{(|X|>a)}\right) \leq \varepsilon$ 每个 $a>\eta(\varepsilon)$, 对于每个 $\varepsilon>0$. 然后操作 $\delta$ 被定义为 $\delta(\varepsilon) \equiv \frac{\varepsilon}{2} / \eta\left(\frac{\varepsilon}{2}\right)$ 满 足断言 2 中的条件。
证明。1. 让 $n \geq 1$ 是任意的。然后 $n 1_A$ 是可积的。因 此，根据命题 4.3.4 的断言 1 ，函数 $|X| \wedge\left(n 1_A\right)$ 是可 积的。现在考虑每个
$\omega \in \operatorname{domain}(X) \cap \operatorname{domain}\left(1_A\right)$. 认为
$$\left|X(\omega) 1_A(\omega)\right| \wedge n \neq|X(\omega)| \wedge\left(n 1_A(\omega)\right)$$然后 $1_A(\omega) \neq 0$. 因此 $1_A(\omega)=1$. 它遵循
$|X(\omega)| \wedge n \neq|X(\omega)| \wedge n)$ ，这是矛盾的。因此
$$\left|X(\omega) 1_A(\omega)\right| \wedge n=|X(\omega)| \wedge\left(n 1_A(\omega)\right)$$
每个 $\omega \in \operatorname{domain}(X) \cap \operatorname{domain}\left(1_A\right)$. 换句话说， $\left|X 1_A\right| \wedge n=|X| \wedge\left(n 1_A\right)$. 我们之前看到
$|X| \wedge\left(n 1_A\right)$ 是可积的。因此 $\left|X 1_A\right| \wedge n$ 是可积的。

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