# 统计代写|概率论代写Probability theory代考|MATHS7103

#### Doug I. Jones

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## 统计代写|概率论代写Probability theory代考|Supplement: Signed Measures

In this section, we show the decomposition theorems for signed measures (Hahn, Jordan) and deliver an alternative proof for Lebesgue’s decomposition theorem. We owe some of the proofs to [89].

Definition 7.35 Let $\mu$ and $v$ be two measures on $(\Omega, \mathcal{A}) . v$ is called totally continuous with respect to $\mu$ if, for any $\varepsilon>0$, there exists a $\delta>0$ such that for all $A \in \mathcal{A}$
$$\mu(A)<\delta \quad \text { implies } \quad v(A)<\varepsilon .$$
Remark 7.36 The definition of total continuity is similar to that of uniform integrability (see Theorem 6.24(iii)), at least for finite $\mu$. We will come back to this connection in the framework of the martingale convergence theorem that will provide an alternative proof of the Radon-Nikodym theorem (Corollary 7.34). $\diamond$
Theorem $7.37$ Let $\mu$ and $v$ be measures on $(\Omega, \mathcal{A})$. If $\nu$ is totally continuous with respect to $\mu$, then $v \ll \mu$. If $v(\Omega)<\infty$, then the converse also holds.

Proof $” \Longrightarrow “$ Let $v$ be totally continuous with respect to $\mu$. Let $A \in \mathcal{A}$ with $\mu(A)=0$. For all $\varepsilon>0$, by assumption, $v(A)<\varepsilon$; hence $v(A)=0$ and thus $v \ll \mu$. exist an $\varepsilon>0$ and sets $A_n \in \mathcal{A}$ with $\mu\left(A_n\right)<2^{-n}$ but $v\left(A_n\right) \geq \varepsilon$ for all $n \in \mathbb{N}$. Define $A:=\limsup {n \rightarrow \infty} A_n=\bigcap{n=1}^{\infty} \bigcup_{k=n}^{\infty} A_k$. Then
$$\mu(A)=\lim {n \rightarrow \infty} \mu\left(\bigcup{k=n}^{\infty} A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} \mu\left(A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} 2^{-k}=0$$

Since $\nu$ is finite and upper semicontinuous (Theorem 1.36), we have
$$v(A)=\lim {n \rightarrow \infty} v\left(\bigcup{k=n}^{\infty} A_k\right) \geq \inf {n \in \mathbb{N}} v\left(A_n\right) \geq \varepsilon>0 .$$ Thus $v \nless \mu$. Example 7.38 In the converse implication of the theorem, the assumption of finiteness is essential. For example, let $\mu=\mathcal{N}{0,1}$ be the standard normal distribution on $\mathbb{R}$ and let $v$ be the Lebesgue measure on $\mathbb{R}$. Then $v$ has the density $f(x)=$ $\sqrt{2 \pi} e^{x^2 / 2}$ with respect to $\mu$. In particular, we have $v \ll \mu$. On the other hand, $\mu([n, \infty)) \stackrel{n \rightarrow \infty}{\longrightarrow} 0$ and $v([n, \infty))=\infty$ for any $n \in \mathbb{N}$. Hence $v$ is not totally continuous with respect to $\mu$. $\diamond$

## 统计代写|概率论代写Probability theory代考|Supplement: Dual Spaces

By the Riesz-Fréchet theorem (Theorem 7.26), every continuous linear functional $F: L^2(\mu) \rightarrow \mathbb{R}$ has a representation $F(g)=\langle f, g\rangle$ for some $f \in L^2(\mu)$. On the other hand, for any $f \in L^2(\mu)$, the map $L^2(\mu) \rightarrow \mathbb{R}, g \mapsto\langle f, g\rangle$ is continuous and linear. Hence $L^2(\mu)$ is canonically isomorphic to its topological dual space $\left(L^2(\mu)\right)^{\prime}$. This dual space is defined as follows.

Definition 7.47 (Dual space) Let $(V,|\cdot|)$ be a Banach space. The dual space $V^{\prime}$ of $V$ is defined by
$V^{\prime}:={F: V \rightarrow \mathbb{R}$ is continuous and linear $}$.
For $F \in V^{\prime}$, we define $|F|^{\prime}:=\sup {|F(f)|:|f|=1}$.
Remark 7.48 As $F$ is continuous, for any $\delta>0$, there exists an $\varepsilon>0$ such that $|F(f)|<\delta$ for all $f \in V$ with $|f|<\varepsilon$. Hence $|F|^{\prime} \leq \delta / \varepsilon<\infty . \diamond$

We are interested in the case $V=L^p(\mu)$ for $p \in[1, \infty]$ and write $|F|_p^{\prime}$ for the norm of $F \in V^{\prime}$. In the particular case $V=L^2(\mu)$, by the Cauchy-Schwarz inequality, we have $|F|_2^{\prime}=|f|_2$. This can be generalized:
Lemma 7.49 Let $p, q \in[1, \infty]$ with $\frac{1}{p}+\frac{1}{q}=1$. The canonical map
\begin{aligned} \kappa: L^q(\mu) & \rightarrow\left(L^p(\mu)\right)^{\prime} \ \kappa(f)(g) &=\int f g d \mu \quad \text { for } f \in L^q(\mu), g \in L^p(\mu) \end{aligned}
is an isometry; that is, $|\kappa(f)|_p^{\prime}=|f|_q$.
Proof We show equality by showing the two inequalities separately.
” $\leq$ “, This follows from Hölder’s inequality.
” $\geq$ “, For any admissible pair $p, q$ and all $f \in \mathcal{L}^q(\mu), g \in \mathcal{L}^p(\mu)$, by the definition of the operator norm, $|\kappa(f)|_p^{\prime}|g|_p \geq\left|\int f g d \mu\right|$. Define the sign function $\operatorname{sign}(x)=\mathbb{1}{(0, \infty)}(x)-\mathbb{1}{(-\infty, 0)}(x)$. Replacing $g$ by $\tilde{g}:=|g| \operatorname{sign}(f)$ (note that $\left.|\tilde{g}|_p=|g|_p\right)$, we obtain
$$|\kappa(f)|_p^{\prime}|g|_p \geq\left|\int f \tilde{g} d \mu\right|=|f g|_1 .$$
First consider the case $q=1$ and $f \in \mathcal{L}^1(\mu)$. Applying (7.12) with $g \equiv 1 \in \mathcal{L}^{\infty}(\mu)$ yields $|\kappa(f)|_{\infty}^{\prime} \geq|f|_1$.
Now let $q \in(1, \infty)$. Let $g:=|f|^{q-1}$. Since $\frac{q-1}{q}=\frac{1}{p}$, we have
$$|\kappa(f)|_p^{\prime} \cdot|g|_p \geq|f g|_1=\left||f|^q\right|_1=|f|_q^q=|f|_q \cdot|f|_q^{q-1}=|f|_q \cdot|g|_p$$

# 概率论代考

## 统计代写|概率论代写概率论代考|补充:签署办法

$$\mu(A)<\delta \quad \text { implies } \quad v(A)<\varepsilon .$$总的连续性的定义类似于一致可积性的定义(见定理6.24(iii))，至少对于有限的是这样 $\mu$。我们将在鞅收敛定理的框架中回到这个联系，它将提供Radon-Nikodym定理的另一种证明(推论7.34)。 $\diamond$

$$\mu(A)=\lim {n \rightarrow \infty} \mu\left(\bigcup{k=n}^{\infty} A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} \mu\left(A_k\right) \leq \lim {n \rightarrow \infty} \sum{k=n}^{\infty} 2^{-k}=0$$

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