Real analysis

未分类

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Q2

Problem

$f(x,y) \in C([0,1]*(0,+\infty))$，$F(x)=\limsup\limits_{y \to \infty}f(x,y)$，Show that $F$ is Borel measurable.

Proof

$$F(x)=\inf\limits_{n}\sup\limits_{y\ge n}f(x,y)$$
First show that $F_n(x)=\sup\limits_{y\ge n}f(x,y)$ is Borel measurable.
$$\{F_n(x)>a\}=\bigcup_{y\ge n}\{f(x,y)>a\}$$
Note that $f(.,y)$ is continuous for any $y \in \mathbb{R}^{+}$, so $\{f(x,y)>a\}$ is open.
Therefore, $\{F_n(x)>a\}$ is open, thus Borel.

$F(x)=\inf\limits_{n} F_n(x)$ is Borel measurable.

Q3

Q4

Q5

Problem

$$g_n(x)=\int_{0}^{\infty}e^{-t}f(nt, x)dt$$ and $f\in L^{1}((0,\infty) \times \mathbb{R})$. Show that
(i) $||g_n||_{L^{1}} \to 0$
(ii) $g_n \to 0$, a.e.

Proof

(i)

$$\begin{aligned} ||g_n||_{L^{1}}&=\int_{\mathbb{R}}|\int_{0}^{\infty}e^{-t}f(nt, x)dt| d\mu(x) \\ &\leq\int_{\mathbb{R}}\int_{0}^{\infty}\frac{1}{n}e^{-u/n}|f(u, x)|du dx \end{aligned}$$

By Tonelli Theorem,

$$\begin{aligned} ||g_n||_{L^{1}}&\leq\int_{\mathbb{R}}\int_{0}^{\infty}\frac{1}{n}e^{-u/n}|f(u, x)|du dx \\ &=\int_{0}^{\infty}\frac{1}{n}e^{-u/n}\int_{\mathbb{R}}|f(u, x)|dx\ du \\ &=\frac{1}{n}\int_{0}^{\infty}e^{-u/n}G(u)\ du \\ \end{aligned}$$

where $G\in L^{1}(0,\infty)$.

By Dominated Convergence Theorem, $\lim_{n\to \infty}{||g_n||_{L^{1}}}=0$

(ii)

$$|g_n(x)|\leq\int_{0}^{\infty}\frac{1}{n}e^{-u/n}|f(u, x)|du\leq\frac{1}{n}\int_{0}^{\infty}|f(u, x)|du$$
Apply Fubini Theorem to $f$, we get $\int|f(u, x)| du\leq \infty$ for a.e. $x\in \mathbb{R}$
Let n approaching $\infty$, we get $\lim_{n\to\infty}g_n(x)=0$ for a.e $x\in \mathbb{R}$.

Q6

Problem

$X=[0,1]$, $\mathcal{L}_0$ is an $\sigma$-algebra on $I$, $\mathcal{A}$ a sub $\sigma$-algebra of $L_0$.
Show that: For any $f \in L^{1}(X, \mathcal{L}_0, m)$, there exists a unique $g \in L^{1}(X, \mathcal{A}, m)$, such that $\int_{A}f \mathbb{d}m=\int_{A}g \mathbb{d}m$ for any $A \in \mathcal{A}$

Proof

Uniqueness
If there is another $h$, $\int_{A}g \mathbb{d}m=\int_{A}h dm$ for any $A \in \mathcal{A}$.
Then $g=h$ a.e. on $[0,1]$.
Existence
Let $\mu(E)=\int_{E}f\mathbb{d}m$ for any $E\in \mathcal{L}_{0}$.
Then $\mu$ is a well-defined complex measure on $X$, and $\mu \ll m$.
$\mu$, $m$ are finite measure on $(X,\mathcal{A})$, and $\mu \ll m$. By Radon-Nikonym Theorem, $\exists\ g \in L^1(X, \mathcal{A}, m)$, s.t. $\mathbb{d}\mu=g\ \mathbb{d}m$.