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这是香港中文大学实分析课程的代写成功案例。
MATH5011课程简介
Real and Complex Analysis, 3rd ed. W. Rudin, McGrawHill, New York 1966.
Measure Theory and Fine Properties of Functions, L.C. Evans and R.F. Gariepy, CRC Press 1992.
Real Analysis: Measure Theory, Integration and Hilbert Spaces, E.M. Stein and R. Shakarchi, Princeton Lectures in Analysis, Princeton 2005.
Real and Abstract Analysis, E. Hewitt and K. Stromberg, Graduate Texts in Mathematics, Springer-Verlag, New York 1975.
Prerequisites
Announcement
Here is the course outline Download file
The mid-term exam of Math 5011 will be held at LSB, LT6 next Friday, Oct 21, 2022, 18:30 pm-21:30pm. There will be no lecture in the morning of next Friday (Oct 21).
Course Description
Abstract integration theory; outer measures and Caratheodory’s construction, Borel, Radon and Hausdorff measures; positive linear functionals and Riesz representation theorem; Lp-spaces and their functional properties; and signed measures, Radon-Nikodym theorem and the dual of the space of continuous functions.
MATH5011 Real Analysis HELP(EXAM HELP, ONLINE TUTOR)
Let $B$ be the set defined in (1). Let $\mu$ be a measure on $(X, \mathcal{M})$. Show that
$$
\mu(B) \leq \liminf _{k \rightarrow \infty} \mu\left(A_k\right) .
$$
Solution Using the characterization $$ B=\bigcup{k=1}^{\infty} \bigcap_{j \geq k} A_j,
$$
and the fact that $\left{\cap_{j \geq k} A_j\right}$ is ascending in $k$, we have
$$
\begin{aligned}
\mu(B) & =\lim {k \rightarrow \infty} \mu\left(\bigcap{j \geq k} A_j\right) \
& =\liminf {k \rightarrow \infty} \mu\left(\bigcap{j \geq k} A_j\right) \
& \leq \liminf _{k \rightarrow \infty} \mu\left(A_k\right)
\end{aligned}
$$
Here we review Riemann integral. Let $f$ be a bounded function defined on $[a, b], a, b \in \mathbb{R}$. Given any partition $P=\left{a=x_0<x_1<\cdots<x_n=b\right}$ on $[a, b]$ and tags $z_j \in\left[x_j, x_{j+1}\right]$, there corresponds a Riemann sum of $f$ given by $R(f, P, \mathbf{z})=\sum_{j=0}^{n-1} f\left(z_j\right)\left(x_{j+1}-x_j\right)$. The function $f$ is called Riemann integrable with integral $L$ if for every $\varepsilon>0$ there exists some $\delta$ such that
$$
|R(f, P, \mathbf{z})-L|<\varepsilon
$$
whenever $|P|<\delta$ and $\mathbf{z}$ is any tag on $P$. (Here $|P|=\max {j=0}^{n-1}\left|x{j+1}-x_j\right|$ is the length of the partition.) Show that
- For any partition $P$, define its Darboux upper and lower sums by
$$
\bar{R}(f, P)=\sum_j \sup \left{f(x): x \in\left[x_j, x_{j+1}\right]\right}\left(x_{j+1}-x_j\right)
$$
and
$$
R(f, P)=\sum_j \inf \left{f(x): x \in\left[x_j, x_{j+1}\right]\right}\left(x_{j+1}-x_j\right)
$$
respectively. Show that for any sequence of partitions $\left{P_n\right}$ satisfying $\left|P_n\right| \rightarrow 0$ as $n \rightarrow \infty, \lim {n \rightarrow \infty} \bar{R}\left(f, P_n\right)$ and $\lim {n \rightarrow \infty} R\left(f, P_n\right)$ exist. - $\left{P_n\right}$ as above. Show that $f$ is Riemann integrable if and only if
$$
\lim {n \rightarrow \infty} \bar{R}\left(f, P_n\right)=\lim {n \rightarrow \infty} R\left(f, P_n\right)=L
$$ - A set $E$ in $[a, b]$ is called of measure zero if for every $\varepsilon>0$, there exists a countable subintervals $J_n$ satisfying $\sum_n\left|J_n\right|<\varepsilon$ such that $E \subset \bigcup_n J_n$. Prove Lebsegue’s theorem which asserts that $f$ is Riemann integrable if and only if the set consisting of all discontinuity points of $f$ is a set of measure zero. Google for help if necessary.
Solution:
(a) It suffices to show: For every $\varepsilon>0$, there exists some $\delta$ such that
$$
0 \leq \bar{R}(f, P)-\bar{R}(f)<\varepsilon
$$
and
$$
0 \leq R(f)-R(f, P)<\varepsilon $$ for any partition $P,|P|<\delta$, where $$ \bar{R}(f)=\inf P \bar{R}(f, P) $$ and $$ R(f)=\sup _P R(f, P) $$ If it is true, then $\lim {n \rightarrow \infty} \bar{R}\left(f, P_n\right)$ and $\lim _{n \rightarrow \infty} R\left(f, P_n\right)$ exist and equal to $\bar{R}(f)$ and $R(f)$ respectively. Given $\varepsilon>0$, there exists a partition $Q$ such that
$$
\bar{R}(f)+\varepsilon / 2>\bar{R}(f, Q)
$$
Let $m$ be the number of partition points of $Q$ (excluding the endpoints). Consider any partition $P$ and let $R$ be the partition by putting together $P$ and $Q$. Note that the number of subintervals in $P$ which contain some partition points of $Q$ in its interior must be less than or equal to $m$. Denote the indices of the collection of these subintervals in $P$ by $J$. We have
$$
0 \leq \bar{R}(f, P)-\bar{R}(f, R) \leq \sum_{j \in J} 2 M \Delta x_j \leq 2 M \times m|P|,
$$
where $M=\sup _{[a, b]}|f|$, because the contributions of $\bar{R}(f, P)$ and $\bar{R}(f, Q)$ from the subintervals not in $J$ cancel out. Hence, by the fact that $R$ is a refinement of $\mathrm{Q}$,
$$
\bar{R}(f)+\varepsilon / 2>\bar{R}(f, Q) \geq \bar{R}(f, R) \geq \bar{R}(f, P)-2 M m|P|,
$$
i.e.,
$$
0 \leq \bar{R}(f, P)-\bar{R}(f)<\varepsilon / 2+2 M m|P|
$$
Now, we choose
$$
\delta<\frac{\varepsilon}{1+4 M m}
$$
Then for $P,|P|<\delta$,
$$
0 \leq \bar{R}(f, P)-\bar{R}(f)<\varepsilon
$$
Similarly, one can prove the second inequality.
Let $g$ be a measurable function in $[0, \infty]$. Show that
$$
m(E)=\int_E g d \mu
$$
defines a measure on $\mathcal{M}$. Moreover,
$$
\int_X f d m=\int_X f g d \mu, \quad \forall f \text { measurable in }[0, \infty] .
$$
Solution: We readily check that
(1) $m(\phi)=0$
(2) $m(E) \geq 0, \forall E \in M$
(3) For mutually disjoint $A_k \in M$,
$$
m\left(\bigcup_{k=1}^{\infty} A_k\right)=\int_X \sum_{k=1}^{\infty} \chi_{A_k} g d \mu=\sum_{k=1}^{\infty} \int \chi_{A_k} g d \mu=\sum_{k=1}^{\infty} m\left(A_k\right)
$$
by monotone convergence theorem, since $\sum_{k=1}^n \chi_{A_k} g \uparrow \sum_{k=1}^{\infty} \chi_{A_k} g$.
To prove the last assertion, consider the following cases:
(a) $f=\chi_E$ for some $E \in M$.
$$
\int_X f d m=\int_E d m=m(E)=\int_E g d \mu=\int_X \chi_E g d \mu=\int_X f g d \mu .
$$
(b) $f$ is a non-negative simple function.
This follows from (a).
(c) $f$ is a non-negative measurable function.
Pick a sequence $s_n \geq 0$ of simple functions such that $s_n \uparrow f$ pointwisely.
Then $0 \leq s_n g \uparrow g$ pointwisely. From (b),
$$
\int_X s_n d m=\int_X s_n g d \mu
$$
Taking $n \rightarrow \infty$, by monotone convergence theorem, we have
$$
\int_X f d m=\int_X f g d \mu .
$$
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