数学代写|Math540 Real Analysis

MY-ASSIGNMENTEXPERT™可以为您提供 math.illinois.edu Math540 Real Analysis实分析的代写代考和辅导服务！

Math540课程简介

Real analysis, as we now understand it, acquired a different emphasis from classical analysis in the second half of the nineteenth century. In his famous remark to Thomas Stieltjes, Charles Hermite wrote: “I turn with terror and horror from this lamentable scourge of continuous functions with no derivatives”. Nowadays, non-regular functions and their generalizations are of fundamental importance in mathematics and physics. The study of non-regular functions requires a better theory of integration; historically, that was the main motivation for the development of measure theory.

Prerequisites 

The course will cover the prescribed syllabus:

Abstract measure theory, outer measure, Lebesgue measure on the real line, measurable functions, Egorov and Lusin theorems, the Lebesgue integral, differentiation, Lp -spaces, elementary Hilbert space theory and trigonometric series. Other topics will be included as time permits.

Text (required): H.L. Royden and Patrick Fitzpatrick, Real Analysis (4th edition), Pearson, 2010.

Additional texts (not required; do not have to buy):

G. B. Folland, Real Analysis, John Wiley & Sons.
W. Rudin, Real and Complex Analysis, McGraw-Hill.

Math540 Real Analysis HELP（EXAM HELP， ONLINE TUTOR）

First, we show that for any $B \in \Sigma_\mu$, there exists $A \in \mathcal{R}_{\sigma,\delta}$ such that $B \subset A$ and $\mu^*(A\setminus B)=0$.

Since $\mu$ is $\sigma$-finite, we can write the space as a countable union of sets of finite $\mu$ measure, i.e., $X = \bigcup_{n=1}^\infty X_n$ where $\mu(X_n) < \infty$. Then, for any $B \in \Sigma_\mu$, we have $B = \bigcup_{n=1}^\infty (B \cap X_n)$, where each $B \cap X_n$ has finite outer measure since $\mu$ is additive on $\mathcal{R}$.

For each $n$, let $A_n$ be a set in $\mathcal{R}$ such that $B \cap X_n \subset A_n$ and $\mu(A_n\setminus (B \cap X_n))=0$ (this is possible since $\mu$ is an additive measure on $\mathcal{R}$). Then, we define $A = \bigcup_{n=1}^\infty A_n \in \mathcal{R}{\sigma,\delta}$. Note that $B \subset A$ since $B \subset \bigcup{n=1}^\infty (B \cap X_n) \subset \bigcup_{n=1}^\infty A_n = A$. Moreover, we have \begin{align*} \mu^(A\setminus B) &= \mu^\left(\bigcup_{n=1}^\infty A_n \setminus B\right) \ &\leq \sum_{n=1}^\infty \mu^(A_n \setminus (B \cap X_n)) \ &= \sum_{n=1}^\infty \mu(A_n \setminus (B \cap X_n)) \ &= 0, \end{align} where the second inequality follows from the subadditivity of outer measure and the last equality follows from the fact that $\mu$ is an additive measure on $\mathcal{R}$.

Therefore, we have shown that for any $B \in \Sigma_\mu$, there exists $A \in \mathcal{R}_{\sigma,\delta}$ such that $B \subset A$ and $\mu^*(A\setminus B)=0$.

The assumption that $\mu$ is $\sigma$-finite is necessary for the argument above, since we used the fact that the space $X$ can be written as a countable union of sets of finite $\mu$ measure. Without this assumption, it is possible that the space $X$ itself has infinite $\mu$ measure, in which case the argument breaks down.

Let $B$ be a subset of $\mathbb{R}$ such that $m^(B)>0$, where $m^$ denotes the Lebesgue outer measure. Since $m^(B)>0$, there exists an open interval $I\subset\mathbb{R}$ such that $I\cap B\neq\varnothing$ and $m^(I)<\infty$. Without loss of generality, we may assume that $I$ is a bounded interval, since unbounded intervals can be expressed as a countable union of bounded intervals.

Let $E$ be a nonmeasurable subset of $I\cap B$. Such a set exists by the Hahn-Banach theorem, which states that there exist nonmeasurable subsets of any set of positive measure. Since $E$ is a subset of $I$, we have $m^(E)\leq m^(I)<\infty$. Also, since $E$ is nonmeasurable, it follows that $m^(E\cap A)+m^(E\cap A^c)\neq m^(A)$ for some measurable set $A\subseteq I$. Letting $B’=(B\setminus I)\cup A$ yields a set $B’\subseteq \mathbb{R}$ such that $m^(B’)=m^(B)$ and $E\subseteq B’$. Moreover, $A$ is a measurable subset of $B’$, so if $E\subseteq A$, we would have $m^(E\cap A)+m^(E\cap A^c)=m^(E)<\infty=m^*(A)$, which is a contradiction. Therefore, $E$ is a nonmeasurable subset of $B’$.

We can repeat this process with $B’$ and $E$ to obtain a nonmeasurable set $E\subseteq B$ as desired. Thus, for every set $B\subseteq\mathbb{R}$ with $m^*(B)>0$, there exists a nonmeasurable set $E\subseteq B$.

MY-ASSIGNMENTEXPERT™可以为您提供UNIVERSITY OF ILLINOIS URBANA-CHAMPAIGN MATH2940 linear algebra线性代数课程的代写代考和辅导服务！ 请认准MY-ASSIGNMENTEXPERT™. MY-ASSIGNMENTEXPERT™为您的留学生涯保驾护航。