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这是麻州大学波士顿分校实分析课程的代写成功案例。
MATH450课程简介
An Introduction to Real Analysis
Course #: MATH 450
Description:
A rigorous treatment of the calculus of functions of one real variable. Emphasis is on proofs. Includes discussion of topology of real line, limits, continuity, differentiation, integration and series.
The real number system consists of an uncountable set (R), together with two binary operations denoted + and ⋅, and an order denoted <. The operations make the real numbers a field, and, along with the order, an ordered field.
Prerequisites
In mathematics, the branch of real analysis studies the behavior of real numbers, sequences and series of real numbers, and real functions.Some particular properties of real-valued sequences and functions that real analysis studies include convergence, limits, continuity, smoothness, differentiability and integrability.
Real analysis is distinguished from complex analysis, which deals with the study of complex numbers and their functions.The theorems of real analysis rely on the properties of the real number system, which must be established.
MATH450 Real Analysis HELP（EXAM HELP， ONLINE TUTOR）
Let $f, g, f_k, k \geq 1$, be measurable functions from $X$ to $\overline{\mathbb{R}}$.
(a) Show that ${x: f(x)<g(x)}$ and ${x: f(x)=g(x)}$ are measurable sets.
(b) Show that $\left{x: \lim _{k \rightarrow \infty} f_k(x)\right.$ exists and is finite $}$ is measurable.
Solution
(a) Suffice to show ${x: F(x)>0}$ and ${x: F(x)=0}$ are measurable. If $F$ is measurable, use
$$
\begin{gathered}
{x: F(x)>0}=F^{-1}(0, \infty] \
{x: F(x)=0}=F^{-1}[0, \infty] \cap F^{-1}[-\infty, 0]
\end{gathered}
$$
Alternatively, one may consider
$$
{x \in X: f(x)<g(x)}=\bigcup_{r \in \mathbb{Q}}\left(f^{-1}[-\infty, r) \cap g^{-1}(r, \infty]\right)
$$
$$
{x \in X: f(x)=g(x)}={x \in X: f(x)g(x)}^c
$$
(b) Since $g(x)=\limsup {k \rightarrow \infty} f_k(x)$ and $\liminf {k \rightarrow \infty} f_k(x)$ are measurable.
$$
\left{x: \lim {k \rightarrow \infty} f_k(x) \text { exists }\right}=\left{x: \liminf {k \rightarrow \infty} f_k(x)=\limsup _{k \rightarrow \infty} f_k(x)\right}
$$
On the other hand, the $\operatorname{set}{x: g(x)<+\infty}$ is also measurable, so is their intersection.
There are two conditions (i) and (ii) in the definition of a measure $\mu$ on $(X, \mathcal{M})$. Show that (i) can be replaced by the “nontriviality condition”: There exists some $E \in \mathcal{M}$ with $\mu(E)<\infty$.
Solution If $\mu$ is a measure satisfying the nontriviality condition and (ii), let $A_1=E, A_i=\phi$ for $i \geq 2$ in ii)
$$
\infty>\mu(E)=\sum_{i=1}^{\infty} \mu\left(A_i\right) \geq \mu\left(A_1\right)+\mu\left(A_2\right)=\mu(E)+\mu(\phi)
$$
so $0 \geq \mu(\phi) \geq 0$. We have $\mu$ is a measure satisfying (i) and (ii).
If $\mu$ is a measure satisfying (i) and (ii), taking $E=\phi$, we have the nontriviality condition.
Let $\left{A_k\right}$ be measurable and $\sum_{k=1}^{\infty} \mu\left(A_k\right)<\infty$ and $A=\left{x \in X: x \in A_k\right.$ for infinitely many $\left.k\right}$.
We know that $A$ is measurable from (1). Show that $\mu(A)=0$.
Solution Since $\sum_{k=1}^{\infty} \mu\left(A_k\right)<\infty$, we have $\sum_{k=n}^{\infty} \mu\left(A_k\right) \rightarrow 0$ as $n \rightarrow \infty$. For any
$\mathrm{n} \in N$, we have
$$
A \subset \bigcup_{k \geq n} A_k
$$
and so
$$
\mu(A) \leq \sum_{k=n}^{\infty} \mu\left(A_k\right)
$$
Taking $n \rightarrow \infty$, we have $\mu(A)=0$.
This result is called Borel-Cantelli lemma.
MY-ASSIGNMENTEXPERT™可以为您提供 UMASS BOSTON MATH450 REAL ANALYSIS实分析的代写代考和辅导服务！
