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# 数学代写|微积分代考calculus assignment|Measure Theory

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## 数学代写|微积分代考calculus assignment|Duality for full covering relations

We add a dual notion to that of full cover. This should naturally belong in Chapter 3 but, to keep the presentation entirely elementary, this more subtle idea and its exploitation was delayed. The definitions are close parallels to Definitions $2.1$ and 2.2.

A covering relation $\beta$ is fine at a point $x_{0}$ if for every $\delta>0$, the relation $\beta$ contains at least one pair $\left([c, d], x_{0}\right)$ for which $c \leq x_{0} \leq d$ and $0<d-c<\delta$.

A covering relation $\beta$ is fine cover of a set $E$ if $\beta$ is fine at each point $x$ belonging to the set $E$.

No discussion of full covers can proceed for long without fine covers making their appearance. This arises simply from the process of negation. The following elementary exercise should clarify.

EXERCISE 44. Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function, let $\epsilon>0$ and define the covering relations
$$\beta_{1}={([c, d], x): x \in[c, d] \text { and }|f(d)-f(c)|<\epsilon}$$

and
$$\beta_{2}={([c, d], x): x \in[c, d] \text { and }|f(d)-f(c)| \geq \epsilon}$$

## 数学代写|微积分代考calculus assignment|Measures

Our goal is to provide three different definitions of Lebesgue’s measure. The fact that all three are equivalent is known as the Vitali covering theorem. The three measures will be denoted ${ }^{1}$ as
$\mathcal{L}, \mathcal{L}_{}$, and $\mathcal{L}^{}$.
By a measure we mean a function defined for all sets and assuming real values (including $+\infty$ ) and satisfying the following two properties:

Definition 6.3. A set function $\mathcal{M}$ defined for all sets of reals numbers is a measure $^{2}$ on $\mathbb{R}$ if it has the following properties:
(a) $\mathcal{M}(\emptyset)=0$.
(b) For any sequence of sets $E, E_{1}, E_{2}, E_{3}, \ldots$ for which
$$E \subset \bigcup_{n=1}^{\infty} E_{n}$$
the inequality
$$\mathcal{M}(E) \leq \sum_{n=1}^{\infty} \mathcal{M}\left(E_{n}\right)$$
must hold.

## 数学代写|微积分代考CALCULUS ASSIGNMENT|Lebesgue’s measure

The measure $\mathcal{L}$ is known as Lebesgue’s measure and assigns a natural length to every set. There are numerous presentations of Lebesgue’s measure and any one of these can be consulted. We offer no proofs of the statements in this section. We simply list what we need.

One traditional construction take place in four steps. For the first step define
$$\mathcal{L}(\emptyset)=0$$
as required by Definition 6.3. For the second step take an arbitrary open interval $(a, b)$ where $-\infty \leq a<b \leq \infty$ and assign
$$\mathcal{L}((a, b))=b-a .$$
For the third step we take an arbitrary open set $G$ and display $G$ as a sequence of component intervals
$$G=\bigcup_{k=1}^{\infty}\left(a_{k}, b_{k}\right) .$$
(There may be only finitely many components but the argument is the same.)
Then we simply define the measure of $G$ to be the sum of the lengths of the component intervals:
$$\mathcal{L}(G)=\bigcup_{k=1}^{\infty} \mathcal{L}\left(\left(a_{k}, b_{k}\right)\right)$$
This is consistent with step 2 since an open interval $(a, b)$ would be considered an open set with a single component. It is traditional to consider the empty set as an open set having no components; to be consistent with step 1 then merely requires us to consider an empty sum to be zero.

## 数学代写|微积分代考CALCULUS ASSIGNMENT|DUALITY FOR FULL COVERING RELATIONS

b1=([C,d],X):X∈[C,d] 和 |F(d)−F(C)|<ε

$$\beta_{2}={([c, d], x): x \in[c, d] \text { and }|f(d)-f(c)| \geq \epsilon}$$

## 数学代写|微积分代考CALCULUS ASSIGNMENT|MEASURES

$\ mathcal {L}, \ mathcal {L} _ {,一种nd\数学{L}^{ }.乙是一种米和一种s在r和在和米和一种n一种F在nC吨一世这nd和F一世n和dF这r一种lls和吨s一种nd一种ss在米一世nGr和一种l在一种l在和s(一世nCl在d一世nG+\infty$ ) 并满足以下两个性质：

$$E \subset \bigcup_{n=1}^{\infty} E_{n}$$
the inequality
$$\mathcal{M}(E) \leq \sum_{n=1}^{\infty} \mathcal{M}\left(E_{n}\right)$$

## 数学代写|微积分代考CALCULUS ASSIGNMENT|LEBESGUE’S MEASURE

$$\mathcal{L}(\emptyset)=0$$
as required by Definition 6.3. For the second step take an arbitrary open interval $(a, b)$ where $-\infty \leq a<b \leq \infty$ and assign
$$\mathcal{L}((a, b))=b-a .$$
For the third step we take an arbitrary open set $G$ and display $G$ as a sequence of component intervals
$$G=\bigcup_{k=1}^{\infty}\left(a_{k}, b_{k}\right) .$$
(There may be only finitely many components but the argument is the same.)
Then we simply define the measure of $G$ to be the sum of the lengths of the component intervals:
$$\mathcal{L}(G)=\bigcup_{k=1}^{\infty} \mathcal{L}\left(\left(a_{k}, b_{k}\right)\right)$$

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