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# 数学代写|勒贝格积分代写Lebesgue Integration代考|BSMA2003 Lebesgue Measure

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## 数学代写|勒贝格积分代写Lebesgue Integration代考|Lebesgue Measure

The $\sigma$-algebra of primary interest to us is the one generated by Borel sets and null sets. Alternatively, as a consequence of Exercise 2.3.3 (5), it is the $\sigma$-algebra of subsets of $\mathbb{R}$ generated by open intervals, and null sets, or the one generated by closed intervals and null sets.

Definition 2.4.1. (Lebesgue measurable set). The $\sigma$-algebra of subsets of $\mathbb{R}$ generated by open intervals and null sets will be denoted by $\mathcal{M}$. Sets in $\mathcal{M}$ will be called Lebesgue measurable, or measurable for short. If I is a closed interval, then $\mathcal{M}(I)$ will denote the Lebesgue measurable subsets of $I$.

For simplicity we will focus on subsets of $I=[0,1]$ though we could just as well use any other interval. Notice that it is a consequence of part (4) of Exercise 2.3.3 that $\mathcal{M}(I)$ is a $\sigma$-algebra of subsets of $I$. It is by no means obvious that $\mathcal{M}$ is not the $\sigma$-algebra of all subsets of $\mathbb{R}$. However, in Appendix $\mathrm{C}$ we will construct a subset of $I$ which is not in $\mathcal{M}$.
We are now ready to state the main theorem of this chapter.

## 数学代写|勒贝格积分代写Lebesgue Integration代考|The Lebesgue Density Theorem

The following theorem asserts that if a subset of an interval $I$ is “equally distributed” throughout the interval, then it must be a null set or a set of full measure, i.e., the complement of a null set. For example, it is not possible to have a set $A \subset[0,1]$ which contains half of each subinterval, i.e., it is impossible to have
$$\mu(A \cap[a, b])=\mu([a, b]) / 2$$
for all $0<a<b<1$. There will always be small intervals with a “high concentration” of points of $A$ and other subintervals with a low concentration. Put another way, it asserts that given any $p<1$ there is an interval $U$ such that a point in $U$ has “probability” at least $p$ of being in $A$.

Theorem 2.5.1. If $A$ is a Lebesgue measurable set and $\mu(A)>0$ and if $0<p<1$, then there is an open interval $U=(a, b)$ such that $\mu(A \cap U) \geq p \mu(U)=p(b-a)$.

Proof. Let $p \in(0,1)$ be given. We know from Proposition 2.4.4 that for any $\varepsilon>0$ there is an open set $V$ which contains $A$ such that $\mu(V)<\mu(A)+\varepsilon$ and that we can express $V$ as $V=\bigcup_{n=1}^{\infty} U_n$ where $\left{U_n\right}_{n=1}^{\infty}$ is a countable collection of pairwise disjoint open intervals.
Then
$$\mu(A) \leq \mu(V)=\sum_{n=1}^{\infty} \operatorname{len}\left(U_n\right)<\mu(A)+\varepsilon .$$

## 数学代写|勒贝格积分代写|勒贝格积分代考|勒贝格密度定理

$$\mu(A \cap[a, b])=\mu([a, b]) / 2$$

$$\慕(A)=sum_{n=1}^{infty}。\operatorname{len}\left(U_n\right)<\mu(A)+\varepsilon 。$$

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