陶哲轩实分析（上）11.1及习题-Analysis I 11.1

这一节开始讲Riemann积分了，整个Riemann积分这一章的处理是比较新的，用相对好理解的piecewise constant function来定义上积分和下积分，二者相等即为可积，和以往比较经典的达布和其实是一回事，但接受起来就更容易。第一节主要说分法partition。

Exercise 11.1.1

If $X=∅$ then the statement is obvious, since $X$ is bounded and connected and is an interval for trivial reasons.
Now suppose $X≠∅$, then (b) implies (a) is clear, by the definition of an interval. Now suppose (a) holds, then $∃M>0,X⊂[-M,M]$, since $X$ is non-empty and bounded, we can say $\sup ⁡X$ and $\inf ⁡X$ both exist in $\mathbf R$. Now choose any $x∈\mathbf R$ such that
$\inf⁡X<x<\sup⁡X$
By the definition of supremum and infimum, we can find $a,b∈X$, s.t.
$\inf⁡X≤a<x<b≤\sup⁡X$
Since $X$ is connected, $[a,b]⊂X$, thus $x∈X$. So we have the relation $(\inf⁡X,\sup⁡X )⊂X$, also we can easily have $X⊂[\inf⁡X,\sup⁡X ]$, then $X$ is equal to $(\inf⁡X,\sup⁡X )$ plus possibly two points $\inf⁡X$ and $\sup⁡X$, anyway $X$ is an interval.

Exercise 11.1.2

By Lemma 11.1.4, $I$ and $J$ are bounded and connected, we have $M_1>0,M_2>0$, s.t.
$(I⊂[-M_1,M_1 ])∧(J⊂[-M_2,M_2 ])⇒I∩J⊂[-\max⁡(M_1,M_2 ),\max⁡(M_1,M_2 )]$
Thus $I∩J$ is bounded.
Let $∀x,y∈I∩J$, then as $x,y∈I$, we have $[x,y]⊆J$, similarly we have $[x,y]⊆J$, so $[x,y]⊆I∩J$, this means $I∩J$ is connected.
We now can conclude, using Lemma 11.1.4 again, that $I∩J$ is a bounded interval.

Exercise 11.1.3

First, if $I_j$ is not of the form $(c,b)$ or $[c.b)$ for any $a≤c≤b$, then we can have some $0<ϵ_j<b-a$ s.t. $b-ϵ∉I_j$, as $I_j$ is an interval, it’s connected, thus we have
$\sup⁡I_j≤b-ϵ_j<b$
Now assume all intervals $I_j,1≤j≤n$ is not of the form $(c,b)$ or $[c.b)$ for any $a≤c≤b$, then we have $\sup⁡I_j≤b-ϵ_j<b,1≤j≤n$. Then we can conclude
$b-\frac{\min_{1≤j≤n}⁡(ϵ_j )}{2}∉I_i,\quad∀1≤i≤n$
so we have $b-\min_{1≤j≤n}⁡(ϵ_j )/2∉I$, this is a contradiction since $I_1,…,I_n$ form a partition of $I$.
Thus there must exists at least one of $I_j$ in the form of $(c,b)$ or $[c.b)$. By the definition of partition, such $I_j$ can only appear once if $I_1,…,I_n$ form a partition of $I$.

Exercise 11.1.4

$\mathbf P\#\mathbf P'$ is a partition of $I$:
Choose $∀x∈I$, then there’s exactly one $K∈\mathbf P$ and $J∈\mathbf P'$ s.t. $x∈K,x∈J$, thus $x∈K∩J$, and any other element belongs to $\mathbf P\#\mathbf P'$ can’t contain $x$, otherwise $K$ or $J$ can’t be unique. This means $\mathbf P\#\mathbf P'$ is a partition of $I$.
$\mathbf P\#\mathbf P'$ is finer than $\mathbf P$ or $\mathbf P'$:
For every $M∈\mathbf P\#\mathbf P'$, by definition we shall be able to find $K∈\mathbf P$ and $J∈\mathbf P'$ s.t.
$M=K∩J$
This means $M⊆K,K∈\mathbf P$ and $M⊆J,J∈\mathbf P'$, thus we have $\mathbf P\#\mathbf P'$ is finer than $\mathbf P$ or $\mathbf P'$.