陶哲轩实分析（上）11.6及习题-Analysis I 11.6

Exerise 11.6.1

If $I$ is of the form $[a,b]$ then by Proposition 11.6.1 $f$ can be Riemann integrable. So we suppose $I$ is of the form $(a,b),(a,b],[a,b)$.
Since $f$ is bounded, $\exists M>0,-M\leq f(x)\leq M,\forall x\in I$, choose $0<{\varepsilon}<(b-a)/2$ to be a small number, then by Proposition 11.6.1, $f$ is Riemann integrable on $[a+{\varepsilon},b-{\varepsilon}]$, in particular we can find a piecewise constant function $h$ on $[a+{\varepsilon},b-{\varepsilon}]$ which majorizes $f$ and
$\int_{[a+{\varepsilon},b-{\varepsilon}]}h<\int_{[a+{\varepsilon},b-{\varepsilon}]}f+{\varepsilon}$
Define
$H(x)=\begin{cases}h(x),&x\in [a+{\varepsilon},b-{\varepsilon}]\\M,&x\in I\backslash[a+{\varepsilon},b-{\varepsilon}] \end{cases}$
Then $H$ majorizes $f$ on $I$ and is piecewise constant, we have
$\overline{\int}_I f\leq \int_IH=\int_{[a+{\varepsilon},b-{\varepsilon}]}h+2M{\varepsilon}<\int_{[a+{\varepsilon},b-{\varepsilon}]}f+(2M+1){\varepsilon}$
By the same logic we have
$\underline{\int}_I f>\int_{[a+{\varepsilon},b-{\varepsilon}]}f-(2M+1){\varepsilon}$
Thus
$\overline{\int}_I f-\underline{\int}_I f<(4M+2){\varepsilon}$
This means $f$ is Riemann integrable on $I$.

Exerise 11.6.2

Definition: Let $I$ be a bounded interval, and let $f:I→\mathbf R$. We say $f$ is piecewise monotonic on $I$ iff there exists a partition $\mathbf P$ such that $f|_J$ is monotonic on $J$ for all $J\in \mathbf P$.
If $f$ is a bounded piecewise monotone function, then there exists a partition $\mathbf P$ such that $f|_J$ is bounded and monotonic on $J$ for all $J\in \mathbf P$. By Corollary 11.6.3 $f|_J$ is Riemann integrable on $J$ for all $J\in \mathbf P$.
We define
$F_J (x)=\begin{cases}f|_J (x),&x\in J\\0,&x\in I\backslash J \end{cases}$
By Theorem 11.4.1(g), $F_J$ is Riemann integrable on $I$, and we further have
$f(x)=\sum_{J\in P}F_J(x)$
So by Theorem 11.4.1(a), $f$ is Riemann integrable on $I$.

Exerise 11.6.3

We consider $\int_{[0,N]}f$, by Proposition 11.6.1, $\int_{[0,N]}f$ exists for all $N\in \mathbf N$. We let
$\mathbf P=\{[n,n+1):n=0,1,\dots,N-1\}\cup \{N\}$
be a partition of $[0,N]$, and define $\overline{f}(x)=f(n),x\in [n,n+1),\overline{f} (N)=f(N)$, then $\overline{f}$ majorizes $f$, thus
$\int_{[0,N]}f\leq \int_{[0,N]}\overline{f}=\sum_{J\in P}\overline{f} (x) =\sum_{n=0}^{N-1}f(n)$
Similarly if we define $\underline{f}(x)=f(n+1),x\in [n,n+1),\underline{f}(N)=f(N)$, then $\underline{f}$ minorizes $f$, so
$\int_{[0,N]}f\geq \int_{[0,N]}\underline{f}=\sum_{J\in P}\underline{f}(x) =\sum_{n=1}^Nf(n)$
Thus we have $\sum_{n=1}^Nf(n)\leq \int_{[0,N]}f\leq \sum_{n=0}^{N-1}f(n)$. Thus if $\sum_{n=0}^{\infty}f(n)$ converges, we can see that $\sum_{n=0}^{\infty}f(n)$ is an upper bound for $\int_{[0,N]}f,\forall N>0$, so $\sup_{N>0}⁡\int_{[0,N]}f\leq \sum_{n=0}^{\infty}f(n)$. On the other hand, if $\sup_{N>0}⁡\int_{[0,N]}f$ is finite, and assume $\sum_{n=0}^{\infty}f(n)$ diverges, since $f(n)\geq 0,\forall n$, we must have $\sum_{n=0}^{\infty}f(n)=+{\infty}$, thus there is $N'$ such that $\sum_{n=0}^{N'}f(n)>\sup(N>0)⁡\int_{[0,N]}f+f(0)$, this means $\sum_{n=1}^{N'}f(n)>\sup_{N>0} ⁡\int_{[0,N]} f\geq \int_{[0,N']}f\geq \sum_{n=1}^{N'}f(n)$, a contradiction.

Exerise 11.6.4

We can define
$f(x)=\begin{cases}1,&x\in [n+1/4,n+3/4],n\in \mathbf N\\0,&x\in [0,+{\infty}),x\notin [n+1/4,n+3/4],n\in \mathbf N\end{cases}$
then $\sum_{n=0}^{\infty}f(n)=0$, but $\int_{[0,N]}f=N/2$, thus $\sup_{N>0}⁡\int_{[0,N]}f$ is not finite.
Conversely we can define
$f(x)=\begin{cases}1,&x=n,n\in \mathbf N\\0,&x\in [0,+{\infty}),x\notin n,n\in \mathbf N\end{cases}$
Then $\sum_{n=0}^{\infty}f(n)=+{\infty}$, but $\int_{[0,N]}f=0$, thus $\sup_{N>0}⁡\int_{[0,N]}f$ is finite.

##Exerise 11.6.5
When $p\leq 0$, we have $1/n^p\nrightarrow0$, thus $\sum_{n=1}^{\infty}(1/n^p)$ diverges.
When $p>0$, the function $f(x)=1/x^p$ is monotonic decreasing on $[1,+{\infty})$, and $f(x)\geq 0,\forall x\in [1,+{\infty})$, thus the condition of Proposition 11.6.4 is satisfied.
Consider
$\int_{[1,N]}f=\int_{[1,N]}\frac{1}{x^p}$
Use Exercise 10.4.3 and some integral calculation we can show that
$\int_{[1,N]}f=\int_{[1,N]}\frac{1}{x^p} =\begin{cases}\frac{1}{1-p} (N^{1-p}-1),&p≠1\\ \ln⁡N,&p=1\end{cases}$
Thus if $0<p<1$, we have $N^{1-p}→{\infty}$ as $N→{\infty}$, so $\sup_{N>0}⁡\int_{[0,N]}f$ is not finite.
if $p=1$, then $\ln⁡N→{\infty}$ as $N→{\infty}$, so $\sup_{N>0}⁡\int_{[0,N]}f$ is not finite.
If $p>1$, then for any $N$ we have
$\int_{[1,N]}f=\frac{1}{1-p} \left(\frac{1}{N^{p-1}} -1\right)=\frac{1}{p-1} \left(1-\frac{1}{N^{p-1}} \right)<\frac{1}{p-1}$
so $\sup_{N>0}⁡\int_{[0,N]}f$ is finite.
Thus we can conclude $\sum_{n=1}^{\infty}(1/n^p)$ only converges when $p>1$.