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陶哲轩实分析11.8及习题-Analysis I 11.8

2019年6月4日2020年1月13日christangdt3条评论

Riemann-Stieltjes积分，实质是用α-length代替实数轴长度，很多积分的性质没有变化.

Exercise 11.8.1. Prove Lemma 11.8.4.
Lemma 11.8.4 Let $I$ be a bounded interval, let ${\alpha}:X\to\mathbf R$ be a function defined on some domain $X$ which contains $I$ , and let $\mathbf P$ be a partition of $I$ . Then we have

${\alpha}[I]=\sum\limits_{J\in \mathbf P}{\alpha}[J]$ .

Solution: We prove this by induction on $n$ . More precisely, we let $P(n)$ be the property that whenever $I$ is a bounded interval, and whenever $\mathbf P$ is a partition of $I$ with cardinality $n$ , that ${\alpha}[I]=\sum_{J\in \mathbf P}{\alpha}[J]$ .
The case $P(0)$ and $P(1)$ are all easy to prove. Now suppose inductively that $P(n)$ is true for $n\geq 1$ , let $I$ be a bounded interval, and $\mathbf P$ be a partition of $I$ with cardinality $n+1$ .
If $I$ is the empty set or a point, then all the intervals in $\mathbf P$ must also be either the empty set or a point and so every interval has length zero and the claim is trivial. Thus we will assume that $I$ is an interval of the form $(a,b],[a,b),(a,b)$ or $[a,b]$ .
Let us first suppose that $b\in I$ , i.e., $I$ is either $(a,b]$ or $[a,b]$ . Since $b\in I$ , we know that one of the intervals $K$ in $\mathbf P$ contains $b$ . Since $K$ is contained in $I$ , it must therefore be of the form $(c,b],[c,b]$ , or ${b}$ for some real number $c$ , with $a\leq c\leq b$ (in the latter case of $K={b}$ , we set $c:=b$ ). In particular, this means that the set $I-K$ is also an interval of the form $[a,c],(a,c),(a,c],[a,c)$ when $c>a$ , or a point or empty set when $a=c$ . Either way, we easily see that

$\displaystyle{{\alpha}[I]={\alpha}(b)-{\alpha}(a)={\alpha}(b)-{\alpha}(c)+{\alpha}(c)-{\alpha}(a)={\alpha}[K]+{\alpha}[I-K]}$

On the other hand, since $\mathbf P$ forms a partition of $I$ , we see that $\mathbf P-{K}$ forms a partition of $I-K$ , By the induction hypothesis, we thus have

$\displaystyle{{\alpha}[I-K]=\sum_{J\in \mathbf P-{K}}{\alpha}[J]}$

Combining these two identities, we obtain

$\displaystyle{{\alpha}[I]=\sum_{J\in \mathbf P}{\alpha}[J] }$

Now suppose that $b\notin I$ , i.e., I is either $(a,b)$ or $[a,b)$ , Then one of the intervals $K$ also is of the form $(c,b)$ or $[c,b)$ . In particular, this means that the set $I-K$ is also an interval of the form $[a,c],(a,c),(a,c],[a,c)$ when $c>a$ , or a point or empty set when $a=c$ . The rest of the argument then proceeds as above.

$\blacksquare$

Exercise 11.8.2. State and prove a version of Proposition 11.2.13 for the Riemann-Stieltjes integral.

Solution: Proposition: Let $I$ be a bounded interval, and let $f:I\to\mathbf R$ be a function. Suppose that $\mathbf P$ and $\mathbf P'$ are partitions of $I$ such that $f$ is piecewise constant both with respect to $\mathbf P$ and with respect to $\mathbf P'$ . Then $p.c.\int_{[\mathbf P]} fd{\alpha}=p.c.\int_{[\mathbf P']}fd{\alpha}$ .

Proof: By Lemma 11.2.7, we know $f$ is piecewise constant with respect to $\mathbf P\#\mathbf P'$ , thus the value

$\displaystyle{p.c.\int_{[\mathbf P\#\mathbf P']}fd{\alpha}=\sum_{J\in \mathbf P\# \mathbf P'}c_J {\alpha}[J] }$

is well defined. Now choose any $K\in \mathbf P$ , then $\mathbf P_K=\{J\in \mathbf P\# \mathbf P':J\subseteq K\}$ is a partition of $K$ , and $f$ is constant with constant value $c_K$ on both $K$ and all elements of $\mathbf P_K$ , thus by Theorem 11.1.13 we have $c_J=c_K,\forall J\in \mathbf P_K$ , and

$\displaystyle{{\alpha}[K]=\sum_{J\in \mathbf P_K}{\alpha}[J] \implies c_K {\alpha}[K]=\sum_{J\in \mathbf P_K}c_K {\alpha}[J]=\sum_{J\in \mathbf P_K}c_J {\alpha}[J] }$

Also, consider the set $\{J\in \mathbf P\# \mathbf P':J\subseteq K \text{ for some }K\in \mathbf P\}\subseteq \mathbf P\# \mathbf P'$ , for any $J\in \mathbf P\# \mathbf P'$ , by definition we can find a $K\in \mathbf P$ and a $K'\in P'$ s.t. $J=K\cap K'$ , so the two sets are equal. Thus

$\displaystyle{\begin{aligned}p.c.\int_{[\mathbf P]}fd{\alpha}&=\sum_{K\in \mathbf P}c_K {\alpha}[K]=\sum_{K\in \mathbf P}\sum_{J\in P_K}c_J {\alpha}[J]=\sum_{J\in \mathbf P\# \mathbf P'}c_J {\alpha}[J]\\&=p.c.\int_{[\mathbf P\# \mathbf P']}fd{\alpha}\end{aligned}}$

Similarly we can prove $p.c.\int_{[\mathbf P' ]}fd{\alpha}=p.c.\int_{[\mathbf P\# \mathbf P' ]} fd{\alpha}$ , and the statement is proved.

$\blacksquare$

Exercise 11.8.3. State and prove a version of Theorem 11.2.16 for the Riemann-Stieltjes integral.

Solution: Proposition: Let $I$ be a bounded interval, and let $f:I\to \mathbf R$ and $g:I\to \mathbf R$ be piecewise constant functions on $I$ . Let ${\alpha}:X\to \mathbf R$ be defined on a domain containing $I$ which is monotone increasing, then
( a ) We have $p.c.\int_I (f+g)d{\alpha}=p.c.\int_I fd{\alpha}+p.c.\int_I gd{\alpha}$ .
( b ) For any real number $c$ , we have $p.c.\int_I (cf)d{\alpha}=c(p.c.\int_I fd{\alpha})$ .
( c ) We have $p.c.\int_I (f-g)d{\alpha}=p.c.\int_I fd{\alpha}-p.c.\int_I gd{\alpha}$ .
( d ) If $f(x)\geq 0,\forall x\in I$ , then $p.c.\int_I fd{\alpha}\geq 0$ .
( e ) If $f(x)\geq g(x),\forall x\in I$ , then $p.c.\int_I fd{\alpha}\geq p.c.\int_I gd{\alpha}$ .
( f ) If $f(x)=c,\forall x\in I$ , then $p.c.\int_I fd{\alpha}=c{\alpha}[I]$ .
( g ) Let $J$ be a bounded interval containing $I$ (i.e. $I\subseteq J$ ), and let $F:J\to \mathbf R$ be the function

$\displaystyle{F(x)=\begin{cases}f(x),&x\in I\\0,&x\notin I\end{cases}}$

Then $F$ is piecewise constant on $J$ , and $p.c.\int_J Fd{\alpha}=p.c.\int_I fd{\alpha}$ .
( h ) Suppose that $\{J,K\}$ is a partition of $I$ into two intervals $J$ and $K$ . Then the functions $f|_J:J\to \mathbf R$ and $f|_K:K\to \mathbf R$ are piecewise constant on $J$ and $K$ respectively, and we have

$\displaystyle{p.c.\int_I fd{\alpha}=p.c.\int_Jf|_J d{\alpha}+p.c.\int_Kf|_K d{\alpha}}$

Proof:
We choose partitions of $I: \mathbf P'$ and $\mathbf P''$ such that $f$ is piecewise constant with respect to $\mathbf P'$ and $g$ is piecewise constant with respect to $\mathbf P''$ . Then let $\mathbf P=\mathbf P'\# \mathbf P''$ , we can see $f$ and $g$ are piecewise constant with respect to $\mathbf P$ . For any $K\in \mathbf P$ , let $c_K,d_K$ denote the constant value of $f$ and $g$ on $K$ .

( a ) $f+g$ is piecewise constant with respect to $\mathbf P$ , with constant value $c_K+d_K$ on $K\in \mathbf P$ ,

$\displaystyle{\begin{aligned}p.c.\int_I (f+g)d{\alpha}&=p.c.\int_{[\mathbf P]} (f+g)d{\alpha}=\sum_{K\in \mathbf P}(c_K+d_K){\alpha}[K]\\&=\sum_{K\in \mathbf P}c_K {\alpha}[K]+\sum_{K\in \mathbf P}d_K {\alpha}[K]\\&=p.c.\int_{[\mathbf P]} fd{\alpha}+p.c.\int_{[\mathbf P]} gd{\alpha}\\&=p.c.\int_I fd{\alpha}+p.c.\int_I gd{\alpha}\end{aligned}}$

( b ) $cf$ is piecewise constant with respect to $\mathbf P$ , with constant value $cc_K$ on $K\in \mathbf P$ ,

$\displaystyle{\begin{aligned}p.c.\int_I (cf)d{\alpha}&=p.c.\int_{[\mathbf P]} (cf)d{\alpha}=\sum_{K\in \mathbf P}(cc_K){\alpha}[K]\\&=c\sum_{K\in \mathbf P}c_K {\alpha}[K]=c\left(p.c.\int_{[\mathbf P]} fd{\alpha}\right)\\&=c\left(p.c.\int_I fd{\alpha}\right)\end{aligned}}$

( c ) Use ( b ) we have $p.c.\int_I (-g)d{\alpha}=-p.c.\int_I gd{\alpha}$ , then use ( a ) we get

$\displaystyle{\begin{aligned}p.c.\int_I(f-g)d{\alpha}&=p.c.\int_I (f+(-g))d{\alpha}\\&=p.c.\int_I fd{\alpha}+p.c.\int_I (-g)d{\alpha}\\&=p.c.\int_Ifd{\alpha}-p.c.\int_Igd{\alpha}\end{aligned}}$

( d ) If $f(x)\geq 0,\forall x\in I$ , then $c_K\geq 0,\forall K\in \mathbf P$ , since ${\alpha}$ is monotone increasing we have $c_K {\alpha}[K]\geq 0,\forall K\in \mathbf P$ , thus

$\displaystyle{p.c.\int_I fd{\alpha}=p.c.\int_{[\mathbf P]} fd{\alpha}=\sum_{K\in P}c_K {\alpha}[K]\geq 0}$

( e ) We have $f(x)-g(x)\geq 0,\forall x\in I$ , so use ( d ) we have

$\displaystyle{p.c.\int_I (f-g)d{\alpha}=p.c.\int_I fd{\alpha}-p.c.\int_I gd{\alpha}\geq 0 \\ \implies p.c.\int_I fd{\alpha}\geq p.c.\int_I gd{\alpha}}$

( f ) If $f(x)=c,\forall x\in I$ , then $c_K=c,\forall K\in P$ , so we have by Lemma 11.8.4

$\displaystyle{p.c.\int_I fd{\alpha}=p.c.\int_{[\mathbf P]} fd{\alpha}=\sum_{K\in \mathbf P}c_K {\alpha}[K]=c\sum_{K\in \mathbf P}{\alpha}[K] =c{\alpha}[I]}$

(g) The set $\mathbf P\cup \{J-I\}$ is a partition of $J$ , and $F$ is piecewise constant on $\mathbf P\cup \{J-I\}$ , with additional constant value $c_{J-I}=0$ , thus

$\displaystyle{\begin{aligned}p.c.\int_J Fd{\alpha}&=p.c.\int_{[\mathbf P\cup \{J-I\}]} Fd{\alpha}=\sum_{K\in \mathbf P\cup \{J-I\}}c_K {\alpha}[K]\\&=\sum_{K\in \mathbf P}c_K {\alpha}[K]+0\cdot {\alpha}[J-I]=\sum_{K\in \mathbf P}c_K {\alpha}[K]\\&=p.c.\int_I fd{\alpha}\end{aligned}}$

( h ) We have $\mathbf P_J=\{L\cap J:L\in P\}$ and $\mathbf P_K=\{L\cap K:L\in P\}$ partitions of $J$ and $K$ , and since $f$ is piecewise constant with $\mathbf P$ , it’s easy to see $f|_J$ is piecewise constant with $\mathbf P_J$ on $J$ , $f|_K$ is piecewise constant with $\mathbf P_K$ on $K$ . As we have $J\subseteq I$ and $K\subseteq I$ , we define

$\displaystyle{F(x)=\begin{cases}f|_J (x),&x\in J\\0,&x\notin J\end{cases},\quad G(x)=\begin{cases}f|_K (x),&x\in K\\0,&x\notin K\end{cases}}$

Since $\{J,K\}$ is a partition of $I$ , we have $f=f|_J+f|_K=F+G$ , thus use ( a ) ,( g ) we have

$\displaystyle{\begin{aligned}p.c.\int_I fd{\alpha}&=p.c.\int_I(F+G)d{\alpha}=p.c.\int_I Fd{\alpha}+p.c.\int_I Gd{\alpha}\\&=p.c.\int_Jf|_J d{\alpha}+p.c.\int_K^ f|_K d{\alpha}\end{aligned}}$

$\blacksquare$

Exercise 11.8.4. State and prove a version of Theorem 11.5.1 for the Riemann-Stieltjes integral.

Solution: Proposition: Let $I$ be a bounded interval, and let $f$ be a function which is uniformly continuous on $I$ . Let ${\alpha}:X\to \mathbf R$ be defined on a domain containing $I$ which is monotone increasing, Then $f$ is Riemann-Stieltjes integrable on $I$ with respect to ${\alpha}$ .

Proof: From Proposition 9.9.15 we see that $f$ is bounded. Now we have to show that $\underline{\int}_I fd{\alpha}=\overline{\int}_I fd{\alpha}$ .
If $I$ is a point or the empty set then the theorem is trivial, so let us assume that $I$ is one of the four intervals $[a,b],(a,b),(a,b]$ , or $[a,b)$ for some real numbers $a0$ be arbitrary. By uniform continuity, there exists a ${\delta}>0$ such that $|f(x)-f(y)|<{\varepsilon}$ whenever $x,y\in I,|x-y|<{\delta}$ . By the Archimedean principle, there exists an integer $N>0$ such that $(b-a)/N<{\delta}$ , Note that we can partition $I$ into $N$ intervals $J_1,\dots,J_N$ , each of length $(b-a)/N$ , we thus have

$\displaystyle{\overline{\int}_I fd{\alpha}\leq \sum_{k=1}^N\left(\sup_{x\in J_k} f(x) \right){\alpha}[J_k],\quad \underline{\int}_I fd{\alpha}\geq \sum_{k=1}^N\left(\inf_{x\in J_k} f(x)\right){\alpha}[J_k] }$

So we have

$\displaystyle{\begin{aligned}\overline{\int}_I fd{\alpha}-\underline{\int}_I fd{\alpha}&\leq \sum_{k=1}^N\left(\sup_{x\in J_k} f(x)-\inf_{x\in J_k} f(x) \right){\alpha}[J_k] \\&\leq {\varepsilon}\sum_{k=1}^N{\alpha}[J_k] ={\varepsilon}{\alpha}[I]={\varepsilon}({\alpha}(b)-{\alpha}(a))\end{aligned}}$

This shows $\underline{\int}_I fd{\alpha}-\overline{\int}_I fd{\alpha}$ cannot be positive.

$\blacksquare$

Exercise 11.8.5. Let $\text{sgn} :\mathbf R\to \mathbf R$ be the signum function

$\text{sgn} (x):=\begin{cases}1&\text{when }x>0\\0&\text{when }x=0\\-1&\text{when }x<0\end{cases}$

Let $f:[-1,1]\to\mathbf R$ be a continuous function. Show that $f$ is Riemann-Stieltjes integrable with respect to $\text{sgn}$ , and that

$\displaystyle{\int_{[-1,1]}fd \text{sgn} =2f(0)}$

Solution: We let ${\varepsilon}>0$ be arbitrary. Then since $f$ is continuous, $\exists {\delta}>0$ such that

$\displaystyle{|f(x)-f(0)|<{\varepsilon},\quad \forall x\in (-{\delta},{\delta})}$

Since $f$ is continuous on $[-1,1]$ , $f$ is bounded, thus $\exists M>0$ , s.t.

$\displaystyle{|f(x)|\leq M,\quad \forall x\in [-1,1]}$

We let $\{[-1,-{\delta}],(-{\delta},{\delta}),[{\delta},1]\}$ be a partition of $[-1,1]$ , then the piecewise constant function

$\displaystyle{g(x)=\begin{cases}f(0)+{\varepsilon},&x\in (-{\delta},{\delta})\\M,&x\in [-1,1]\backslash (-{\delta},{\delta})\end{cases}}$

majorizes $f$ and we have

$\displaystyle{\begin{aligned}&\quad\quad\overline{\int}_{[-1,1]} fd\text{sgn}\leq \int_{[-1,1]}gd\text{sgn}\\&=M(\text{sgn}(-{\delta})-\text{sgn}(-1))+(f(0)+{\varepsilon})(\text{sgn}({\delta})-\text{sgn}(-{\delta}))+M(\text{sgn}(1)-\text{sgn}({\delta}))\\&=2(f(0)+{\varepsilon})\end{aligned}}$

Similarly, the piecewise constant function

$\displaystyle{h(x)=\begin{cases}f(0)-{\varepsilon},&x\in (-{\delta},{\delta})\\-M,&x\in [-1,1]\backslash(-{\delta},{\delta}) \end{cases}}$

minorizes $f$ and we have

$\displaystyle{\underline{\int}_{[-1,1]}fd\text{sgn}\geq \int_{[-1,1]}hd\text{sgn}=2(f(0)-{\varepsilon})}$

Thus we have

$\displaystyle{2(f(0)-{\varepsilon})\leq \underline{\int}_{[-1,1]} fd\text{sgn}\leq \overline{\int}_{[-1,1]}fd\text{sgn}\leq 2(f(0)+{\varepsilon})}$

And the result follows.

$\blacksquare$

陶哲轩实分析11.6及习题-Analysis I 11.6

2019年6月3日2020年1月9日christangdt9条评论

单调函数是Riemann可积的，并且介绍了integral test和重要的 $1/n^{p}$ 级数收敛判别

Exercise 11.6.1. Use Proposition 11.6.1 to prove Corollary 11.6.3.
Corollary 11.6.3 Let $I$ be a bounded interval, and let $f:I\to\mathbf R$ be both monotone and bounded. Then $f$ is Riemann integrable on $I$ .

Solution: 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

$\displaystyle{\int_{[a+{\varepsilon},b-{\varepsilon}]}h<\int_{[a+{\varepsilon},b-{\varepsilon}]}f+{\varepsilon}}$

Define

$\displaystyle{H(x)=\begin{cases}h(x),&x\in [a+{\varepsilon},b-{\varepsilon}]\\M,&x\in I\backslash[a+{\varepsilon},b-{\varepsilon}] \end{cases}}$

Then $H$ majorizes $f$ on $I$ and is piecewise constant, we have

$\displaystyle{\overline{\int}_I f\leq \int_IH=\int_{[a+{\varepsilon},b-{\varepsilon}]}h+2M{\varepsilon}<\int_{[a+{\varepsilon},b-{\varepsilon}]}f+(2M+1){\varepsilon}}$

By the same logic we have

$\displaystyle{\underline{\int}_I f>\int_{[a+{\varepsilon},b-{\varepsilon}]}f-(2M+1){\varepsilon}}$

Thus

$\displaystyle{\overline{\int}_I f-\underline{\int}_I f<(4M+2){\varepsilon}}$

This means $f$ is Riemann integrable on $I$ .

$\blacksquare$

Exercise 11.6.2. Formulate a reasonable notion of a piecewise monotone function, and then show that all bounded piecewise monotone functions are Riemann integrable.

Solution:

Definition: Let $I$ be a bounded interval, and let $f:I\to \mathbf R$ . 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

$\displaystyle{F_J (x)=\begin{cases}f|_J (x),&x\in J\\0,&x\in I\backslash J \end{cases}}$

By Theorem 11.4.1(g), $F_J$ is Riemann integrable on $I$ , and we further have

$\displaystyle{f(x)=\sum_{J\in P}F_J(x)}$

So by Theorem 11.4.1(a), $f$ is Riemann integrable on $I$ .

$\blacksquare$

Exercise 11.6.3. Prove Proposition 11.6.4.
Proposition 11.6.4 (Integral test). Let $f:[0,\infty )\to\mathbf R$ be a monotone increasing function which is non-negative (i.e., $f(x)\geq 0$ for all $x\geq 0$ ). Then the sum $\sum_{n=0}^{\infty}f(n)$ is convergent if and only if $\sup_{N>0}\int_{[0,N]}f$ is finite.

Solution: We consider $\int_{[0,N]}f$ , by Proposition 11.6.1, $\int_{[0,N]}f$ exists for all $N\in \mathbf N$ . We let

$\displaystyle{\mathbf P=\{[n,n+1):n=0,1,\dots,N-1\}\cup \{N\}}$

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

$\displaystyle{\int_{[0,N]}f\leq \int_{[0,N]}\overline{f}=\sum_{J\in P}\overline{f} (x) =\sum_{n=0}^{N-1}f(n)}$

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

$\displaystyle{\int_{[0,N]}f\geq \int_{[0,N]}\underline{f}=\sum_{J\in P}\underline{f}(x) =\sum_{n=1}^Nf(n)}$

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.

$\blacksquare$

Exercise 11.6.4. Give examples to show that both directions of the integral test break down if $f$ is not assumed to be monotone decreasing.

Solution: We can define

$\displaystyle{f(x)=\begin{cases}1,&x\in [n+1/4,n+3/4],n\in \mathbf N\\0,&x\in [0,+{\infty}),x\notin [n+1/4,n+3/4],n\in \mathbf N\end{cases}}$

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

$\displaystyle{f(x)=\begin{cases}1,&x=n,n\in \mathbf N\\0,&x\in [0,+{\infty}),x\notin n,n\in \mathbf N\end{cases}}$

Then $\sum_{n=0}^{\infty}f(n)=+{\infty}$ , but $\int_{[0,N]}f=0$ , thus $\sup_{N>0} \int_{[0,N]}f$ is finite.

$\blacksquare$

Exercise 11.6.5. Use Proposition 11.6.4 to prove Corollary 11.6.5.
Corollary 11.6.5 Let $p$ be a real number. Then $\sum_{n=1}^{\infty}\frac{1}{n^p}$ converges absolutely when $p>1$ and diverges when $p\leq 1$ .

Solution: 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

$\displaystyle{\int_{[1,N]}f=\int_{[1,N]}\frac{1}{x^p} }$

Use Exercise 10.4.3 and some integral calculation we can show that

$\displaystyle{\int_{[1,N]}f=\int_{[1,N]}\frac{1}{x^p} =\begin{cases}\dfrac{1}{1-p} (N^{1-p}-1),&p\neq 1\\ \ln N,&p=1\end{cases}}$

Thus if $0<p<1$ , we have $N^{1-p}\to {\infty}$ as $N\to {\infty}$ , so $\sup_{N>0} \int_{[0,N]}f$ is not finite.
if $p=1$ , then $\ln N\to {\infty}$ as $N\to {\infty}$ , so $\sup_{N>0} \int_{[0,N]}f$ is not finite.
If $p>1$ , then for any $N$ we have

$\displaystyle{\int_{[1,N]}f=\frac{1}{1-p} \left(\frac{1}{N^{p-1}} -1\right)=\frac{1}{p-1} \left(1-\frac{1}{N^{p-1}} \right)<\frac{1}{p-1}}$

so $\sup_{N>0} \int_{[0,N]}f$ is finite.
Thus we can conclude $\sum_{n=1}^{\infty}(1/n^p)$ only converges when $p>1$ .

$\blacksquare$

陶哲轩实分析11.5及习题-Analysis I 11.5

2019年5月31日2020年1月9日christangdt一条评论

有界的逐段连续函数是Riemann可积的

Exercise 11.5.1. Prove Proposition 11.5.6.
Proposition 11.5.6. Let $I$ be a bounded interval, and let $f:I\to\mathbf R$ be both peicewise continuous and bounded. Then $f$ is Riemann integrable.

Solution: Since $f$ is piecewise continuous, we can find a parititon $\mathbf P$ such that $f|_J$ is continuous on $J$ for $\forall J\in \mathbf P$ . By Proposition 11.5.3, we have $f|_J$ is Riemann integrable on $J$ .
We define

$\displaystyle{F_J (x)=\begin{cases}f|_J (x),&x\in J\\0,&x\in I\backslash J\end{cases}}$

By Theorem 11.4.1(g), $F_J$ is Riemann integrable on $I$ , and we further have

$\displaystyle{f(x)=\sum_{J\in \mathbf P}F_J (x)}$

So by Theorem 11.4.1(a), $f$ is Riemann integrable on $I$ .

$\blacksquare$

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