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陶哲轩实分析11.4及习题-Analysis I 11.4

2019年5月30日2020年1月9日christangdt3条评论

Riemann积分的算律，常规运算基本保持Riemann可积性。

Exercise 11.4.1. Prove Theorem 11.4.1.
Theorem 11.4.1 (Laws of Riemann integration). Let $I$ be a bounded interval, and let $f:I\to\mathbf R$ and $g:I\to\mathbf R$ be Riemann integrable functions on $I$ .
( a ) The function $f+g$ is Riemann integrable, and we have $\int_I(f+g)=\int_If+\int_Ig$ .
( b ) For any real number $c$ , the function $cf$ is Riemann integrable, and we have $\int_I(cf)=c(\int_If)$ .
( c ) The function $f-g$ is Riemann integrable, and we have $\int_I(f-g)=\int_If-\int_Ig$ .
( d ) If $f(x)\geq 0$ for all $x\in I$ , then $\int_If\geq 0$ .
( e ) If $f(x)\geq g(x)$ for all $x\in I$ , then $\int_If\geq \int_Ig$ .
( f ) If $f$ is the constant function $f(x)=c$ for all $x$ in $I$ , then $\int_If=c|I|$ .
( g ) Let $J$ be a bounded interval containing $I$ (i.e., $I\subseteq J$ ), and let $F:J\to\mathbf R$ be the function

$F(x):=\begin{cases}f(x)&\text{if }x\in I\\0&\text{if }x\notin I\end{cases}$

Then $F$ is Riemann integrable on $J$ , and $\int_JF=\int_If$ .
( 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 Riemann integrable on $J$ and $K$ respectively, and we have

$\displaystyle{\int_If=\int_Jf|_J+\int_Kf|_K}$

Solution: Since $f$ and $g$ are both Riemann integrable, we have

$\displaystyle{\overline{\int}_I f=\underline{\int}_I f,\quad \overline{\int}_I g=\underline{\int}_I g}$

So for $\forall {\varepsilon}>0$ , we have piecewise constant function $\overline{f}$ majorizes $f$ , $\underline{f}$ minorizes $f$ , $\overline{g}$ majorizes $g$ , $\underline{g}$ minorizes $g$ , such that

$\displaystyle{{\int}_I f-{\varepsilon}<{\int}_I \underline{f}\leq {\int}_I f\leq {\int}_I \overline{f}<{\int}_I f+{\varepsilon},\\{\int}_I g-{\varepsilon}<{\int}_I \underline{g}\leq {\int}_I g\leq {\int}_I \overline{g}<{\int}_I g+{\varepsilon}}$

( a ) It is easy to prove that $\overline{f}+\overline{g}$ majorizes $f+g$ and $\underline{f}+\underline{g}$ minorizes $f+g$ , so we have

$\displaystyle{{\int}_I (\underline{f}+\underline{g}) \leq \underline{\int}_I (f+g) \leq \overline{\int}_I (f+g)\leq {\int}_I (\overline{f}+\overline{g})}$

Use Theorem 11.2.16 we have

$\displaystyle{{\int}_I (\underline{f}+\underline{g}) ={\int}_I \underline{f}+{\int}_I \underline{g}>{\int}_I f+{\int}_I g-2{\varepsilon} \\{\int}_I (\overline{f}+\overline{g}) ={\int}_I \overline{f}+{\int}_I \overline{g}<{\int}_I f+{\int}_I g+2{\varepsilon}}$

Thus

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

So $f+g$ is Riemann integrable, and

$\displaystyle{{\int}_I f+{\int}_I g-2{\varepsilon}<{\int}_I (\underline{f}+\underline{g}) \leq {\int}_I (f+g) \leq {\int}_I (\overline{f}+\overline{g}) <{\int}_If+{\int}_Ig+2{\varepsilon}}$

which shows ${\int}_I (f+g) ={\int}_I f+{\int}_I g$ .

( b ) First suppose $c>0$ , then $c\overline{f}$ majorizes $cf$ , $c\underline{f}$ minorizes $cf$ , so we have

$\displaystyle{{\int}_I (c\underline{f}) \leq \underline{\int}_I (cf) \leq \overline{\int}_I (cf)\leq {\int}_I(c\overline{f})}$

Use Theorem 11.2.16 we have

$\displaystyle{{\int}_I(c\underline{f}) =c{\int}_I \underline{f}>c{\int}_I f-c{\varepsilon},\quad {\int}_I (c\overline{f}) =c{\int}_I \overline{f}<c{\int}_I f+c{\varepsilon}}$

Thus

$\displaystyle{c{\int}_I f-c{\varepsilon}<{\int}_I (c\underline{f}) \leq \underline{\int}_I (cf) \leq \overline{\int}_I (cf)\leq {\int}_I (c\overline{f}) <c{\int}_I f+c{\varepsilon}}$

which shows $\displaystyle{{\int}_I (cf) =c{\int}_I f}$ .
Now if $c=0$ , then $cf=0$ , which is a piecewise constant function on $I$ , thus

$\displaystyle{{\int}_I (cf) =p.c.{\int}_I (cf) =p.c.{\int}_I (0) =0=0\cdot {\int}_I f}$

Now suppose $c=-1$ , then $-\overline{f}$ minorizes $-f$ , $-\underline{f}$ majorizes $-f$ , so we have

$\displaystyle{{\int}_I (-\overline{f}) \leq \underline{\int}_I (-f) \leq \overline{\int}_I (-f)\leq {\int}_I (-\underline{f}) }$

Use Theorem 11.2.16 we have

$\displaystyle{{\int}_I (-\overline{f}) =-{\int}_I \overline{f}>-{\int}_I f-{\varepsilon},\quad {\int}_I (-\underline{f}) =-{\int}_I \underline{f}<-{\int}_I f+{\varepsilon}}$

Thus

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

which shows ${\int}_I (-f) =-{\int}_I f$ .
Finally for any $c<0$ , we have $c=(-1)(-c)$ and $cf=(-1)(-cf),-c>0$ , so use the result for $c=-1$ and $c>0$ , we have

$\displaystyle{{\int}_I (cf) ={\int}_I (-1)(-cf) =-{\int}_I (-cf) =-(-c) {\int}_I f=c{\int}_I f}$

( c ) Use (a) and (b) we could have

$\displaystyle{{\int}_I(f-g) ={\int}_I (f+(-g)) ={\int}_I f+{\int}_I(-g)={\int}_I f-{\int}_I g}$

( d ) The constant function $h(x)=0$ minorizes $f$ , thus

$\displaystyle{{\int}_I f\geq \underline{\int}_I (f)\geq {\int}_I h={\int}_I0\cdot h=0{\int}_I h=0}$

( e ) The function $f(x)-g(x)\geq 0,\forall x\in I$ , thus use ( d ) and ( c )

$\displaystyle{{\int}_I f-{\int}_I g={\int}_I (f-g) \geq 0 \implies {\int}_I f\geq {\int}_I g}$

( f ) In this case, we can use the piecewise constant function integral formula:

$\displaystyle{{\int}_I f=p.c.{\int}_I f=\sum_{J\in P}c_J |J| =c\sum_{J\in P}|J| =c|I|}$

( g ) We can define $\overline{F}$ and $\underline{F}$ which are piecewise and majorizes/minorizes $F$ as:

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

then use Theorem 11.2.16 we have

$\displaystyle{{\int}_I f-{\varepsilon}<{\int}_I \underline{f}={\int}_J \underline{F}\leq \underline{\int}_J F\leq \overline{\int}_J F\leq {\int}_J \overline{F}={\int}_I \overline{f}<{\int}_I f+{\varepsilon}}$

which shows ${\int}_J F={\int}_I f$ .

( h ) Use Theorem 11.2.16 (h) we have

$\displaystyle{{\int}_I \overline{f}={\int}_J \overline{f}|_J +{\int}_K \overline{f}|_K ,{\int}_I \underline{f}={\int}_J \underline{f}|_J +{\int}_K \underline{f}|_K }$

thus

$\displaystyle{{\int}_I \overline{f}-{\int}_I \underline{f}=({\int}_J\overline{f}|_J -{\int}_J \underline{f}|_J )+({\int}_K\overline{f}|_K -{\int}_K\underline{f}|_K )<2{\varepsilon}}$

Since we have $\overline{f}\geq f\geq \underline{f}$ on $I,J,K$ , this means

$\displaystyle{{\int}_J\overline{f}|_J -{\int}_J\underline{f}|_J \geq 0,\quad {\int}_K\overline{f}|_K -{\int}_K\underline{f}|_K \geq 0}$

Thus we have

$\displaystyle{{\int}_J\overline{f}|_J -{\int}_J\underline{f}|_J <2{\varepsilon},{\int}_K\overline{f}|_K -{\int}_K\underline{f}|_K <2{\varepsilon}}$

Since $\overline{f}|_J$ majorizes $f|_J$ and $\underline{f}|_J$ minorizes $f|_J$ , we can say $f|_J$ is Riemann integrable on $J$ , by the same logic $f|_K$ is Riemann integrable on $K$ . We let

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

Then $f(x)=F(x)+G(x)$ , use (a) and (g) we have

$\displaystyle{{\int}_I f={\int}_I F+{\int}_I G={\int}_Jf|_J +{\int}_Kf|_K }$

$\blacksquare$

Exercise 11.4.2. Let $a<b$ be real numbers, and let $f:[a,b]\to\mathbf R$ be a continuous, non-negative function (so $f(x)\geq 0$ for all $x\in [a,b]$ ). Suppose that $\int_{[a,b]}f=0$ . Show that $f(x)=0$ for all $x\in [a,b]$ .

Solution: Assume $\exists c\in [a,b],f(c)>0$ , since $f$ is continuous, let ${\varepsilon}=f(c)/2$ , then $\exists {\delta}>0$ , s.t.

$\displaystyle{|f(x)-f(c)|<{\varepsilon}=f(c)/2,\quad \forall x\in [a,b]\cap (c-{\delta},c+{\delta})}$

Since the length of $[a,b]\cap (c-{\delta},c+{\delta})$ is longer than ${\delta}/2$ , and we have

$\displaystyle{f(x)>f(c)/2,\quad \forall x\in [a,b]\cap (c-{\delta},c+{\delta})}$

We can define

$\displaystyle{g(x)=\begin{cases}f(c)/2,&x\in [a,b]\cap (c-{\delta},c+{\delta})\\f(x),&x\in [a,b],x\notin (c-{\delta},c+{\delta})\end{cases}}$

Then we can say $g$ minorizes $f$ on $[a,b]$ , since $f$ is Riemann integrable on $[a,b]$ , and ${[a,b]\cap (c-{\delta},c+{\delta}),[a,b]\backslash(c-{\delta},c+{\delta})}$ is a partition of [a,b], we can say the function $f|_{[a,b]\backslash(c-{\delta},c+{\delta})}$ is Riemann integrable by Theorem 11.4.1(h), and constant function is Riemann integrable by Theorem 11.4.1(f), thus $g$ is integrable by repeated use of Theorem 11.4.1(g). Now use Theorem 11.4.1(h), (d),(f) we have

$\displaystyle{0=\int_{[a,b]}f\geq \int_{[a,b]}g=\int_{[a,b]\cap (c-{\delta},c+{\delta})}\left(\frac{f(c)}{2}\right) +{\int}_{[a,b]\backslash(c-{\delta},c+{\delta})} f\geq \frac{f(c)}{2} \left|\frac{{\delta}}{2}\right|=\frac{{\delta}f(c)}{4}>0}$

this is a contradiction.

$\blacksquare$

Exercise 11.4.3. Let $I$ be a bounded interval, let $f:I\to\mathbf R$ be a Riemann integrable function, and let $\mathbf P$ be a partition of $I$ . Show that

$\displaystyle{\int_If=\sum_{J\in \mathbf P}\int_If}$

Solution: Repeatly use Theorem 11.4.1(h), we can get

$\displaystyle{{\int}_If=\sum_{J\in P}{\int}_Jf|_J =\sum_{J\in P}{\int}_Jf}$

$\blacksquare$

Exercise 11.4.4. Without repeating all the computations in the above proofs, give a short explanation as to why the remaining cases of Theorem 11.4.3 and Theorem 11.4.5 follow automatically from the cases presented in the text.

Solution: It is easy to see from Theorem 11.4.1 that if f is Riemann integrable, then so is $-f$ .
As $\min (f,g)=-\max (-f,-g)$ , we settled Theorem 11.4.3.
For Theorem 11.4.5, we already have $f_+ g_+$ Riemann integrable if $f$ and $g$ are integrable, notice that

$f_+ g_-=f_+ (-g)_+ \\ f_- g_+=(-f)_+ g_+ \\f_- g_-=(-f)_+ (-g)_+$

and both $-f$ and $-g$ are Riemann integrable, the conclusion follows.

$\blacksquare$

陶哲轩实分析11.3及习题-Analysis I 11.3

2019年5月29日2020年1月9日christangdt留下评论

这一节用逐段常值函数定义了上积分和下积分，定义与后续的Lebesgue测度精神相通，使用上积分和下积分定义Riemann积分，并且再定义Riemann和，这样Riemann和的引入更自然一些。

Exercise 11.3.1. Let $f:I\to\mathbf R$ , $g:I\to\mathbf R$ and $h:I\to\mathbf R$ be functions. Show that if $f$ majorizes $g$ and $g$ majorizes $h$ , then $f$ majorizes $h$ . Show that if $f$ and $g$ majorize each other, then they must be equal.

Solution: Since $f$ majorizes $g$ we have $f(x)\geq g(x),\forall x\in I$ , and $g$ majorizes $h$ means $g(x)\geq h(x),\forall x\in I$ , thus we have $f(x)\geq h(x),\forall x\in I$ , which means $f$ majorizes $h$ . If $f$ and $g$ majorizes each other, then it means $f(x)\geq g(x),\forall x\in I$ and $g(x)\geq f(x),\forall x\in I$ , so we shall have $f(x)=g(x),\forall x\in I$ .

$\blacksquare$

Exercise 11.3.2. Let $f:I\to\mathbf R$ , $g:I\to\mathbf R$ and $h:I\to\mathbf R$ be functions. If $f$ majorizes $g$ , is it true that $f+h$ majorizes $g+h$ ? Is is true that $f\cdot h$ majorizes $g\cdot h$ ? If $c$ is a real number, is it true that $cf$ majorizes $cg$ ?

Solution: We have

$\begin{aligned}f \text{ majorizes }g &\implies f(x)\geq g(x),\forall x\in I \\&\implies f(x)+h(x)\geq g(x)+h(x),\forall x\in I\\&\implies (f+h)(x)\geq (g+h)(x),\forall x\in I \\&\implies f+h \text{ majorizes }g+h\end{aligned}$

We cannot have $fh$ majorizes $gh$ if $f$ majorizes $g$ , since if we let $h(x)=-1$ on $I$ , we shall have $(fh)(x)=f(x)h(x)=-f(x)\leq -g(x)=g(x)h(x)=(gh)(x)$ .
We cannot have $cf$ majorizes $cg$ if $f$ majorizes $g$ , we could let $c=-1$ and use a similar proof to show $(cf)(x)\leq (cg)(x)$ .

$\blacksquare$

Exercise 11.3.3. Prove Lemma 11.3.7.
Lemma 11.3.7. Let $f:I\to\mathbf R$ be a piecewise constant function on a bounded interval $I$ . Then $f$ is Riemann integrable, and $\int_If=p.c.\int_If$ .

Solution: For $f$ , we see $f(x)\geq f(x),\forall x\in I$ , thus $f$ majorizes $f$ , and $f$ is piecewise constant, thus by definition of upper Riemann integral we have

$\displaystyle{\overline{\int}_I f\leq p.c.\int_If}$

By the same logic we have $f$ minorizes $f$ , thus

$\displaystyle{ \underline{\int}_I f\geq p.c.\int_I f}$

Thus we have $\underline{\int}_I f\geq \overline{\int}_I f$ , use Lemma 11.3.3 we can see $f$ is Riemann integrable, and

$\displaystyle{ \int_If=p.c.\int_If}$

$\blacksquare$

Exercise 11.3.4. Prove Lemma 11.3.11.
Lemma 11.3.11. Let $f:I\to\mathbf R$ be a bounded function on a bounded interval $I$ , and let $g$ be a function which majorized $f$ and which is piecewise constant with respect to some partition $\mathbf P$ of $I$ . Then

$p.c.\displaystyle{\int_Ig\geq U(f,\mathbf P)}$

Similarly, if $h$ is a function which minorizes $f$ and is piecewise constant with respect to $\mathbf P$ , then

$p.c.\displaystyle{\int_Ih\leq L(f,\mathbf P)}$

Solution: If $g$ is piecewise constant with $\mathbf P$ and majorizes $f$ on $I$ , we denote $c_J$ as the constant value of $g$ for each $J\in \mathbf P$ , then $c_J\geq f(x),\forall x\in J$ , thus

$c_J\geq \sup_{x\in J}f(x),\quad \forall J\in P$

so we have

$\displaystyle{ \begin{aligned}p.c.\int_Ig &=\sum\limits_{J\in \mathbf P}c_J|J|=\sum\limits_{J\in \mathbf P,J\neq \emptyset}c_J|J|\\&\geq \sum\limits_{J\in \mathbf P,J\neq \emptyset}\left(\sup_{x\in J} f(x) \right)|J|=U(f,\mathbf P)\end{aligned}}$

A similar proof can show $p.c.\int_Ih\leq L(f,\mathbf P)$ .

$\blacksquare$

Exercise 11.3.5. Prove Proposition 11.3.12.
Proposition 11.3.12. Let $f:I\to\mathbf R$ be a bounded function on a bounded interval $I$ . Then

$\displaystyle{\overline{\int}_If=\inf\{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}}$

and

$\displaystyle{\underline{\int}_If=\sup\{L(f,\mathbf P):\mathbf P\text{ is a partition of }I\}}$

Solution: By Lemma 11.3.11, let $\mathbf P$ be a partition of $I$ . If $g$ is piecewise constant with $\mathbf P$ and majorizes $f$ on $I$ , we have $p.c.\int_Ig\geq U(f,\mathbf P)$ , so

$\displaystyle{ p.c.\int_Ig\geq \inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}}$

then since $g$ is arbitrary as long as it is piecewise constant with $\mathbf P$ and majorizes $f$ on $I$ ,

$\displaystyle{ \overline{\int}_I f\geq \inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}}$

On the other hand, let $\mathbf P$ be a partition of $I$ . We define a new piecewise constant function with $\mathbf P$ and majorizes $f$ on $I$ as follows: $\forall x\in I,\exists J\in \mathbf P,x\in J$ , this $J$ is unique, so we let

$G(x)=\sup_{x\in J} f(x)$

Then $G$ is piecewise constant with $\mathbf P$ and majorizes $f$ on $I$ , further we have

$\displaystyle{ p.c.\int_IG=U(f,\mathbf P)}$

Thus we can say that

$\displaystyle{ \overline{\int}_I f\leq p.c.\int_IG=U(f,\mathbf P)}$

Since $\mathbf P$ could be any partition of $I$ , we can say

$\displaystyle{ \overline{\int}_I f\leq U(f,\mathbf P),\quad \forall \mathbf P\text{ is a partition of }I}$

so it is easy to see

$\displaystyle{ \overline{\int}_I f\leq \inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}}$

Thus we can have

$\displaystyle{ \overline{\int}_I f=\inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}}$

A similar proof can show that

$\displaystyle{ \underline{\int}_I f=\sup \{L(f,\mathbf P):\mathbf P\text{ is a partition of }I\}}$

$\blacksquare$

陶哲轩实分析11.2及习题-Analysis I 11.2

2019年5月28日2020年1月8日christangdt一条评论

这节介绍逐段常值函数。

Exercise 11.2.1. Prove Lemma 11.2.7.
Lemma 11.2.7. Let $I$ be a bounded interval, let $\mathbf P$ be a partition of $I$ , and let $f:I\to\mathbf R$ be a function which is piecewise constant with respect to $\mathbf P$ . Let $\mathbf P'$ be a partition of $I$ which is finer than $\mathbf P$ . Then $f$ is also piecewise constant with respect to $\mathbf P'$ .

Solution: We choose $\forall K\in \mathbf P'$ , then $\exists J\in \mathbf P,K\subseteq J$ , we can know $f|_J$ is constant, thus $f$ is constant on $K$ , this means $f$ is piecewise constant with respect to $\mathbf P'$ .

$\blacksquare$

Exercise 11.2.2. Prove Lemma 11.2.8.
Lemma 11.2.8. Let $I$ be a bounded interval, and let $f:I\to\mathbf R$ and $g:I\to \mathbf R$ be piecewise constant functions on $I$ . Then the functions $f+g,f-g,\max (f,g)$ and $fg$ are also piecewise constant functions on $I$ . Here of course $\max(f,g):I\to\mathbf R$ is the function $\max (f,g)(x):=\max(f(x),g(x))$ . If $g$ does not vanish anywhere on $I$ (i.e., $g(x)\neq 0$ for all $x\in I$ ) then $f/g$ is also a piecewise constant function on $I$ .

Solution: By the condition given, there exists partitions $\mathbf P$ and $\mathbf P'$ , s.t. $f$ is piecewise constant with respect to $\mathbf P$ , and $g$ is piecewise constant with respect to $\mathbf P'$ .
Now we consider any element $K\in \mathbf P\#\mathbf P'$ , by Lemma 11.2.7, $f$ and $g$ are piecewise constant with respect to $\mathbf P\#\mathbf P'$ , thus we can have $c,d$ , s.t. $f(x)=c,g(x)=d,\forall x\in K$ , this means:

$(f+g)(x)=f(x)+g(x)=c+d,\quad \forall x\in K \\ (f-g)(x)=f(x)-g(x)=c-d,\quad \forall x\in K \\ \max (f,g) (x)=\max (f(x),g(x))=\max (c,d),\quad \forall x\in K \\(fg)(x)=f(x)g(x)=cd,\quad \forall x\in K$

If in addition we have $g(x)\neq 0,\forall x\in I$ , we can further have $d\neq 0$ and

$(f/g)(x)=f(x)/g(x)=c/d,\quad \forall x\in K$

From above we can see that $f+g,f-g,\max (f,g),fg$ and if $g$ does not vanish on $I$ , $f/g$ are constant on $K$ , since we choose arbitrary $K$ from $\mathbf P\#\mathbf P'$ , we conclude all the functions are piecewise constant on $I$ .

$\blacksquare$

Exercise 11.2.3. Prove Proposition 11.2.13.
Proposition 11.2.13. (Piecewise constant integral is independent of partition). 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 $p.c.\int_{[\mathbf P]}f=p.c.\int_{[\mathbf P']}f$ .

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

$p.c.\int_{[\mathbf P \#\mathbf P' ]} f=\sum\limits _{J\in \mathbf P \#\mathbf P'}c_J |J|$

is well defined. Now choose any $K\in \mathbf P$ , then $P_K={J\in \mathbf P \#\mathbf P':J\subseteq K}$ 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

$|K|=\sum\limits_{J\in \mathbf P_K}|J| \implies c_K|K|=\sum\limits_{J\in \mathbf P_K}c_K |J|=\sum\limits_{J\in \mathbf P_K}c_J |J|$

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 $K'\in \mathbf P'$ s.t. $J=K\cap K'$ , so the two sets are equal. Thus

$p.c.\int_{[\mathbf P]}f=\sum\limits_{K\in \mathbf P}c_K |K| =\sum\limits_{K\in \mathbf P}\sum_{J\in \mathbf P_K}c_J |J|=\sum\limits_{J\in \mathbf P \#\mathbf P'}c_J |J|=p.c.\int_{[\mathbf P \#\mathbf P']}f$

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

$\blacksquare$

Exercise 11.2.4. Prove Theorem 11.2.16.
Theorem 11.2.16. Let $I$ be a bounded interval, and let $f:I\to\mathbf R$ and $g:I\to \mathbf R$ be piecewise constant functions on $I$ .

Solution: 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 ) We have $p.c.\int_I (f+g)=p.c.\int_If+p.c.\int_Ig$ .
Solution: $f+g$ is piecewise constant with respect to $\mathbf P$ , with constant value $c_K+d_K$ on $K\in \mathbf P$ , thus

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

( b ) For any real number $c$ , we have $p.c.\int_I(cf)=c(p.c.\int_If)$ .
Solution: $cf$ is piecewise constant with respect to $\mathbf P$ , with constant value $cc_K$ on $K\in \mathbf P$

$p.c.\int_I(cf) =p.c.\int_{[\mathbf P]}(cf) =\sum_{K\in \mathbf P}(cc_K)|K| =c\sum_{K\in \mathbf P}c_K |K|=c(p.c.\int_{[\mathbf P]}f)=c(p.c.\int_If)$

( c ) We have $p.c.\int_I (f-g)=p.c.\int_If-p.c.\int_Ig$ .
Solution: Use (b) we have $p.c.\int_I(-g) =-p.c.\int_Ig$ , then use (a) we get

$p.c.\int_I(f-g) =p.c.\int_I(f+(-g)) =p.c.\int_If+p.c.\int_I(-g) =p.c.\int_If-p.c.\int_Ig$

( d ) If $f(x)\geq 0$ for all $x\in I$ , then $p.c.\int_If\geq 0$ .
Solution: If $f(x)\geq 0,\forall x\in I$ , then $c_K\geq 0,\forall K\in \mathbf P$ , so we have $c_K |K|\geq 0,\forall K\in \mathbf P$

$p.c.\int_If=p.c.\int_{[\mathbf P]}f=\sum _{K\in \mathbf P}c_K |K|\geq 0$

( e ) If $f(x)\geq g(x)$ for all $x\in I$ , then $p.c.\int_If\geq p.c.\int_Ig$ .
Solution: We have $f(x)-g(x)\geq 0,\forall x\in I$ , so use (d) we have

$p.c.\int_I(f-g) =p.c.\int_If-p.c.\int_Ig\geq 0 \implies p.c.\int_If\geq p.c.\int_Ig$

( f ) If $f$ is the constant function $f(x)=c$ for all $x\in I$ , then $p.c.\int_If=c|I|$ .
Solution: If $f(x)=c,\forall x\in I$ , then $c_K=c,\forall K\in \mathbf P$ , so we have by Theorem 11.1.13

$p.c.\int_If=p.c.\int_{[\mathbf P]}f=\sum_{K\in \mathbf P}c_K |K|=c\sum_{K\in \mathbf P}|K| =c|I|$

( g ) Let $J$ be a bounded interval containing $I$ (i.e., $I\subseteq J$ ), and let $F:J\to\mathbf R$ be the function

$F(x):=\begin{cases}f(x)&\text{if } x\in I\\0&\text{if }x\notin I\end{cases}$

Then $F$ is piecewise constant on $J$ , and $p.c.\int_JF=p.c.\int_If$ .
Solution: The set $\mathbf P\cup \{J\backslash I\}$ is a partition of $J$ , and $F$ is piecewise constant on $\mathbf P\cup \{J\backslash I\}$ , with additional constant value $c_{J\backslash I}=0$ , thus

$\begin{aligned}p.c.\int_JF&=p.c.\int_{[\mathbf P\cup \{J\backslash I\}]}F=\sum\limits_{K\in \mathbf P\cup \{J\backslash I}\}c_K |K|\\&=\sum\limits_{K\in \mathbf P}c_K |K|+0\cdot |J\backslash I|\\&=\sum\limits_{K\in \mathbf P}c_K |K|=p.c.\int_If\end{aligned}$

( 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

$p.c.\int_If=p.c.\int_Jf|_J+p.c.\int_Kf|_K$ .

Solution: We have $\mathbf P_{\mathbf J}=\{L\cap J:L\in \mathbf P\}$ and $\mathbf P_{\mathbf K}=\{L\cap K:L\in \mathbf P\}$ be 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 $\mathbf P_{\mathbf J}$ on J, $f|_K$ is piecewise constant with $\mathbf P_{\mathbf K}$ on $K$ . As we have $J\subseteq I$ and $K\subseteq I$ , we define

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

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

$\begin{aligned}p.c.\int_If&=p.c.\int_I(F+G)=p.c.\int_IF+p.c.\int_IG\\&=p.c.\int_Jf|_J +p.c.\int_Kf|_K \end{aligned}$

$\blacksquare$

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