陶哲轩实分析（上）11.4及习题-Analysis I 11.4

Exercise 11.4.1

Since $f$ and $g$ are both Riemann integrable, we have
$\overline{\int}_I f=\underline{\int}_I f,\quad \overline{\int}_I g=\underline{\int}_I g$
So for $∀ϵ>0$, we have piecewise constant function $\overline{f}$ majorizes $f$, $\underline{f}$ minorizes $f$, $\overline{g}$ majorizes $g$, $\underline{g}$ minorizes $g$, such that
${\int}_I^ f-ϵ<{\int}_I \underline{f}≤{\int}_I f≤{\int}_I \overline{f}<{\int}_I f+ϵ,\\{\int}_I g-ϵ<{\int}_I \underline{g}≤{\int}_I g≤{\int}_I \overline{g}<{\int}_I g+ϵ$
( a ) It’s easy to prove that $\overline{f}+\overline{g}$ majorizes $f+g$ and $\underline{f}+\underline{g}$ minorizes $f+g$, so we have
${\int}_I (\underline{f}+\underline{g}) ≤\underline{\int}_I (f+g) ≤\overline{\int}_I (f+g)≤{\int}_I (\overline{f}+\overline{g})$
Use Theorem 11.2.16 we have
${\int}_I (\underline{f}+\underline{g}) ={\int}_I \underline{f}+{\int}_I \underline{g}>{\int}_I f+{\int}_I g-2ϵ \\ {\int}_I (\overline{f}+\overline{g}) ={\int}_I \overline{f}+{\int}_I \overline{g}<{\int}_I f+{\int}_I g+2ϵ$
Thus
$\overline{\int}_I (f+g)-\underline{\int}_I (f+g)≤{\int}_I (\overline{f}+\overline{g}) -{\int}_I (\underline{f}+\underline{g}) <4ϵ$
So $f+g$ is Riemann integrable, and
${\int}_I f+{\int}_I g-2ϵ<{\int}_I (\underline{f}+\underline{g}) ≤{\int}_I (f+g) ≤{\int}_I (\overline{f}+\overline{g}) <{\int}_If+{\int}_Ig+2ϵ$
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
${\int}_I (c\underline{f}) ≤\underline{\int}_I (cf) ≤\overline{\int}_I (cf)≤{\int}_I(c\overline{f})$
Use Theorem 11.2.16 we have
${\int}_I(c\underline{f}) =c{\int}_I \underline{f}>c{\int}_I f-cϵ,\quad {\int}_I (c\overline{f}) =c{\int}_I \overline{f}<c{\int}_I f+cϵ$
Thus
$c{\int}_I f-cϵ<{\int}_I (c\underline{f}) ≤\underline{\int}_I (cf) ≤\overline{\int}_I (cf)≤{\int}_I (c\overline{f}) <c{\int}_I f+cϵ$
Which shows ${\int}_I (cf) =c{\int}_I f$.
Now if $c=0$, then $cf=0$, which is a piecewise constant function on $I$, thus
${\int}_I (cf) =p.c.{\int}_I (cf) =p.c.{\int}_I (0) =0=0\cdot {\int}_I f$
Now suppose $c=-1$, then $-\overline{f}$ minorizes $-f$, $-\underline{f}$ majorizes $-f$, so we have
${\int}_I (-\overline{f}) ≤\underline{\int}_I (-f) ≤\overline{\int}_I (-f)≤{\int}_I (-\underline{f})$
Use Theorem 11.2.16 we have
${\int}_I (-\overline{f}) =-{\int}_I \overline{f}>-{\int}_I f-ϵ,\quad {\int}_I (-\underline{f}) =-{\int}_I \underline{f}<-{\int}_I f+ϵ$
Thus
$-{\int}_I f-ϵ<-{\int}_I \overline{f}≤\underline{\int}_I (-f) ≤\overline{\int}_I (-f)≤-{\int}_I \underline{f}<-{\int}_I f+ϵ$
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
${\int}_I (cf) ={\int}_I (-1)(-cf) =-{\int}_I (-cf) =-(-c) {\int}_I f=c{\int}_I f$
( c ) Use (a) and (b) we could have
${\int}_I(f-g) ={\int}_I (f+(-g)) ={\int}_I f+{\int}_I(-g)={\int}_I f-{\int}_I g$
( d ) The constant function $h(x)=0$ minorizes $f$, thus
${\int}_I f≥\underline{\int}_I (f)≥{\int}_I h={\int}_I0\cdot h=0{\int}_I h=0$
( e ) The function $f(x)-g(x)≥0,∀x∈I$, thus use ( d ) and ( c )
${\int}_I f-{\int}_I g={\int}_I (f-g) ≥0 ⇒ {\int}_I f≥{\int}_I g$
( f ) In this case, we can use the piecewise constant function integral formula:
${\int}_I f=p.c.{\int}_I f=∑_{J∈P}c_J |J| =c∑_{J∈P}|J| =c|I|$
( g ) We can define $\overline{F}$ and $\underline{F}$ which are piecewise and majorizes/minorizes $F$ as:
$\overline{F} (x)=\begin{cases}\overline{f} (x),&x∈I\\0,&x∉I\end{cases},\quad \underline{F} (x)=\begin{cases}\underline{f} (x),&x∈I\\0,&x∉I\end{cases}$
Then use Theorem 11.2.16 we have
${\int}_I f-ϵ<{\int}_I \underline{f}={\int}_J \underline{F}≤\underline{\int}_J F≤\overline{\int}_J F≤{\int}_J \overline{F}={\int}_I \overline{f}<{\int}_I f+ϵ$
which shows ${\int}_J F={\int}_I f$.
( h ) Use Theorem 11.2.16 (h) we have
${\int}_I \overline{f}={\int}_J \overline{f}|_J +{\int}_K \overline{f}|_K ,{\int}_I \underline{f}={\int}_J \underline{f}|_J +{\int}_K \underline{f}|_K$
Thus
${\int}_I \overline{f}-{\int}_I \underline{f}=({\int}_J\overline{f}|_J -{\int}_J \underline{f}|_J )+({\int}_K\overline{f}|_K -{\int}_K\underline{f}|_K )<2ϵ$
Since we have $\overline{f}≥f≥\underline{f}$ on $I,J,K$, this means
${\int}_J\overline{f}|_J -{\int}_J\underline{f}|_J ≥0,\quad {\int}_K\overline{f}|_K -{\int}_K\underline{f}|_K ≥0$
Thus we have
${\int}_J\overline{f}|_J -{\int}_J\underline{f}|_J <2ϵ,{\int}_K\overline{f}|_K -{\int}_K\underline{f}|_K <2ϵ$
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
$F(x)=\begin{cases}f(x),&x∈J\\0,&x∉J\end{cases},\quad G(x)=\begin{cases}f(x),&x∈K\\0,&x∉K\end{cases}$
Then $f(x)=F(x)+G(x)$, use (a) and (g) we have
${\int}_I f={\int}_I F+{\int}_I G={\int}_Jf|_J +{\int}_Kf|_K$

Exercise 11.4.2

Assume $∃c∈[a,b],f(c)>0$, since $f$ is continuous, let $ϵ=f(c)/2$, then $∃δ>0$, s.t.
$|f(x)-f(c)|<ϵ=f(c)/2,\quad ∀x∈[a,b]∩(c-δ,c+δ)$
Since the length of $[a,b]∩(c-δ,c+δ)$ is longer than $δ/2$, and we have
$f(x)>f(c)/2,\quad ∀x∈[a,b]∩(c-δ,c+δ)$
We can define
$g(x)=\begin{cases}f(c)/2,&x∈[a,b]∩(c-δ,c+δ)\\f(x),&x∈[a,b],x∉(c-δ,c+δ)\end{cases}$
Then we can say $g$ minorizes $f$ on $[a,b]$, since $f$ is Riemann integrable on $[a,b]$, and $\{[a,b]∩(c-δ,c+δ),[a,b]\backslash(c-δ,c+δ)\}$ is a partition of [a,b], we can say the function $f|_{[a,b]\backslash(c-δ,c+δ)}$ 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
$0=\int_{[a,b]}f≥\int_{[a,b]}g=\int_{[a,b]∩(c-δ,c+δ)}\left(\frac{f(c)}{2}\right) +∫_{[a,b]\backslash(c-δ,c+δ)} f≥\frac{f(c)}{2} \left|\frac{δ}{2}\right|=\frac{δf(c)}{4}>0$
This is a contradiction.

Exercise 11.4.3

Repeated use of Theorem 11.4.1(h), we can get
$∫_If=∑_{J∈P}∫_Jf|_J =\sum_{J∈P}∫_Jf$

Exercise 11.4.4

It’s 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.