陶哲轩实分析（上）11.10及习题-Analysis I 11.10

Exercise 11.10.1

As $F$ and $G$ are differentiable, they are continuous on $[a,b]$ and thus Riemann integrable by Corollary 11.5.2, then $FG'$ and $F'G$ are Riemann integrable by Theorem 11.4.5. Notice that $(FG)'=F' G+FG'$ and $F'G+FG'$ is Riemann integrable on $[a,b]$, we can use the second fundamental theorem of calculus to have
$∫_{[a,b]}(F'G+FG')=(FG)(b)-(FG)(a)$
By some simple calculation we can have the final result.

Exercise 11.10.2

I will prove Lemma 11.10.5 completely.
Let $\mathbf P$ be a partition of $[{\phi}(a),{\phi}(b)]$ such that $f$ is piecewise constant with respect to $\mathbf P$. we may assume that $\mathbf P$ does not contain the empty set. For each $J\in \mathbf P$, let $c_J$ be the constant value of $f$ on $J$, thus
$∫_{[{\phi}(a),{\phi}(b)]} f=∑_{J\in \mathbf P}c_J |J|$
For each interval $J$, let ${\phi}^{-1} (J):=\{x\in [a,b]:{\phi}(x)\in J\}$. Then given any $x,y\in {\phi}^{-1} (J),x<y$, we have ${\phi}(x),{\phi}(y)\in J$ by the definition of ${\phi}^{-1} (J)$, and ${\phi}(x)≤{\phi}(y)$ since ${\phi}$ is monotone increasing, and $∀z\in [x,y]$, we have ${\phi}(x)≤{\phi}(z)≤{\phi}(y)$ again by ${\phi}$ is monotone increasing, this means ${\phi}(z)\in J$ since $J$ is an interval, so ${\phi}^{-1} (J)$ is connected, and thus is an interval. Furthermore, if $x\in {\phi}^{-1} (J)$, then ${\phi}(x)\in J$ and $f({\phi}(x))=c_J$, thus $c_J$ is the constant value of $f\circ {\phi}$ on ${\phi}^{-1}(J)$.
Thus if we define $\mathbf Q:=\{{\phi}^{-1}(J):J\in \mathbf P\}$, then for any $c\in [a,b], {\phi}(c)\in [{\phi}(a),{\phi}(b)]$, so there is a $J\in \mathbf P$ such that ${\phi}(c)\in J$, which means $c\in {\phi}^{-1}(J)$. Assume we can find two sets $Q_1,Q_2\in \mathbf Q$ s.t. $c\in Q_1∩Q_2$, then it means we can find $J_1,J_2\in \mathbf P$ s.t. ${\phi}(c)\in J_1∩J_2$, contradict the fact that $\mathbf P$ is a partition of $[{\phi}(a),{\phi}(b)]$. We can conclude $Q$ is a partition of $[a,b]$.
Given any $Q\in \mathbf Q$, we have $Q={\phi}^{-1}(J)$ for some $J\in \mathbf P$, so $c_J$ is the constant value of $f\circ {\phi}$ on this $Q$, which means $f\circ {\phi}$ piecewise constant with respect to $\mathbf Q$. Thus
$∫_{[a,b]}f\circ {\phi} d{\phi}=∫_{[\mathbf Q]}f\circ {\phi} d{\phi}=∑_{J\in \mathbf P}c_J {\phi}[{\phi}^{-1}(J)]$
To calculate ${\phi}[{\phi}^{-1}(J)]$, we let $J\in \mathbf P$, then $J$ is an interval with endpoints $a<b$, and $|J|=b-a$, as ${\phi}$ is monotone increasing, we can find $a',b'\in {\phi}^{-1} (J)$, s.t. ${\phi}(a' )=a,{\phi}(b' )=b$, and ${\phi}^{-1}(J)$ is an interval which has endpoints $a',b'$, thus ${\phi}[{\phi}^{-1}(J)]={\phi}(b' )-{\phi}(a')=b-a=|J|$, and the claim follows.

Exercise 11.10.3

For $∀ϵ>0$, we can find a partition of $[a,b]$, namely $\mathbf P$, and piecewise constant function $\overline{f}$ which majorizes $f$ and $\underline{f}$ which minorizes $f$, both with respect to $\mathbf P$, such that
$∫_{[a,b]}f-ϵ<∫_{[a,b]}\underline{f}≤∫_{[a,b]}\overline{f}<∫_{[a,b]}f+ϵ$
Now we define $\overline{g}(x)=\overline{f}(-x)$ and $\underline{g}(x)=\underline{f}(-x)$ on $[-b,-a]$, then it is easy to see that $\overline{g}$ majorizes $g$ and $\underline{g}$ minorizes $g$. For any $J∈\mathbf P$, we define $K_J=\{-x:x∈J\}$, then $|K_J |=|J|$, and $\mathbf P'=\{K_J:J∈P\}$ is a partition of $[-b,-a]$, and the constant value of $\overline{g}$ on $K_J$ is the same as the constant value of $\overline{f}$ on $J$, the constant value of $\underline{g}$ on $K_J$ is the same as the constant value of $\underline{f}$ on $J$, so we have
$∫_{[-b,-a]}\overline{g}=p.c.∫_{[-b,-a]}\overline{g}=∑_{K_J∈\mathbf P'}c_J |K_J|=∑_{J∈\mathbf P}c_J |J| =∫_{[a,b]}\overline{f}$
Similarly we have
$∫_{[-b,-a]}\underline{g}=∫_{[a,b]}\underline{f}$
So we have
$∫_{[a,b]}f-ϵ<∫_{[-b,-a]}\underline{g}≤\underline{∫}_{[-b,-a]}g≤\overline{∫}_{[-b,-a]}g≤∫_{[-b,-a]}\overline{g}<∫_{[a,b]}f+ϵ$
and the conclusion follows.

Exercise 11.10.4

Let $[a,b]$ be a closed interval, and let $ϕ:[a,b]→[ϕ(b),ϕ(a)]$ be a differentiable monotone decreasing function such that $ϕ'$ is Riemann integrable. Let $f: [ϕ(b),ϕ(a)]→\mathbf R$ be a Riemann integrable function on $[ϕ(b),ϕ(a)]$. Then $(f\circ ϕ) ϕ':[a,b]→\mathbf R$ is Riemann integrable on $[a,b]$ and
$∫_{[a,b]}(f\circ ϕ) ϕ'=-∫_{[ϕ(b),ϕ(a)]}f$