陶哲轩实分析（上）11.8及习题-Analysis I 11.8

Exercise 11.8.1

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 $α[I]=\sum_{J\in \mathbf P}α[J]$ .
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
$α[I]=α(b)-α(a)=α(b)-α(c)+α(c)-α(a)=α[K]+α[I-K]$
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
$α[I-K]=\sum_{J\in \mathbf P-\{K\}}α[J]$
Combining these two identities, we obtain
$α[I]=\sum_{J\in \mathbf P}α[J]$
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.

Exercise 11.8.2

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 $p.c.\int_{[\mathbf P]} fdα=p.c.\int_{[\mathbf P']}fdα$.
Proof: By Lemma 11.2.7, we know $f$ is piecewise constant with respect to $\mathbf P\#\mathbf P'$, thus the value
$p.c.\int_{[\mathbf P\#\mathbf P']}fdα=\sum_{J\in \mathbf P\# \mathbf P'}c_J α[J]$
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
$α[K]=\sum_{J\in \mathbf P_K}α[J] \implies c_K α[K]=\sum_{J\in \mathbf P_K}c_K α[J]=\sum_{J\in \mathbf P_K}c_J α[J]$
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 $K'∈P'$ s.t. $J=K\cap K'$, so the two sets are equal. Thus
$\begin{aligned}p.c.\int_{[\mathbf P]}fdα&=\sum_{K\in \mathbf P}c_K α[K]=\sum_{K\in \mathbf P}\sum_{J\in P_K}c_J α[J]=\sum_{J\in \mathbf P\# \mathbf P'}c_J α[J]\\&=p.c.\int _{[\mathbf P\# \mathbf P']}fdα\end{aligned}$
Similarly we can prove $p.c.\int_{[\mathbf P' ]}fdα=p.c.\int_{[\mathbf P\# \mathbf P' ]} fdα$, and the statement is proved.

Exercise 11.8.3

Proposition: Let $I$ be a bounded interval, and let $f:I→\mathbf R$ and $g:I→\mathbf R$ be piecewise constant functions on $I$. Let $α:X→\mathbf R$ be defined on a domain containing $I$ which is monotone increasing, then
( a ) We have $p.c.∫_I (f+g)dα=p.c.∫_I fdα+p.c.∫_I gdα$.
( b ) For any real number $c$, we have $p.c.∫_I (cf)dα=c(p.c.∫_I fdα)$.
( c ) We have $p.c.∫_I (f-g)dα=p.c.∫_I fdα-p.c.∫_I gdα$.
( d ) If $f(x)\geq 0,\forall x\in I$, then $p.c.∫_I fdα\geq 0$.
( e ) If $f(x)\geq g(x),\forall x\in I$, then $p.c.∫_I fdα\geq p.c.∫_I gdα$.
( f ) If $f(x)=c,\forall x\in I$, then $p.c.∫_I fdα=cα[I]$.
( g ) Let $J$ be a bounded interval containing $I$ (i.e. $I\subseteq J$), and let $F:J→\mathbf R$ be the function
$F(x)=\begin{cases}f(x),&x\in I\\0,&x\notin I\end{cases}$
Then $F$ is piecewise constant on $J$, and $p.c.∫_J Fdα=p.c.∫_I fdα$.
( h ) Suppose that $\{J,K\}$ is a partition of $I$ into two intervals $J$ and $K$. Then the functions $f|_J:J→\mathbf R$ and $f|_K:K→\mathbf R$ are piecewise constant on $J$ and $K$ respectively, and we have
$p.c.∫_I fdα=p.c.∫_Jf|_J dα+p.c.∫_Kf|_K dα$
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 $K\in \mathbf P$,
$\begin{aligned}p.c.∫_I (f+g)dα&=p.c.∫_{[\mathbf P]} (f+g)dα=\sum_{K\in \mathbf P}(c_K+d_K)α[K]\\&=\sum_{K\in \mathbf P}c_K α[K]+\sum_{K\in \mathbf P}d_K α[K]\\&=p.c.∫_{[\mathbf P]} fdα+p.c.∫_{[\mathbf P]} gdα\\&=p.c.∫_I fdα+p.c.∫_I gdα\end{aligned}$
( b ) $cf$ is piecewise constant with respect to $\mathbf P$, with constant value $cc_K$ on $K∈\mathbf P$,
$\begin{aligned}p.c.∫_I (cf)dα&=p.c.∫_{[\mathbf P]} (cf)dα=∑_{K∈\mathbf P}(cc_K)α[K]\\&=c∑_{K∈\mathbf P}c_K α[K]=c\left(p.c.∫_{[\mathbf P]} fdα\right)\\&=c\left(p.c.∫_I fdα\right)\end{aligned}$
( c ) Use ( b ) we have $p.c.∫_I (-g)dα=-p.c.∫_I gdα$, then use ( a ) we get
$\begin{aligned}p.c.∫_I(f-g)dα&=p.c.∫_I (f+(-g))dα\\&=p.c.∫_I fdα+p.c.∫_I (-g)dα\\&=p.c.∫_Ifdα-p.c.∫_Igdα\end{aligned}$
( d ) If $f(x)≥0,∀x∈I$, then $c_K≥0,∀K∈\mathbf P$, since $α$ is monotone increasing we have $c_K α[K]≥0,∀K∈\mathbf P$, thus
$p.c.∫_I fdα=p.c.∫_{[\mathbf P]} fdα=∑_{K∈P}c_K α[K]≥0$
( e ) We have $f(x)-g(x)≥0,∀x∈I$, so use ( d ) we have
$p.c.∫_I (f-g)dα=p.c.∫_I fdα-p.c.∫_I gdα≥0 \\⇒ p.c.∫_I fdα≥p.c.∫_I gdα$
( f ) If $f(x)=c,∀x∈I$, then $c_K=c,∀K∈P$, so we have by Lemma 11.8.4
$p.c.∫_I fdα=p.c.∫_{[\mathbf P]} fdα=∑_{K∈\mathbf P}c_K α[K]=c∑_{K∈\mathbf P}α[K] =cα[I]$

(g) The set $\mathbf P∪\{J-I\}$ is a partition of $J$, and $F$ is piecewise constant on $\mathbf P∪\{J-I\}$, with additional constant value $c_{J-I}=0$, thus
$\begin{aligned}p.c.∫_J Fdα&=p.c.∫_{[\mathbf P∪\{J-I\}]} Fdα=∑_{K∈\mathbf P∪\{J-I\}}c_K α[K]\\&=∑_{K∈\mathbf P}c_K α[K]+0\cdot α[J-I]=∑_{K∈\mathbf P}c_K α[K]\\&=p.c.∫_I fdα\end{aligned}$

( h ) We have $\mathbf P_J=\{L∩J:L∈P\}$ and $\mathbf P_K=\{L∩K:L∈P\}$ 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 $J⊆I$ and $K⊆I$, we define
$F(x)=\begin{cases}f|_J (x),&x∈J\\0,&x\notin J\end{cases},\quad G(x)=\begin{cases}f|_K (x),&x∈K\\0,&x\notin K\end{cases}$
Since $\{J,K\}$ is a partition of $I$, we have $f=f|_J+f|_K=F+G$, thus use ( a ) ,( g ) we have
$\begin{aligned}p.c.∫_I fdα&=p.c.∫_I(F+G)dα=p.c.∫_I Fdα+p.c.∫_I Gdα\\&=p.c.∫_Jf|_J dα+p.c.∫_K^ f|_K dα\end{aligned}$

Exercise 11.8.4

Proposition: Let $I$ be a bounded interval, and let $f$ be a function which is uniformly continuous on $I$. Let $α:X→\mathbf R$ be defined on a domain containing $I$ which is monotone increasing, Then $f$ is Riemann-Stieltjes integrable on $I$ with respect to $α$.
Proof: From Proposition 9.9.15 we see that $f$ is bounded. Now we have to show that $\underline{∫}_I fdα=\overline{∫}_I fdα$.
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 $a<b$.
Let $ε>0$ be arbitrary. By uniform continuity, there exists a $δ>0$ such that $|f(x)-f(y)|<ε$ whenever $x,y∈I,|x-y|<δ$. By the Archimedean principle, there exists an integer $N>0$ such that $(b-a)/N<δ$, Note that we can partition $I$ into $N$ intervals $J_1,\dots,J_N$, each of length $(b-a)/N$, we thus have
$\overline{\int}_I fdα≤∑_{k=1}^N\left(\sup_{x∈J_k}⁡f(x) \right)α[J_k],\quad \underline{\int}_I fdα≥∑_{k=1}^N\left(\inf_{x∈J_k}⁡f(x)\right)α[J_k]$
So we have
$\begin{aligned}\overline{∫}_I fdα-\underline{∫}_I fdα&≤∑_{k=1}^N\left(\sup_{x∈J_k}⁡f(x)-\inf_{x∈J_k}⁡f(x) \right)α[J_k] \\&≤ε∑_{k=1}^Nα[J_k] =εα[I]=ε(α(b)-α(a))\end{aligned}$
This shows $\underline{∫}_I fdα-\overline{∫}_I fdα$ cannot be positive.

Exercise 11.8.5

We let $ε>0$ be arbitrary. Then since $f$ is continuous, $∃δ>0$ such that
$|f(x)-f(0)|<ϵ,\quad ∀x∈(-δ,δ)$
Since $f$ is continuous on $[-1,1]$, $f$ is bounded, thus $∃M>0$, s.t.
$|f(x)|≤M,\quad ∀x∈[-1,1]$
We let $\{[-1,-δ],(-δ,δ),[δ,1]\}$ be a partition of $[-1,1]$, then the piecewise constant function
$g(x)=\begin{cases}f(0)+ϵ,&x∈(-δ,δ)\\M,&x∈[-1,1]\backslash (-δ,δ)\end{cases}$
majorizes $f$ and we have
$\begin{aligned}&\quad\quad\overline{∫}_{[-1,1]} fd\text{sgn}≤∫_{[-1,1]}gd\text{sgn}\\&=M(\text{sgn}(-δ)-\text{sgn}(-1))+(f(0)+ϵ)(\text{sgn}(δ)-\text{sgn}(-δ))+M(\text{sgn}(1)-\text{sgn}(δ))\\&=2(f(0)+ϵ)\end{aligned}$
Similarly, the piecewise constant function
$h(x)=\begin{cases}f(0)-ϵ,&x∈(-δ,δ)\\-M,&x∈[-1,1]\backslash(-δ,δ) \end{cases}$
minorizes $f$ and we have
$\underline{∫}_{[-1,1]} fd\text{sgn}≥∫_{[-1,1]}hd\text{sgn}=2(f(0)-ϵ)$
Thus we have
$2(f(0)-ϵ)≤\underline{∫}_{[-1,1]} fd\text{sgn}≤\overline{∫}_{[-1,1]}fd\text{sgn}≤2(f(0)+ϵ)$
And the result follows.