陶哲轩实分析（上）11.2及习题-Analysis I 11.2

这节介绍逐段常值函数。

Exercise 11.2.1

We choose $∀K∈\mathbf P'$, then $∃J∈\mathbf P,K⊆J$, we can know $f|_J$ is constant, thus $f$ is constant on $K$, this means $f$ is piecewise constant with respect to $\mathbf P'$.

Exercise 11.2.2

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 $K∈\mathbf P\#\mathbf P'$, 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. $f(x)=c,g(x)=d,∀x∈K$, this means:
$(f+g)(x)=f(x)+g(x)=c+d,\quad ∀x∈K \\ (f-g)(x)=f(x)-g(x)=c-d,\quad ∀x∈K \\ \max⁡(f,g) (x)=\max⁡(f(x),g(x))=\max⁡(c,d),\quad ∀x∈K \\ (fg)(x)=f(x)g(x)=cd,\quad ∀x∈K$
If in addition we have $g(x)≠0,∀x∈I$, we can further have $d≠0$ and
$(f/g)(x)=f(x)/g(x)=c/d,\quad ∀x∈K$
From above we can see that $f+g,f-g,\max⁡(f,g),fg$ and if $g$ doesn’t 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 $I$.

Exercise 11.2.3

By Lemma 11.2.7, we know $f$ is piecewise constant with respect to $\mathbf P\#\mathbf P'$, thus the value
$p.c.∫_{[\mathbf P\#\mathbf P' ]} f=∑_{J∈\mathbf P\#\mathbf P'}c_J |J|$
is well defined. Now choose any $K∈\mathbf P$, then $P_K=\{J∈\mathbf P\#\mathbf P':J⊆K\}$ 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 $c_J=c_K,∀J∈\mathbf P_K$, and
$|K|=∑_{J∈\mathbf P_K}|J| ⇒ c_K |K|=∑_{J∈\mathbf P_K}c_K |J|=∑_{J∈\mathbf P_K}c_J |J|$
Also, consider the set $\{J∈\mathbf P\#\mathbf P':J⊆K\text{ for some }K∈\mathbf P\}⊆\mathbf P\#\mathbf P'$, for any $J∈\mathbf P\#\mathbf P'$, by definition we can find a $K∈\mathbf P$ and a $K'∈\mathbf P'$ s.t. $J=K∩K'$, so the two sets are equal. Thus
$p.c.∫_{[\mathbf P]}f=∑_{K∈\mathbf P}c_K |K| =∑_{K∈\mathbf P}∑_{J∈\mathbf P_K}c_J |J|=∑_{J∈\mathbf P\#\mathbf P'}c_J |J|=p.c.∫_{[\mathbf P\#\mathbf P']}f$
Similarly we can prove $p.c.∫_{[\mathbf P' ]}f=p.c.∫_{[\mathbf P\#\mathbf P']}f$, and the statement is proved.

Exercise 11.2.4

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 $K∈\mathbf P$, 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 $K∈\mathbf P$
$\begin{aligned}p.c.∫_I (f+g) &=p.c.∫_{[\mathbf P]}(f+g) =∑_{K∈\mathbf P}(c_K+d_K)|K|=∑_{K∈\mathbf P}c_K |K|+∑_{K∈\mathbf P}d_K |K|\\&=p.c.∫_{[\mathbf P]}f+p.c.∫_{[\mathbf P]}g=p.c.∫_If+p.c.∫_I g \end{aligned}$

( b ) $cf$ is piecewise constant with respect to $\mathbf P$, with constant value $cc_K$ on $K∈\mathbf P$
$p.c.∫_I(cf) =p.c.∫_{[\mathbf P]}(cf) =∑_{K∈\mathbf P}(cc_K)|K| =c∑_{K∈\mathbf P}c_K |K|=c(p.c.∫_{[\mathbf P]}f)=c(p.c.∫_If)$

( c ) Use (b) we have $p.c.∫_I(-g) =-p.c.∫_Ig$, then use (a) we get
$p.c.∫_I(f-g) =p.c.∫_I(f+(-g)) =p.c.∫_If+p.c.∫_I(-g) =p.c.∫_If-p.c.∫_Ig$

( d ) If $f(x)≥0,∀x∈I$, then $c_K≥0,∀K∈\mathbf P$, so we have $c_K |K|≥0,∀K∈\mathbf P$
$p.c.∫_If=p.c.∫_{[\mathbf P]}f=∑_{K∈\mathbf P}c_K |K|≥0$

( e ) We have $f(x)-g(x)≥0,∀x∈I$, so use (d) we have
$p.c.∫_I(f-g) =p.c.∫_If-p.c.∫_Ig≥0 ⇒ p.c.∫_If≥p.c.∫_Ig$

( f ) If $f(x)=c,∀x∈I$, then $c_K=c,∀K∈\mathbf P$, so we have by Theorem 11.1.13
$p.c.∫_If=p.c.∫_{[\mathbf P]}f=∑_{K∈\mathbf P}c_K |K|=c∑_{K∈\mathbf P}|K| =c|I|$

( g ) The set $\mathbf P∪\{J\backslash I\}$ is a partition of $J$, and $F$ is piecewise constant on $\mathbf P∪\{J\backslash I\}$, with additional constant value $c_{J\backslash I}=0$, thus
$p.c.∫_JF=p.c.∫_{[\mathbf P∪\{J\backslash I\}]}F=∑_{K∈\mathbf P∪\{J\backslash I\}}c_K |K|\\=∑_{K∈\mathbf P}c_K |K|+0⋅|J\backslash I|=∑_{K∈\mathbf P}c_K |K|=p.c.∫_If$

( h ) We have $\mathbf P_{\mathbf J}=\{L∩J:L∈\mathbf P\}$ and $\mathbf P_{\mathbf K}=\{L∩K:L∈\mathbf P\}$ 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 $J⊆I$ and $K⊆I$, we define
$F(x)=\begin{cases}f|_J (x),& x∈J\\0, & x∉J\end{cases},\quad G(x)=\begin{cases}f|_K (x),& x∈K\\0, & x∉K\end{cases}$
Since $\{J,K\}$ is a partition of $I$, we have $f=f|_J+f|_K=F+G$, thus use (a) ,(g) we have
$p.c.∫_If=p.c.∫_I(F+G)=p.c.∫_IF+p.c.∫_IG\\=p.c.∫_Jf|_J +p.c.∫_Kf|_K$