陶哲轩实分析（上）11.3及习题-Analysis I 11.3

Exercise 11.3.1

Since $f$ majorizes $g$ we have $f(x)≥g(x),∀x∈I$, and $g$ majorizes $h$ means $g(x)≥h(x),∀x∈I$, thus we have $f(x)≥h(x),∀x∈I$, which means $f$ majorizes $h$.
If $f$ and $g$ majorize each other, then it means $f(x)≥g(x),∀x∈I$ and $g(x)≥f(x),∀x∈I$, so we shall have $f(x)=g(x),∀x∈I$.

Exercise 11.3.2

We have
$\begin{aligned}f \text{ majorizes }g &⇒f(x)≥g(x),∀x∈I \\&⇒f(x)+h(x)≥g(x)+h(x),∀x∈I \\&⇒(f+h)(x)≥(g+h)(x),∀x∈I \\&⇒ f+h \text{ majorizes }g+h\end{aligned}$
We cannot have $fh$ majorizes $gh$ if $f$ majorizes $g$, since if we let $h(x)=-1$ on $I$, we shall have $(fh)(x)=f(x)h(x)=-f(x)≤-g(x)=g(x)h(x)=(gh)(x)$.
We cannot have $cf$ majorizes $cg$ if $f$ majorizes $g$, we could let $c=-1$ and use a similar proof to show $(cf)(x)≤(cg)(x)$.

Exercise 11.3.3

For $f$, we see $f(x)≥f(x),∀x∈I$, thus $f$ majorizes $f$, and $f$ is piecewise constant, thus by definition of upper Riemann integral we have
$\overline{\int}_I f≤p.c.\int_If$
By the same logic we have $f$ minorizes $f$, thus
$\underline{\int}_I f≥p.c.\int_I f$
Thus we have $\underline{\int}_I f≥\overline{\int}_I f$, use Lemma 11.3.3 we can see $f$ is Riemann integrable, and
$\int_If=p.c.\int_If$

Exercise 11.3.4

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 $J∈\mathbf P$, then $c_J≥f(x),∀x∈J$, thus
$c_J≥\sup_{x∈J}f(x),∀J∈P$
So we have
$p.c.\int_Ig=\sum\limits_{J∈\mathbf P}c_J |J|=\sum\limits_{J∈\mathbf P,J\neq \emptyset}c_J |J|≥\sum\limits_{J∈\mathbf P,J\neq \emptyset}\left(\sup_{x∈J}⁡f(x) \right) |J|=U(f,\mathbf P)$
A similar proof can show $p.c.\int_Ih≤L(f,\mathbf P)$.

Exercise 11.3.5

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 $p.c.\int_Ig≥U(f,\mathbf P)$, so
$p.c.\int_Ig≥\inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}$
then since $g$ is arbitrary as long as it is piecewise constant with $\mathbf P$ and majorizes $f$ on $I$,
$\overline{\int}_I f≥\inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}$
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: $∀x∈I,∃J∈\mathbf P,x∈J$, this $J$ is unique, so we let
$G(x)=\sup_{x∈J}⁡f(x)$
Then $G$ is piecewise constant with $\mathbf P$ and majorizes $f$ on $I$, further we have
$p.c.\int_IG=U(f,\mathbf P)$
Thus we can say that
$\overline{\int}_I f≤p.c.∫_IG=U(f,\mathbf P)$
Since $\mathbf P$ could be any partition of $I$, we can say
$\overline{\int}_I f≤U(f,\mathbf P),\quad ∀\mathbf P\text{ is a partition of }I$
So it’s easy to see
$\overline{\int}_I f≤\inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}$
Thus we can have
$\overline{\int}_I f=\inf \{U(f,\mathbf P):\mathbf P\text{ is a partition of }I\}$
A similar proof can show that
$\underline{\int}_I f=\sup \{L(f,\mathbf P):\mathbf P\text{ is a partition of }I\}$