陶哲轩实分析（上）11.9及习题-Analysis I 11.9

Exercise 11.9.1

Since $f$ is monotonic increasing, $f$ is Riemann ingegrable and $F$ is well-defined.
We assume $F(x)$ is differentiable at some $q∈Q∩[0,1]$, then if $q≠0,q≠1$, $F$ is differentiable at $q$ iff we have
$\lim_{x→q;x∈[0,q)}\frac{F(x)-F(q)}{x-q}=\lim_{x→q;x∈(q,1]}\frac{F(x)-F(q)}{x-q}$
From Exercise 9.8.5 we know that $f$ is not continuous at $q$ and if we locate $q=q(n)$ for some $n∈\mathbf N$ as in Exercise 9.8.5, we shall have $f(x)≥f(q)+2^{-n},x>q$, so
$\lim_{x→q;x∈[0,q)}\frac{F(x)-F(q)}{x-q}=\lim_{x→q;x∈[0,q)}\frac{∫_x^qf(y)dy}{x-q}≤\lim_{x→q;x∈[0,q)}\frac{∫_x^qf(q)dy}{x-q}=f(q) \\ \lim_{x→q;x∈(q,1]}\frac{F(x)-F(q)}{x-q}=\lim_{x→q;x∈(q,1]}\frac{\int_q^xf(y)dy}{x-q}≥\lim_{x→q;x∈(q,1]}\frac{\int_q^x(f(q)+2^{-n})dy}{x-q}=f(q)+2^{-n}$
Thus
$\lim_{x→q;x∈(q,1]}\frac{F(x)-F(q)}{x-q}-\lim_{x→q;x∈[0,q)}\frac{F(x)-F(q)}{x-q}≥2^{-n}$
contradicts the result that the two limits are equal.
If $q=0$ or $q=1$, $F$ may be differentiable at $q$ by definition, since $f$ is continuous at 0 or 1 when restricted on $[0,1]$.

Exercise 11.9.2

When $I$ is empty or a single point then the statement is trivial. Now suppose $I$ is a nonempty interval, choose some $x_0∈I$, then for any $x∈I,x≠x_0$, the function $F-G$ is differentiable on $[x_0,x]$ or $[x,x_0]$, thus we have some $ξ$ between $x$ and $x_0$, s.t.
$\frac{(F-G)(x)-(F-G)(x_0)}{x-x_0}=(F-G)' (ξ)=f(ξ)-f(ξ)=0$
So we have
$(F-G)(x)=F(x)-G(x)=F(x_0 )-G(x_0 )$
This means if we let $C=F(x_0)-G(x_0)$, we will have $F(x)=G(x)+C,∀x∈I,x≠x_0$, since we also have $F(x_0 )=G(x_0 )+C$ valid, the conclusion is true.
Alternatively, we can use the second Fundamental theorem of calculus, since
$(F-G)' (x)=0,\quad ∀x∈I$
the function $F-G:I→\mathbf R$ is an antiderivative of the function $g(x)=0,x∈I$. Since $g$ is constant, it is Riemann integrable, so choose some $x_0∈I$, then for any $x∈I,x≠x_0$, we can use the second Fundamental theorem of calculus on $[x_0,x]$ or $[x,x_0]$ and have
$∫_{[x_0,x]}g=0=(F-G)(x)-(F-G)(x_0)$
or
$∫_{[x,x_0]}g=0=(F-G)(x_0 )-(F-G)(x)$
In either case we have $(F-G)(x_0)=(F-G)(x)$, This means if we let $C=F(x_0)-G(x_0)$, we will have $F(x)=G(x)+C,∀x∈I,x≠x_0$. Since we also have $F(x_0)=G(x_0)+C$ valid, the conclusion is true.

Exercise 11.9.3

Since $f$ is monotone increasing, $f$ is Riemann integrable, thus if $f$ is continuous at $x_0$, then $F$ is differentiable at $x_0$ by the first Fundamental theorem of calculus.
Conversely we let $F$ is differentiable at $x_0$ and assume $f$ is not continuous at $x_0$, then we shall have $f(x_0-)<f(x_0+)$, let $η=(f(x_0-)+f(x_0+))/2$, then we can find $ϵ>0$, s.t.
$f(x_0-)<η-ϵ<η<η+ϵ<f(x_0+)$
As $f$ is monotone increasing, we know $f(x)<η-ϵ$ if $x<x_0$ and $f(x)>η+ϵ$ if $x>x_0$. Then for $y>x_0$, we have
$F(y)-F(x_0 )=∫_{[x_0,y]}f>∫_{[x_0,y]}(η+ϵ) =(η+ϵ)(y-x_0 )$
And for $y<x_0$, we have
$F(x_0 )-F(y)=∫_{[y,x_0]}f<∫_{[y,x_0]}(η-ϵ) =(η-ϵ)(x_0-y)$
So we have
$\lim_{y→x_0+}\frac{F(y)-F(x_0)}{y-x_0}≥η+ϵ>η \\ \lim_{y→x_0-}\frac{F(y)-F(x_0)}{y-x_0}=\lim_{y→x_0-}\frac{F(x_0 )-F(y)}{x_0-y}≤η-ϵ<η$
Thus $F' (x_0+)>F' (x_0-)$, a contradiction to $F$ being differentiable at $x_0$.