Real Analysis, Folland 3.25 Exampes Functions of Bounded Variation Background Information:
Taking $a = -\infty$ and considering the total variation as a function of $b$. To with $F:\mathbb{R}\rightarrow \mathbb{C}$ and $x\in\mathbb{R}$, we define $$T_F(x) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1}|:n\in\mathbb{N},-\infty < x_0 < \ldots < x_n = x\}$$ $T_F$ is called the total variation of $F$.
We observe that the sums in the definition of $T_F$ are made bigger if the additional subdivision points $x_j$ are added. Hence, if $a < b$, the deinition of $T_F(b)$ is unaffected if we assume that $a$ is always one of the subdivision points. It follows that $$T_F(b) - T_F(a) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1}|:n\in\mathbb{N},a = x_0 < \ldots < x_n = b\}$$
Thus $T_F$ is an increasing function with values in $[0,\infty]$. If $T_F(\infty) = \lim_{x\rightarrow \infty}T_F(x)$ is finite, we say that $F$ is of bounded variation on $\mathbb{R}$, and we denote the space of all such $F$ by $BV$.
More generally, the supremum on the right side is called the total variation of $F$ on $[a,b]$. It depends only on the values of $F$ on $[a,b]$, so we may define $BV([a,b])$ to be the set of all functions on $[a,b]$ whose total variation on $[a,b]$ is finite.
Question:
I have been working on these examples today and I am sort of stuck at for c,d,and e. I will provide the proofs I made for a and b just because.

3.25 Examples:
a.) If $F:\mathbb{R}\rightarrow \mathbb{R}$ is bounded and increasing, then $F\in BV$ (in fact, $T_F(x) = F(x) - F(-\infty)$).
b.) If $F,G\in BV$ and $a,b\in\mathbb{C}$, then $aF + bG\in BV$.
c.) If $F$ is differentiable on $\mathbb{R}$ and $F'$ is bounded, then $F\in BV([a,b])$ for $-\infty < a < b < \infty$(by MVT).
d.) If $F(x) = \sin x$, then $F\in BV([a,b])$ for $-\infty < a < b < \infty$, but $F\notin BV$.
e.) If $F(x) = x\sin(x^-1)$ for $x\neq 0$ and $F(0) = 0$, then $F\notin BV([a,b])$ for $a\leq 0 < b$ or $a < 0\leq b$.

Proof a.) - If $F:\mathbb{R} \rightarrow\mathbb{R}$ and $x\in\mathbb{R}$, then $$T_F(x) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1}|):n\in\mathbb{N}, -\infty < x_0 < \ldots < x_n = x \}$$
Since $F$ is increasing, the intervals $(F(x^-1),F(x^+) (x\in\mathbb{R})$ are disjoint, and for $|x| < N$ they lie in the interval $(F(-N),F(N))$. Hence $$\sum_{|x| < N}|F(x^{+} - F(x^{-1}| \leq F(N) - F(-N) < \infty$$
so now we know $F$ is finite, therefore the total variation of $F$ is also finite, therefore $$\lim\limits_{x\rightarrow \infty}T_F(x) = T_F(\infty)\Rightarrow F\in BV$$
Proof b.) - Since $F,G\in BV \Rightarrow \lim_{x\rightarrow \infty}T_F(x) = T_F(\infty)$ and $\lim_{x\rightarrow \infty}T_G(x) = T_G(\infty)$ since $a,b$ are just scalars it follows that $$\lim_{x\rightarrow \infty} aT_F(x) + bT_G(x) = aT_F(\infty) + bT_G(\infty) \Rightarrow aF + bG\in BV$$
Questions for c.) Since we know $F$ is differentiable on $\mathbb{R}$ and $F'$ is bounded we can conclude that $F$ is uniformly continuous. Therefore, it seems to me to complete this proof we have to assume that $F$ is uniformly continuous on $[a,b]$ and differentiable on $(a,b)$ then we can apply the Mean Value theorem. But I am not sure how we show finiteness to conclude that $F\in BV([a,b])$.
Questions for d.) Since $F(x) = sin(x)$ we know that $F$ is then continuous and I guess based on c.) I need to show $F\in BV([a,b])$ but I am not sure how to show that $F\notin BV$
I am still thinking more about e.) and d.) I will re-edit if I get any further.
 A: For (c), let $x,y \in [a,b]$ with $x\neq y$. By the MVT, there is some $z$ between $x$ and $y$ with $|F(x) - F(y)| =|F'(z)||x-y| \leq M|z-y|$ where $M$ is a bound for $|F'|$ on $[a,b]$. (Note, this shows that $F$ is Lipschitz, which is even stronger than uniform continuity.) Now let $\{x_0,x_1,\ldots,x_n\}$ be any partition of $[a,b]$. We have
$$\begin{aligned}
\sum_{k=1}^{n}|F(x_k)-F(x_{k-1})| &\leq M\sum_{k=1}^{n}|x_k - x_{k-1}| \\
&= \sum_{k=1}^{n}(x_k - x_{k-1}) \\
& = M(x_n - x_0) \\
&= M(b-a) \\ \end{aligned}$$ so the variation of $F$ on $[a,b]$ is bounded by $M(b-a)$.
For (d), we have $|F'(x)| = |\cos(x)| \leq 1$, so we can apply (c) with $M=1$ to conclude that $F$ has bounded variation on any finite-length interval $[a,b]$. But $F$ does not have bounded variation on all of $\mathbb R$. To see this, consider the points $x_k = (2k+1)\pi/2$ for $k=0,1,\ldots N$. The corresponding values of $F$ at these points are $1,-1,1,-1,\ldots$, so each $|F(x_k) - F(x_{k-1})|$ is $2$. Therefore the variation of $F$ on $\mathbb{R}$ is at least $2N$. As $N$ can be as large as we like, this means that $F$ does not have bounded variation on $\mathbb R$.
For (e), consider the points $x_k = 2/((2k+1)\pi)$ for $k=0,1,\ldots,N$. For $k \geq 1$ we have
$$\begin{aligned}
|F(x_k) - F(x_{k-1})|
&= \frac{2}{\pi}\left| \frac{\sin\left(\frac{\pi(2k+1)}{2}\right)}{2k+1} -  \frac{\sin\left(\frac{\pi(2k-1)}{2}\right)}{2k-1} \right| \\
&= \frac{2}{\pi}\left| \frac{\pm 1}{2k+1} - \frac{\mp 1}{2k-1}\right| \\
&= \frac{2}{\pi}\left|\frac{1}{2k+1} + \frac{1}{2k-1}\right| \\
&\geq \frac{2}{\pi}\frac{1}{2k+1}
\end{aligned}$$
so
$$\sum_{k=1}^{N}|F(x_k) - F(x_{k-1})|$$
becomes arbitrarily large as $N$ increases.
A: 
3.25 Examples: 
a.) If $F:\mathbb{R}\rightarrow \mathbb{R}$ is bounded and increasing, then $F\in BV$ (in fact, $T_F(x) = F(x) - F(-\infty)$).
b.) If $F,G\in BV$ and $a,b\in\mathbb{C}$, then $aF + bG\in BV$.
c.) If $F$ is differentiable on $\mathbb{R}$ and $F'$ is bounded, then $F\in BV([a,b])$ for $-\infty < a < b < \infty$(by MVT).
d.) If $F(x) = \sin x$, then $F\in BV([a,b])$ for $-\infty < a < b < \infty$, but $F\notin BV$.
e.) If $F(x) = x\sin(x^{-1})$ for $x\neq 0$ and $F(0) = 0$, then $F\notin BV([a,b])$ for $a\leq 0 < b$ or $a < 0\leq b$.

Proof a.) - If $F:\mathbb{R} \rightarrow\mathbb{R}$ and $x\in\mathbb{R}$, then 
$$T_F(x) = \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N}, -\infty < x_0 < \ldots < x_n = x \}$$
Since $F$ is increasing, for $j\geq 1$ we have $|F(x_j) - F(x_{j-1})|= F(x_j) - F(x_{j-1})$. So we have 
\begin{align*}
T_F(x) &= \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1}|:n\in\mathbb{N}, -\infty < x_0 < \ldots < x_n = x \} \\
& =\sup\{\sum_{1}^{n}(F(x_j) - F(x_{j-1}):n\in\mathbb{N}, -\infty < x_0 < \ldots < x_n = x \} \\
& =\sup\{ F(x) -F(x_0): -\infty < x_0 <  x \}
\end{align*}
because 
$\sum_{1}^{n}(F(x_j) - F(x_{j-1})=F(x_n)-F(x_0)$ and $x_n=x$. 
Since $F$ is increasing, we have 
$$T_F(x)= \sup\{ F(x) -F(x_0): -\infty < x_0 <  x \}= F(x) -\lim_{y\to -\infty}F(y)$$
Notating $\lim_{y\to -\infty}F(y)$ as $F(-\infty)$, we have 
$$T_F(x)= F(x) -\lim_{y\to -\infty}F(y)= F(x) - F(-\infty)$$
Finally 
$$T_F(+\infty)= \lim_{x \to +\infty} F(x) - F(-\infty)$$
Since $F$ is bounded, we have $\lim_{x \to +\infty} F(x)<+\infty$ and $F(-\infty)> -\infty$, so $T_F(+\infty)<+\infty$, which means that $F \in BV$.
Proof b.) - Since $F,G\in BV \Rightarrow \lim_{x\rightarrow \infty}T_F(x) = T_F(\infty)<+\infty$ and $\lim_{x\rightarrow \infty}T_G(x) = T_G(\infty)<+\infty$ since $a,b$ are just scalars it follows that $$\lim_{x\rightarrow \infty} aT_F(x) + bT_G(x) = aT_F(\infty) + bT_G(\infty)<+\infty \Rightarrow aF + bG\in BV$$
Proof c.) - $F$ is differentiable on $\mathbb{R}$. Then, by the Mean Value Theorem (for differentiable functions), for any $p, q \in \mathbb{R}$, $p<q$, there is $r\in [p,q]$ such that 
$$ F(p) - F(q) = F'(r) (p-q)$$
In particular, we have 
$$ |F(p) - F(q)| = |F'(r)|(p-q)$$
Now, since $F'$ is bounded, we know that there is $C>0$ such that, for all $x\in \mathbb{R}$, $|F'(r)|\leq C$. So we have, for all  $p, q \in \mathbb{R}$, $p<q$, 
$$ |F(p) - F(q)| = C(p-q)$$
So, we have for any $[a,b]\subset \mathbb{R}$ 
\begin{align*}
T_F(b)-T_F(a) &= \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N}, a = x_0 < \ldots < x_n = b \} \\
& \leq\sup\{\sum_{1}^{n}C(x_j - x_{j-1}):n\in\mathbb{N}, a = x_0 < \ldots < x_n = b \} \\
& =C\sup\{\sum_{1}^{n}(x_j - x_{j-1}):n\in\mathbb{N}, a = x_0 < \ldots < x_n = b \} \\
& =C\sup\{ (b-a) \}\\
&= C(b-a) <+\infty
\end{align*}
So $F\in BV([a,b])$ for $-\infty < a < b < \infty$.
Proof d.) - Let $F(x) = \sin x$. Then $F$ is differentiable on $\mathbb{R}$ and $F'$ is bounded. In fact, for all $x \in \mathbb{R}$,
$F'(x)=\cos x$ and $|F'(x)|=|\cos x| \leq 1$. So, by the item c.) above,  $F\in BV([a,b])$ for $-\infty < a < b < \infty$.
Now, let us prove $F\notin BV$. In fact let us prove a little more. let us prove that for any $ x \in \mathbb{R}$, $T_F(x)=+\infty$.
Given any $ x \in \mathbb{R}$, and any $m\in \mathbb{N}$, $m>0$, let $k\in \mathbb{Z}$ such that 
$k\pi +\frac{\pi}{2} \leq x$ (such $k$ always exist). 
\begin{align*}
T_F(x) & = \sup\left\{\sum_{1}^{n}|F(x_j) - F(x_{j-1}|:n\in\mathbb{N}, -\infty < x_0 < \ldots < x_m = x \right\} \\
& \geq \sup\left\{\sum_{1}^{m}|F(x_j) - F(x_{j-1}|:  x_0=(k-m)\pi +\frac{\pi}{2} <x_1=(k-m+1)\pi +\frac{\pi}{2}\\< \ldots < x_m =k\pi +\frac{\pi}{2} \right\} \\
& =\sup\left\{\sum_{1}^{m}|\sin(x_j) - \sin(x_{j-1})|:  x_0=(k-m)\pi +\frac{\pi}{2} <x_1=(k-m+1)\pi +\frac{\pi}{2}\\< \ldots < x_m =k\pi +\frac{\pi}{2} \right\} \\
& =\sup\left\{2m :  x_0=(k-m)\pi +\frac{\pi}{2} <x_1=(k-m+1)\pi +\frac{\pi}{2} < \ldots < x_m =k\pi +\frac{\pi}{2} \right\} \\
& =2m  \\
\end{align*}
So, any $x \in \mathbb{R}$, and any $m\in \mathbb{N}$, $m>0$, we have  $T_F(x)\geq 2m$. So $T_F(x) =+\infty$. As a consequence, 
$$T_F(+\infty)=\lim_{x \to +\infty}T_F(x) =+\infty$$
Proof e.) - Case 1. Suppose  $a\leq 0 < b$. We will adapt the same sequence we have used in second part of d.). 
Let $k\in \mathbb{N}$ such that 
$(k\pi +\frac{\pi}{2})^{-1} \leq b$. Such $k$ always exists (note that it will be a fixed $k$ for the rest of this proof).
\begin{align*}
T_F(b)-T_F(a) &= \sup\{\sum_{1}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N}, a = x_0 < \ldots < x_n = b \} \\
& \geq \sup\{\sum_{1}^{n+1}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N}, 
a = x_0 < x_1=((k+n-1)\pi +\frac{\pi}{2})^{-1} <\ldots \\
&\phantom{\sup\{\sum_{1}^{n+1}|F(x_j) - F(x_{j-1})|:}< x_n=(k\pi +\frac{\pi}{2})^{-1} < x_{n+1} = b \}\\
& \geq \sup\{\sum_{2}^{n}|F(x_j) - F(x_{j-1})|:n\in\mathbb{N}, 
x_1=((k+n-1)\pi +\frac{\pi}{2})^{-1} <\ldots \\
&\phantom{\sup\{\sum_{1}^{n+1}|F(x_j) - F(x_{j-1})|:}< x_n=(k\pi +\frac{\pi}{2})^{-1}  \}
\end{align*}
For $j\in \{1,\ldots n\}$, $x_j= ((k+n-j)\pi +\frac{\pi}{2})^{-1}=\frac{2}{(2(k+n-j)+1)\pi}$ and we have 
$$F(x_j)=\frac{2}{(2(k+n-j)+1)\pi}\sin \frac{(2(k+n-j)+1)\pi}{2} = \frac{ 2(-1)^{k+n-j}}{(2(k+n-j)+1)\pi}$$
So, for $j\in \{2,\ldots n\}$,
\begin{align*}
|F(x_j) - F(x_{j-1})| & = \left |\frac{\pm 2}{(2(k+n-j)+1)\pi} -\frac{\mp 2}{(2(k+n-j+1)+1)\pi}\right | \\
& = \left |\frac{\pm 2}{(2(k+n-j)+1)\pi} +\frac{\pm 2}{(2(k+n-j+1)+1)\pi}\right | \\
& = \frac{ 2}{(2(k+n-j)+1)\pi} +\frac{ 2}{(2(k+n-j+1)+1)\pi} \\
& \geq \frac{ 2}{(2(k+n-j)+1)\pi} \\
& = \frac{ 2}{\pi} \frac{ 1}{(2(k+n-j)+1)} \\
\end{align*}
So 
$$\sum_{2}^{n}|F(x_j) - F(x_{j-1})|\geq \frac{ 2}{\pi} \sum_{j=2}^{n} \frac{ 1}{(2(k+n-j)+1)}=\frac{ 2}{\pi}\sum_{r=k}^{k+n-2} \frac{ 1}{(2r+1)}$$
So $\sum_{2}^{n}|F(x_j) - F(x_{j-1})|$ can be make arbitrarily large, as we make $n$ enough large. So $T_F(b)-T_F(a) =+\infty$, which means $F\notin BV([a,b])$
Case 2.  Suppose $a < 0\leq b$. It completely similar to the case 1, using the sequence $\{-(k\pi +\frac{\pi}{2})^{-1}\}_k$. 
