Folland Chapter 3: Bounded Variation Difficulty Understanding Folland Proves in Lemma 3.26 that if $F \in BV$, then $T_F+F$ and $T_F-F$ are increasing, where: 
I can sort of follow the proof in Folland, but I don't understand why $T_F+F$ and $T_F-F$ are increasing for a BV function visually or graphically.  Say $sin(x)$ on $[0,2 \pi]$ is a BV function.  I know that $T_F$ is increasing, but why are $T_F +sin(x)$ and $T_F-sin(x)$ increasing?  I can't visualize why such is true.  A picture would really help me understand.  Thanks. I just need more visual help for understanding the definition of total variation. 
 A: For a function that is both $BV$ and Riemann integrable, $T_F(x)$ is equivalent to
$$T_F(x) = \int_{x_0}^x |F'(t)|\;dt,$$
the total absolute change in the function on the interval $[x_0, x]$.
Given that $F(x) = \sin(x)$ satisfies these, then considering it on the interval $[0,2\pi]$, we get a few cases:
On $[0,\tfrac{\pi}{2}]$
$$T_{\sin}(x) = \int_0^x |\cos(t)|\;dt = \big[\sin(t)\big]_0^x = \sin(x)$$
On $[\tfrac{\pi}{2}, \tfrac{3\pi}{2}]$
\begin{align}
T_{\sin}(x)
&= \int_0^x |\cos(t)|\;dt
= T_{\sin}(\tfrac{\pi}{2})
+ \int_{\tfrac{\pi}{2}}^x (-\cos(t))\;dt\\
&= 1 + \big[-\sin(t)\big]_{\tfrac{\pi}{2}}^x = 2 - \sin(x)
\end{align}
On $[\tfrac{3\pi}{2},2\pi]$
\begin{align}
T_{\sin}(x)
&= \int_0^x |\cos(t)|\;dt
= T_{\sin}(\tfrac{3\pi}{2})
+ \int_{\tfrac{3\pi}{2}}^x \cos(t)\;dt\\
&= 3 + \big[\sin(t)\big]_{\tfrac{3\pi}{2}}^x
= 4+\sin(x)
\end{align}
You can then readily see that $T_{\sin}(x) \pm \sin(x)$ is non-decreasing on each of these subinterval.
Visualizing It
The special case of Riemann differentiable functions is quite easy to understand:
\begin{align}
T_F(x) \pm F(x)
&= \int_0^x |F'(t)|\;dt \pm \left(\int_0^x F'(t)\;dt + F(0)\right)\\
&= \int_0^x \big(|F'(t)| \pm F'(t)\big) \;dt \pm F(0)
\end{align}
Clearly $|F'(t)| \pm F'(t) \geq 0$, so the corresponding integral is non-decreasing in $x$.
Note that this also shows that, for a particular $x$, at most one of $T_F(x) + F(x)$ and $T_F(x) - F(x)$ is strictly increasing.
A: I hope this helps

Lemma 3.26 - If $F\in BV$ is real-valued, then $T_F + F$ and $T_F - F$ are increasing.

Proof:
Suppose $x < y$ and $\epsilon > 0$. Choose $x_0 < x_1 < \ldots < x_n = x$ such that $$\sum_{1}^{n}[F(x_j) - F(x_{j-1}] > T_F(x) - \epsilon$$
Note that, $$|F(y) - F(x)| + \sum_{1}^{n}[F(x_j) - F(x_{j-1}] \leq T_F(y)$$ and $$F(y) = [F(y) - F(x)] + F(x)$$ So we have
\begin{align*}
T_F(y) \pm F(y) &\geq \sum_{1}^{n} [F(x_j) - F(x_{j-1}] + |F(y) - F(x)| \pm [F(y) - F(x)] \pm F(x)\\
&\geq T_F(x) - \epsilon \pm F(x)
\end{align*}
Since $\epsilon$ is arbitrary we are done.
A: Adding to adfriedman's answer, the special case when $F$ is piecewise monotone also gives a lot of intuition. I will use the definition of total variation for functions that are also defined on a finite interval $[a,b],$ in which case you simply restrict your partitions to lie in said interval.

Claim 1: If $F : [0,1] \rightarrow \mathbb R$ is non-decreasing, then $T_F = F - F(0).$

Proof of claim: For any $0 \leq x_0 < x_1 < \dots < x_n = x,$ then for each $i$ we have,
$$ F(x_i) - F(x_{i-1}) = |F(x_i) - F(x_{i-1})| \geq 0.$$
Hence summing over these gives,
$$ \sum_{i=1}^n |F(x_i)-F(x_{i-1})| = \sum_{i=1}^n F(x_i) - \sum_{i=1}^n F(x_{i-1}) = F(x) - F(x_0). $$
Taking the supremum over all such partitions and noting that $F(0) \geq F(x_0)$ for all $x_0 \in [0,1],$ we get $T_F(x) = F(x) - F(0)$ as required.
It is also easy to see that if $F$ is non-increasing, then $T_F = F(0) - F$ (notice the sign change comes from the absolute value, and we now maximise $F$ over $x_0 \in [0,1]$). So up to a constant, if $F$ is monotone then $T_F$ equals $F$ but with a sign change to make it increasing.
The next step is to notice that $T_F$ has a locality-type property.

Claim 2: If $F : [0,1] \rightarrow \mathbb R$ has bounded total variation, then for any $0<x <y < 1$ we have,
  $$ T_F(y) = T_F(x) + T_{F|_{[x,y]}}(y). $$

By this rather ungodly piece of notation I mean,
$$ T_{F|_{[x,y]}}(z) = \sup\left\{ \sum_{i=1}^n |F(x_i)-F(x_{i-1})|\ \middle|\ x \leq x_0 \leq \dots \leq x_n = z \right\}. $$
I'll leave this as a technical exercise; working from first principles you can show that both sides are less than or equal to each other up to an arbitrary $\varepsilon>0.$
Now we put these things together. Suppose $F : [0,1] \rightarrow \mathbb R$ is piecewise monotone, so there exists $0 = t_0 < t_1 < t_2 < \dots < t_k = 1$ such that $F$ is monotone on each $[t_i,t_{i+1}].$ We will further assume $F$ is non-decreasing on each $[t_{2i},t_{2i+1}]$ and non-decreasing on each $[t_{2i+1},t_{2i+2}]$; that is $F$ is initially increasing, and switches between increasing and decreasing on each interval.

Claim 3: If $x \in [t_{2i},t_{2i+1}]$ we have
  \begin{align*} 
  T_F(x) &= (f(x) - f(t_{2i})) - (f(t_{2i}) - f(t_{2i-1})) + (f(t_{2i-1})-f(t_{2i-2}) + \dots \\
  &= f(x) - 2f(t_{2i}) + 2f(t_{2i-1}) - 2f(t_{2i-2}) + \dots - f(t_0),
\end{align*}
  and if $x \in [t_{2i+1},t_{2i+2}]$ we have
  \begin{align*} 
  T_F(x) &= -(f(x) - f(t_{2i+1})) + (f(t_{2i+1}) - f(t_{2i})) - (f(t_{2i})-f(t_{2i-1}) + \dots \\
  &= -f(x) + 2f(t_{2i+1}) - 2f(t_{2i}) + 2f(t_{2i-1}) - \dots - f(t_0),
\end{align*}

The proof is straightforward given the above two claims; note that if $x \in [t_{2i},t_{2i+1}]$ we have
$$ T_F(x) = T_F(t_{2i}) + f(x) - f(t_{2i}),$$
and if $x \in [t_{2i+1},t_{2i+2}]$ we have
$$ T_F(x) = T_F(t_{2i+1}) - f(x) + f(t_{2i+1}). $$
So we just apply this iteratively.

So what does this mean geometrically? If you try sketching $T_F$ for such a curve $F,$ you'll find you (up to adding constants) you follow $F$ if it is increasing, and reflect it if it is decreasing. For the case $F(x) = \sin x$ on $[0,2\pi]$ have the rough sketch:

The dotted part shows what $y = \sin x$ and $y = 2 + \sin x$ would normally look like if it wasn't reflected. In this case we have four different regions on $[0,2\pi]$ where $F$ alternates between increasing and decreasing.

As a final comment, this is consistent with the formula,
$$ T_F(x) = \int_0^x |F'(x)|\,\mathrm{d}x, $$
which holds for any $F : [0,1] \rightarrow \mathbb R$ absolutely continuous. If we additionally assume that $F'$ exists everywhere, is continuous and changes sign finitely many times on $[0,1],$ then it is piecewise monotone. In this case $|F'| = \pm F'$ on these subintervals, and you can arrive at the above formula using the fundamental theorem of calculus.
