Show that $\int_a^b |f'| \leq TV(f)$, and show that this is equality $iff$ $f$ is absolutely continuous. Let  $f$ be of bounded variation on $[a,b]$ and define $v(x) = TV(f_{[a,x]}$ for all $x \in [a,b]$.
Show that $\int_a^b |f'| \leq TV(f)$, and show that this is equality $iff$ $f$ is absolutely continuous.
$proof$:
proof:
Let $P$ be the trivial partition of $[a,x]$ for all $x \in (a,b]$. Then we have that:
$v(x+h) \geq V(f,P) = |f(x+h)-f(a)|$ 
$v(x) \geq V(f,P) = |f(x)-f(a)|$
And thus we have:
$v'(x) = \lim_{h \rightarrow 0^+} \frac{v(x+h)-v(x)}{h}$ 
$\geq \lim_{h \rightarrow 0^+} \frac{|f(x+h)-f(a) - f(x) + f(a)|}{h}$
$\geq \lim_{h \rightarrow 0^+} \frac{|f(x+h)-f(x)|}{h}$
$=f'(x)$
Furthermore, $f$ can be written as a sum of monotonic function, and $v$ is a monotonic function, thus both are differential almost everywhere, and so the previous inequality holds almost everywhere. 
Furthermore, by the monotonicty of the integral we have:
$\int_a^b |f'| \leq \int_a^b v' = TV(f,[a,b]) - TV(f,[a,a]) = TV(f,[a,b])$
$\\$

Is this correct? Furthermore, how do I prove that equality holds if $f$ is absolutely continuous? Thanks!!!
 A: For any partition $a=t_{0}<t_{1}<\cdots<t_{n}=b$, we have 
\begin{align*}
\sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|&=\sum_{i=1}^{n}\left|\int_{t_{i-1}}^{t_{i}}f'(x)dx\right|\\
&\leq\sum_{i=1}^{n}\int_{t_{i-1}}^{t_{i}}|f'(x)|dx\\
&=\int_{a}^{b}|f'(x)|dx,
\end{align*}
so $TV(f)\leq\displaystyle\int_{a}^{b}|f'(x)|dx$ is established.
Next we denote $\text{sgn}f'(x)=1$ for $f'(x)\geq 0$ and $\text{sgn}f'(x)=0$ for $f'(x)<0$, then $\text{sgn}f'\in L^{1}[a,b]$ and hence can be approximated by step functions $\varphi$ on $[a,b]$ in $L^{1}$ sense.
Note that for 
\begin{align*}
\varphi(x)=\sum_{i=1}^{n}a_{i}\chi_{(t_{i-1},t_{i})}(x),
\end{align*}
we have
\begin{align*}
\max(\min(\varphi(x),1),-1)=\sum_{i=1}^{n}\max(\min(a_{i},1),-1)\chi_{(t_{i-1},t_{i})}(x),
\end{align*}
and 
\begin{align*}
|\max(\min(\varphi(x),1),-1)-\text{sgn}f'(x)|\leq|\varphi(x)-\text{sgn}f'(x)|,
\end{align*}
so we can assume that all $a_{i}$ are such that $|a_{i}|\leq 1$.
As a consequence,
\begin{align*}
\int_{a}^{b}|f'(x)|dx&=\int_{a}^{b}f'(x)\text{sgn}f'(x)dx\\
&=\int_{a}^{b}f'(x)(\text{sgn}f'(x)-\varphi(x))dx+\int_{a}^{b}f'(x)\varphi(x)dx.
\end{align*}
We can use Lebesgue Dominated Convergence Theorem to make the term
\begin{align*}
\int_{a}^{b}f'(x)(\text{sgn}f'(x)-\varphi(x))dx
\end{align*}
to be arbitrarily small.
Now we estimate that
\begin{align*}
\left|\int_{a}^{b}f'(x)\varphi(x)dx\right|&\leq\sum_{i=1}^{n}|a_{i}|\left|\int_{a}^{b}f'(x)\chi_{(t_{i-1},t_{i})}(x)dx\right|\\
&\leq\sum_{i=1}^{n}\left|\int_{t_{i-1}}^{t_{i}}f'(x)dx\right|\\
&=\sum_{i=1}^{n}\left|f(t_{i})-f(t_{i-1})\right|\\
&\leq TV(f),
\end{align*}
we are done.
