L1 norm less than BV norm I will appreciate any hint on this

Prove that if $f$ is a function of bounded variation then
  $\|f'\|_{L_1} \leq \|f\|_{BV}$.

When $f$ is differentiable just by the Fundamental Theorem of Calculus we can get that the $L_1$ norm equals the total variation and then of course I would be less than its $BV-$norm but the general case I don't know how to proceed (considering that $f'$ always exists a.e. since $f$ is of bounded variation).
Thank you

I assume you know that a function $F$ of bounded variation on $[a,b]$ can be written as the difference of monotone functions $F=F_{+}-F_{-}$. And I'll assume you also know that $F_{\pm}$ have derivatives a.e., and these derivatives are non-negative where they exist. One pair of functions $F_{\pm}$ can be defined using the variation function $V_{a}^{x}(F)$:
$$
                F_{+}(x)=(V_{a}^{x}(F)+F(x))/2,\\ F_{-}(x)=(V_{a}^{x}(F)-F(x))/2.
$$
These are convenient choices for the monotone decomposition because $F_{-}(x)+F_{+}(x)=V_{a}^{x}(F)$.
Now assume that $f$ is non-decreasing on $[a,b]$, and extend $f$ for $x > b$ to be $f(b)$. Define
$$
              g_{n}(x)=n\{f(x+1/n)-f(x)\},\;\;\; n = 1,2,3,\cdots,\;\; x\in [a,b].
$$
The function $g_{n}$ is non-negative and converges pointwise a.e. to $f'(x)$ as $n\rightarrow\infty$. By Fatou's lemma,
$$
\begin{align}
    \int_{a}^{b}f'(x)dx
        & \le \liminf_{n}\int_{a}^{b}g_{n}(x)dx \\
        & = \liminf_{n} \int_{a}^{b}n(f(x+1/n)-f(x))dx \\
        & = \liminf_{n} \left(n\int_{b}^{b+1/n}fdx-n\int_{a}^{a+1/n}fdx\right) \\
        & = \liminf_{n} \left(f(b)-n\int_{a}^{a+1/n}fdx\right) \\
        & \le \liminf_{n} \left(f(b)-n\int_{a}^{a+1/n}f(a)dx\right) \\
        & = f(b)-f(a).
\end{align}
$$
Therefore,
$$
\begin{align}
     \int_{a}^{b}|F'|dx & \le \int_{a}^{b}\{ F_{+}'+F_{-}'\}dx \\
                & \le F_{+}(b)-F_{+}(a) + F_{-}(b)-F_{-}(a) \\
                & = V_{a}^{b}(F).
\end{align}
$$
