Let $[a,b]$ be a real bounded interval and let $\mathfrak{P}([a,b])$ be the set of all partitions of $[a,b]$, i.e. $\mathfrak{P}([a,b])=\{\{t_0,\dotsc,t_n\}:a=t_0<t_1<\dotso<t_n=b\}$. Define:
$$\mathrm{Var}_f[a,b]:=\sup_{\{t_0,\dotsc,t_n\}\in\mathfrak{P}([a,b])}\sum_{i=1}^n|f(t_i)-f(t_{i-1})|,$$
for an arbitrary $f:[a,b]\to\mathbb{R}$, to be the total variation of $f$ over $[a,b]$. Let $\mathrm{BV}([a,b])$ be the space of all functions $f$ such that $\mathrm{Var}_f[a,b]$ is finite. Such functions are called bounded variation functions (whence the name BV for the space), and it is known that the following hold.
Fact 1
$f\in\mathrm{BV}([a,b])\implies f=f_1-f_2$ with $f_i$ being bounded nondecreasing functions. In particular, since nondecreasing functions are a.e. differentiable, a BV function is always a.e. differentiable.
Fact 2
If $f$ is a bounded nondecreasing function, we have:
$$f(x)-f(y)\geq\int_x^yf'(t)\mathrm{d}t,$$
for any $x\geq y$.
A proof of these should be found here.
Consider the definition in the title:
$$\|f\|:=\|f\|_{L^1}+\|f'\|_{L^1},$$
where $f'$ is the a.e. derivative of $f$ which exists by the above Fact.
If there is a single point $c\in[a,b]$ where $|f(c)|<\infty$ and $f$ is BV, then $f$ is bounded, since, for $d\in[a,b]$, we have:
\begin{align*} |f(b)-f(d)|+|f(d)-f(c)|+|f(c)-f(a)|<\infty && \text{if }d\geq c \\ |f(b)-f(c)|+|f(c)-f(d)|+|f(d)-f(a)|<\infty && \text{if }d\leq c, \end{align*}
and hence $|f(c)-f(d)|<\infty$, in fact $|f(c)-f(d)|<\mathrm{Var}_f[a,b]$, which yields $\|f\|_\infty\leq|f(c)|+\mathrm{Var}_f[a,b]$, where $\|\cdot\|_\infty$ is the sup norm. But since constants are integrable on bouded intervals, we conclude that if $f$ is BV on $[a,b]$ and $[a,b]$ is bouded, then $f\in L^1([a,b])$.
If $f$ is BV, consider $f_1,f_2$ as in Fact 1. By Fact 2, $f_1(b)-f_1(a)\geq\int_a^bf_1'(x)\mathrm{d}x$, and similarly for $f_2$. Hence $f_1',f_2'$ are integrable. Since $\|f'\|_{L^1}\leq\|f_1'\|_{L^1}+\|f_2'\|_{L^1}<\infty$, we have $f'\in L^1([a,b])$.
So we have that the above definition yields a finite quantity. Now, this is evidently not a norm on BV, since there is the usual problem of a.e. equality. But suppose we take the quotient space of BV w.r.t. identifying all a.e. equal functions. Then this should give us a norm. Do we get a Banach space for the quotient? Take $f_n$ which is Cauchy. This easily implies $f_n,f_n'$ are Cauchy in $L^1$, hence converge to some $f,g$: $f_n\to f,f_n'\to g$ in $L^1$. How do I prove that $f'=g$?