it is well known that if a function $f:[0,T]\to\mathbb{R}$ satisfies the inequality 
$$\vert f(t)-f(s)\vert\leq \int\_s^t{m(r) dr},$$
for $s<t$ and some $m\in L^1([0,T])$ then $f$ is absolutely continuous.
On the other hand, if the function satisfies the inequality
$$\vert f(t)-f(s)\vert\leq (g(s)+g(t))\vert t-s\vert,$$
for some $g\in L^1([0,T])$, then $f\in W^{1,1}([0,T])$, indeed both conclusions are the same. 


My question is: if the inequality 
$$\vert f(t)-f(s)\vert\leq (g(s)+g(t))\vert t-s\vert+\int\_s^t{m(r) dr}$$
is valid, do we have the same conclusion? I think that it is true, but I don't know how to show it.



Absolutely continuous functions
I think the answer to this question is ``yes" if in the last inequality you assume that $g\in L^1$ and $m\in L^1$. There are two simple ways to see this:


1. In one dimensional case, $f\in W^{1,1}$ if and only if $f$ is absolutely continuous. Both the first inequality and the second inequality implies $f$ is absolutely continuous, essentially just by definition using absolutely continuity of integral (when you fix a division of the interval, the right-hand side of the second inequality can be seen as a discrete approximation of the integral over the interval).
2. The second convenient way is to use the “differential quotient", namely for $h\in (0,T)$, consider 
\begin{equation\*}
\Delta\_hf(t)=\frac{|f(t+h)-f(t)|}{h}, \quad t\in [0,T-h].
\end{equation\*}
Then the Sobolev space $W^{1‚1}$ can be characterized as 
$$W^{1,1}([0,T])=\Big\{f\in L^1:\sup\_{0<h<T}\int\_0^{T-h}\Delta\_hf(t)dt<\infty\Big\}.$$
It is straightforward to verify this from the last inequality you have.


