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.