If $f$ is of bounded variation on $[a,b]$ then $\int_a^b |f'| \leq T_{a}^{b}(f)$ [duplicate]

I am trying to prove the following statement:

If $f$ is of bounded variation on $[a,b]$, then $\int_a^b |f'|dm \leq T_{a}^{b}(f)$

I am able to show this for $f'$ but not for $|f'|$ using the fact that $f$ can be written as the difference of two monotonically non-decreasing functions $h,g$:

$\int_a^b f' \leq \int_a^b g'-h' \leq g(b)-g(a)+h(a)-h(b) \leq |g(b)-g(a)+h(a)-h(b)|$

and since the latter is, by definition, the total variation of $f$ on the 'trivial' partition that consists only of the interval: $[a,b] $, then the supremum over all partitions must be larger, and so we obtain:

$\int_a^b f' \leq T_a^b(f)$

I cannot seem to extend this to $|f'|$. I tried using triangle-inequalities, but unless I am missing something, they're not very helpful here.

I saw an attempt to solve this using the following claim: $T_a^b(f) = T_a^b(g)+T_a^b(h)$ but it seems incorrect since I found a counter example to this: $f(x)=-x, g(x)=x, h(x)=2x$ on $[0,1]$.

Any guidance here would be much appreciated!

Setting $g=f^+$, $h=f^-$ we do get that:

$T_a^b(f) = T_a^b(g)+T_a^b(h)$

and therefore:

$\int_a^b |f'|dm \leq \int_a^bg' + \int_a^b h' \leq T_a^b(g)+T_a^b(h)=T_a^b(f)$

Thanks to Kavi Rama Murthy for the useful comment.