For $n \geq 1$ let $f, f_n : [0, 1] \to \mathbb{R}$ be monotone nonincreasing functions. Suppose that $f_n \nearrow f$ pointwise monotonically as $n \to \infty$.
Is then $\mathrm{TV}(f - f_n) \to 0$ as $n \to \infty$, where $\mathrm{TV}$ is the total variation?
Note that assuming $f_n$ to be merely of bounded variation is not enough: If $f_n$ is the negative of the indicator function of $(0, 1/n)$ then $f_n \nearrow 0$ pointwise monotonically, but $\mathrm{TV}(f_n) = 2$ for each $n$.
If the answer is "yes", I would appreciate a proof without the Lebesgue monotone convergence theorem.
Edit. $f := 1_{[0, 1)}$ and $f_n := 1_{[0, 1 - 1/n)}$ is a counterexample.