I am trying to work through Exercise 2.29 of Leoni's A First Course in Sobolev Spaces, which gives an alternative proof of the theorem in the title in the real-valued case. The exercise is broken up into multiple parts, and it is the last part I am struggling to complete.
For $x \in [a,b]$, let $V(x) = \text{Var}_{[a,x]} u$.
For $n \in \mathbb{N}$, let $P_n$ be a partition $a = x_0 < x_1 < \dots < x_{N_n} = b$ such that $\text{Var}_{[a,b],P} u = \text{Var}_{[a,b]} u - \frac{1}{2^n}$. (The subscript of $P$ means I am taking the variation with respect to a given partition, while no subscript indicates the supremum over all partitions).
Inductively define the function $v_n$ as follows:
If $u(x_0) \leq u(x_1)$, define $v_n(x) = u(x) - u(x_0)$ on $[x_0, x_1]$. Otherwise if $u(x_0) > u(x_1)$, define $v_n(x) = u(x_0) - u(x)$.
For $1 \leq k < N_n$, suppose $v_n$ is defined on $[x_0,x_k]$. If $u(x_k) \leq u(x_{k+1})$, set $v_n(x) = u(x) + v(x_k) - u(x_k)$ on $(x_k,x_{k+1}]$. Otherwise if $u(x_k) > u(x_{k+1})$, set $v(x) = -u(x) + v(x_k) + u(x_k)$.
We have the following results:
i) On each subinterval, either $u + v_n$ or $u - v_n$ is constant.
ii) $V(x) = Var_{[a,x]} v_n(x)$ and $v_n(b) = Var_{[a,b],P} u$.
iii) The functions $V - v_n$ is increasing, with $|v_n'(x)| = |u'(x)|$ almost everywhere.
What remains to show is that $|V'(x)| = |u'(x)|$ almost everywhere. My strategy is to use the inequality $||V'| - |v_n'|| \leq |(V - v_n)'|$. However, while we get uniform convergence of $v_n$ to $V$, we don't know anything about uniform convergence on the derivatives. The fact that all the derivatives of $V - v_n$ seems like it could lead somewhere, but I'm not sure how to use this inequality. I think I'm missing something relatively simple.