Why does the equality $\sum_i^k |f(x_i)-f(x_{i-1})|=\lim_{n\to\infty}\sum_i^k |f_n(x_i)-f_n(x_{i-1})|$ holds?
Shouldn't the limit be inside the sum? And the by the convergence of $f_n$ to f, the equality holds. Is it correct?
Theorem: Let ${f_k}$ be a sequence of bounded variation on $[a,b]$ such that for each $x\in [a,b]$ $f(x)=\lim_{k\to \infty}f_k(x)$. If there exist $K>0$ such that $V(f_n,[a,b])\le K$ for all $n\in \mathbb N$, then $f$ is bounded variation on $[a,b]$.
Proof:
Let $a=x_0<\cdots<x_k=b$ be a partition. Then $$\sum_i^k |f(x_i)-f(x_{i-1})|=\lim_{n\to\infty}\sum_i^k |f_n(x_i)-f_n(x_{i-1})|.$$ Since the variation is the supremum you have $$\sum_i^k |f_n(x_i)-f_n(x_{i-1})|\le V(f_n,[a,b])\le K$$ for each $n$ and so $$\lim_{n\to\infty}\sum_i^k |f_n(x_i)-f_n(x_{i-1})|\le K.$$ This shows that $$\sum_i^k |f(x_i)-f(x_{i-1})|\le K$$ for every partition $a=x_0<\cdots<x_k=b$ of $[a,b]$. Hence, $K$ is an upper bound and so by the definition of supremum of a set you get $V(f,[a,b])\le K$.