Total variation of a sequence of functions I have this problem, 
Let $\{f_n\}$ be a sequence of real valued functions on $[a,b]$ that converges at each point of $[a,b]$ to a function $f$. Then $T_a^b(f)≤\liminf T_a^b(f_n)$.
It is solved in, Pointwise convergent and total variation.
But I have a question about the answer of Nana (the last one) she states:

Hint:
Let $a=x_0\lt \ldots \lt x_n=b$ be a subdivision of $[a,b]$. Let $ε>0$. Then there is an $N$ such that
  $$|f_n(x_k) - f(x_k)|\lt \varepsilon /2,\qquad |f_n(x_{k-1}) - f(x_{k-1})| \lt \varepsilon /2$$
Then consider
$$\sum_{k=1}^n |f(x_k) - f(x_{k-1})|\leq  \sum_{k=1}^n |f(x_k) - f_n(x_{k-1})| +\sum_{k=1}^n |f(x_{k-1}) - f_n(x_{k-1})| \\+\sum_{k=1}^n |f_n(x_k) - f_n(x_{k-1})|$$
Can you continue?

My question is about how to choose of the epsilon, because if 
$$|f_n(x_k) - f(x_k)|\lt \varepsilon /2,\qquad |f_n(x_{k-1}) - f(x_{k-1})| \lt \varepsilon /2$$
then 
$$\sum_{k=1}^n |f(x_k) - f(x_{k-1})|\leq  \sum_{k=1}^n \varepsilon +\sum_{k=1}^n |f_n(x_k) - f_n(x_{k-1})| \\ = n\varepsilon +\sum_{k=1}^n |f_n(x_k) - f_n(x_{k-1})|
\leq n\varepsilon +T_a^b(fn)$$
and the what? I know we want to get $T_a^b(f)\leq \varepsilon +T_a^b(fn)$, but I don't understand how, please help... 
 A: The problem is that the index $n$ of the function $f_n$ shouldn't be the same as that of the number of elements of the partition. We assume that the partition is $a=x_0<\dots <x_K=b$, and the inequality should read 
$$\sum_{k=1}^K |f(x_k) - f(x_{k-1})|\leq  \sum_{k=1}^K |f(x_k) - f_n(x_{k-1})| +\sum_{k=1}^K |f(x_{k-1}) - f_n(x_{k-1})| \\+\sum_{k=1}^K |f_n(x_k) - f_n(x_{k-1})|.$$
A: Let $f_n$ be a sequence of functions on $[a,b]$ that converges at each point of $[a,b]$ to $f$. Let $a=x_0<x_1<...x_n=b$ be a partition of $[a,b]$ and let $\epsilon > 0$. Then there exists $N$ such that $$\sum\limits_{i=1}^n |f(x_i) -f(x_{i-1})| \leq \sum\limits_{i=1}^n |f(x_i) - f_n(x_i)| +\sum\limits_{i=1}^n |f(x_{i-1}) - f_n(x_{i-1})| +\sum\limits_{i=1}^n |f_n(x_i) -f_n(x_{i-1})| < \epsilon +T_a^b(f_n)$$ for $n \geq N$. Thus $$T_a^b(f) \leq \epsilon +T_a^b(f) \ for \ n \geq N$$ so $$T_a^b(f) \leq \epsilon + \underline{\lim} T_a^b(f_n)$$ Since $\epsilon $ is arbitrary, $$T_a^b(f) \leq \underline{\lim} T_a^b(f_n)$$
