I have the following question:
Is it true that $V(f)=\lim_{||P||\rightarrow0}V_P(f)$?
To make clear what I mean: Let $f:[a,b]\subset\mathbb{R}\rightarrow \mathbb{R}$, and $P=\{ t_0=a<\cdots<t_m=b \} $ a partition of $[a,b]\subset\mathbb{R}$, then $ V_P(f)=\displaystyle\sum_{i=1}^{m}| f(t_i)-f(t_{i-1}) |. $
Then $V(f)=$sup$ \{ V_P(f):P\mbox{ is a partition of } [a,b] \}$
Now, if f is not of bounded variation, I know this to be false. For example, take $f=\begin{cases} 0 & x\in\mathbb{Q}\\ 1 & x\in\mathbb{R-Q} \end{cases}$
Then I can find succesions $(P_n)_{n\in\mathbb{N}}$,$(Q_n)_{n\in\mathbb{N}}$ so that $V_{P_n}(f)\rightarrow 0$, $V_{Q_n}(f)\rightarrow \infty$. Showing that in particular, $\lim_{||P||\rightarrow0}V_P(f)$ does not exists(not even $\pm\infty$)
However, I feel like this is cheating. Because $f$ is not of bounded variation.(Thomae's function fails in the same manner) So:
Is it possible to find a counterexample to the question where f is of bounded variation? If not, how does one prove it?
PS: I hope I made myself clear, any help is appreciated and very welcome. Feel free to edit anything to might think will help to make the question clearer. Thanks everyone
Take as a counterexample the function $f:[-1,1] \to \Bbb{R}$ such that $f(x)=\begin{cases}\ 0 & x \neq 0\\ 1 & x=0\\ \end{cases}$