Let $u,w\in BV(0,1)$ be given. Since we are in dimension 1 $u$ is continuous almost everywhere and has a representation $u^{l}$ and $u^{r}$ (left, right hand side continuous). Thus we can consider the convex combination $$ \delta u^{l}(x)+(1-\delta)u^{r}(x) $$ for $\delta\in[0,1]$ and $x\in[0,1]$.
Furthermore the derivative of $w$ is a finite signed Radon measure and I ask myself if we can define the integral $$ \int_{0}^{1}u(x)\ dw'(\cdot)(x). $$ In general this is not possible for $u\in BV(0,1)$ since we must be able to evaluate $u$ for all $x\in [0,1]$. That is why I came up with the following definition, which depends on the convex combination stated above: $$ \delta\bullet\int_{0}^{1}u(x)\ dw'(\cdot)(x):=\int_{0}^{1} \delta u^{l}(x)+(1-\delta)u^{r}(x)\ dw'(\cdot)(x) $$ Is this expression well defined? Of course for the dirac part of $w'$ as well as the Lebesgue part there is no problem. But I do not know enough of the Cantor part of $w'$ to conclude that the $\delta\bullet\int$ is well defined, and my knowledge about measure theory seems unsufficient :-(! I am thankful for every hint or comment and thank You very much in advance!
Alex
Not really. The derivative induces a finite signed Radon measure, but it has more structure than the measure. Consider two functions $$w_1(x)=\begin{cases} 0, \quad & x\le 0 \\ 1, \quad &x>0 \end{cases}$$ $$w_2(x)=\begin{cases} 0, \quad & x<0 \\ 1, \quad &x\ge 0 \end{cases}$$ The corresponding Radon measures are the same: Dirac delta at $0$, because integration against either $dw_j$ gives the same result for continuous functions. But an attempt to integrate discontinuous functions reveals the difference: $\int f\,dw_1$ is defined when $f(0+)=f(0)$, while $\int f\,dw_2$ is defined when $f(0-)=f(0)$. BV functions behave like every point has "left half" and "right half": $dw_1$ puts its weight on the right half of $0$, while $dw_2$ puts its weight on the left.
If $u$ is a bounded variation function on $[0,1]$, then it is a function, and functions can be evaluated at any point of their domain. Do you consider BV as a space of equivalence classes like $L^1$? (I don't.)
Anyway: you can integrate any bounded Borel measurable function $f$ with respect to any finite signed Radon measure $\mu$. The construction is standard: split $\mu=\mu^+-\mu^-$ into positive and negative parts. The integrals $\int f\,d\mu^+$ and $\int f\,d\mu^-$ are standard Lebesgue integrals, and both are finite.
But if $u$ is not really a function (and $w'$ is not really a Radon measure), then we have to get creative.
Never mind the Cantor part. The set of discontinuities of $u$ is at most countable, and therefore the issues of "representative" of $u$ do not affect integration with respect to the Cantor part of $w'$. The jumps of $w$ are the only problem, and one may have to worry about left vs right jumps, as I noted above.
The following looks most natural to me: both integrals $\int u^l\,dw^r$ and $\int u^r\,dw^l$ make sense for any BV functions $u,w$. You can define the integral to be $$\delta \int u^l\,dw^r+(1-\delta)\int u^r\,dw^l $$