In a nutshell: how can we integrate rough paths using Lebesgue integration?
What I mean by a rough path is a continuous map $[a,b]\to\mathbb{R}^n$, or $[a,b]\to\mathbb{C}$ in the complex case.
(The real case) Let $\gamma:[a,b]\to\mathbb{R}^n$ be a path and let $f:\mathbb{R}^n\to\mathbb{R}$ be a function. If $\gamma$ is smooth, we often define $$\int_\gamma f := \int_a^b(f\circ\gamma)|\gamma'|.$$ The integral on the right may be defined as a Riemann or Lebesgue integral (the latter case being useful whenever $f$ is ill-behaved e.g. $f(\textbf{x}) = 1 \iff \textbf{x}\in\mathbb{Q}^n$). However, in the former (Riemann) case, the definition is merely a special case of the definition by Riemann sums $$\int_\gamma f := \lim_{\Delta\gamma_k\to 0}\sum_{k=1}^mf(\gamma(t_k))|\Delta\gamma_k|,$$ a definition which includes rough paths as well, begging the question: how can we integrate arbitrary paths on $[a,b]\to\mathbb{R}^n$ using Lebesgue integration?
(The complex case) Given a $\mathbb{R}\to\mathbb{C}$ function $f = u+iv$, we define $$\int_a^b f := \int_a^bu + i\int_a^bv,$$ which again allows us to use either Riemann or Lebesgue integration. The choice carries over once a smooth path $\gamma:\mathbb{R}\to\mathbb{C}$ is introduced, as then $$\int_\gamma f := \int_a^b(f\circ\gamma)\gamma'$$ which can again be computed using either kind of integral. As in the real case, the above is, when using Riemann integration, a special case of the definition by Riemann sums $$\int_\gamma f := \lim_{\Delta\gamma_k\to 0}\sum_{k=1}^mf(\gamma(t_k))\Delta\gamma_k,$$ begging, again, the question: how can we integrate rough paths on $[a,b]\to\mathbb{C}$ using Lebesgue integration?
Attempts at a definition: tackling the complex case, we may note that $\gamma = \alpha + i\beta$ for paths $\alpha,\beta:\mathbb{R}\to\mathbb{R}$, and as $$f(\gamma_k)\Delta\gamma_k = \bigg(u(\gamma_k)\Delta\alpha_k-v(\gamma_k)\Delta\beta_k\bigg) + i\bigg(u(\gamma_k)\Delta\beta_k + v(\gamma_k)\Delta\alpha_k\bigg)$$ it seems reasonable to define $$\int_\gamma f := \left(\int_\alpha u - \int_\beta v\right) + i\left(\int_\beta u + \int_\alpha v\right),$$
but I get stuck trying to Lebesgue integrate $\int_\alpha u$.
There are similar questions on the website, so allow me to explain why I still found it worthwhile to write this post.
Regarding this post and this post: the answer/question assumes that the path is differentiable.
The only post that may answer my questions is this one, but I hardly understand anything as I know not what a rough path lift nor a $p$-variation path is.
This post may provide some partial answers. In particular, it shows that the graph of a nowhere differentiable function $[a,b]\to\mathbb{R}$ has infinite length, but need that imply its integral is infinite? Perhaps it implies that integrating over a nowhere differentiable contour yields an infinite integral, but I'm still unsure that needs to follow (it seems possible to me that, depending on how the integral is defined, if the function is everywhere zero or alternates between positive and negative values, then the integral yields a finite result).