Let $f,\alpha$ be two bounded complex functions on $[0,1]$. We say that $f$ is integrable w.r.t. $\alpha$ iff the Riemann sum $$\sum f(t_i)(\alpha(x_i)-\alpha(x_{i-1}))$$ converges to a fixed number $I\in\mathbb{C}$ as the partition $P=\left\{0=x_0<x_1<\cdots<x_n=1\right\}$ gets finer. When this is the case, we write $f\in\mathcal{R}(\alpha)$ and that $\int f\,d\alpha=I$. It is known that when $\alpha$ is of bounded variation, $\mathcal{R}(\alpha)$ contains all continuous functions.
I have two questions related to this.
Q1. If $f\in\mathcal{R}(\alpha)$, do we have $\overline{f}\in\mathcal{R}(\alpha)$ too?
Q2. Does the answer change if we further assume that $\alpha$ is of bounded variation?
There is an example of a pair of sequences $(a_n)$ and $(b_n)$ such that $\sum a_n b_n$ converges while $\sum \overline{a_n} b_n$ diverges. From this I get a feeling that the answers to my questions are both negative, but I could not find counterexamples by myself.
Similarly to sequences $(a_n)$ and $(b_n)$ such that $\sum a_nb_n$ converges and $\sum \overline{a_n}b_n$ diverges, you can find bounded functions $f$ and $\alpha$ such that $\int f\alpha'$ converges but $\int \overline{f}\alpha'$ diverges. The idea is to choose $f$ and $\alpha$ of opposite argument.
For example, if $$ f : x \mapsto e^{i/x}, \qquad \alpha(x) = \int_0^x \frac{e^{i/t}}{t} \, dt. $$ $\alpha$ is well defined because, $$ \int_\varepsilon^1 \frac{e^{i/t}}{t} \, dt = \int_1^{1/\varepsilon} \frac{e^{iu}}{u} \, du = \int_1^{1/\varepsilon} \frac{\cos(u)}{u} \, du + i\int_1^{1/\varepsilon} \frac{\sin(u)}{u} \, du, $$ which is known to be convergent when $\varepsilon \rightarrow 0$. Notice that $\alpha(0) = 0$, $\alpha$ is absolutely continuous but has unbounded variation since its derivative is unbounded. Similarly, $$ \int_\varepsilon^1 f(t)\alpha'(t) \, dt = \int_1^{1/\varepsilon} \frac{\cos(2u)}{u} \, du + i\int_1^{1/\varepsilon} \frac{\sin(2u)}{u} \, du $$ is also convergent. However, $$ \int_\varepsilon^1 \overline{f(t)}\alpha'(t) \, dt = \int_\varepsilon^1 \frac{1}{t} \, dt = -\ln(\varepsilon) $$ diverges.
Howver, I haven't been able to prove that $\int f d\alpha$ is well defined neither that $\int \overline{f} d\alpha$ is not. Here is what the computation gives given a partition of $[0,1]$, $$ \begin{align*} \sum_{i = 1}^n f(t_i)(\alpha(x_i) - \alpha(x_{i - 1})) & = f(t_1)\alpha(x_1) + \sum_{i = 2}^n f(t_i)\int_{x_{i - 1}}^{x_i} \alpha'(t) \, dt\\ & = \sum_{i = 2}^n \int_{x_{i - 1}}^{x_i} f(t)\alpha'(t) \, dt + \sum_{i = 2}^n \int_{x_{i - 1}}^{x_i} (f(t_i) - f(t))\alpha'(t) \, dt + f(t_1)\alpha(x_1)\\ & = \int_{x_1}^1 f(t)\alpha'(t) \, dt + \sum_{i = 2}^n \int_{x_{i - 1}}^{x_i} (f(t_i) - f(t))\alpha'(t) \, dt + f(t_1)\alpha(x_1), \end{align*} $$ and the very same computation holds if you replace $f$ by $\overline{f}$. By boundedeness of $f$ and continuity of $\alpha$, $f(t_1)\alpha(x_1) \rightarrow 0$ and $\overline{f(t_1)}\alpha(x_1) \rightarrow 0$ so it remains to prove that, $$ \sum_{i = 2}^n \int_{x_{i - 1}}^{x_i} (f(t_i) - f(t))\alpha'(t) \, dt \rightarrow 0, \qquad \sum_{i = 2}^n \int_{x_{i - 1}}^{x_i} (\overline{f(t_i)} - \overline{f(t)})\alpha'(t) \, dt \rightarrow 0. $$ The first assumtion imply the existence of $\int f d\alpha$ with $\int f d\alpha = \int_0^1 f(t)\alpha'(t) \, dt$ and the second one imply the inexistence of $\int \overline{f} d\alpha$. However, I am struggling to prove them because of the divergence of the derivative of $f$ in $0$, which makes $f(t_i) - f(t)$ hard to bound efficiently. I hope it helped you a bit still !