I just did an exercise in a some lecture notes, my result implied that the Fourier coefficients of an harmonic $L^{1}$ function $u$ was the integer values of the borel measure in the Poisson integral representation of $u$. I been trying to find some literature regarding this but settings are not really the same in the books I found, is this is a correct observation?
The exercise was to establish uniqueness of the measure in the Poisson integral and I did it via the Fourier series which the Poisson integral at lest as far as I know is.
These functions are $L^{1}$ in the disk in the sense that its dialations are $L^{1}$ on each circle for $r \in (0,1)$
Yes, you are correct. If let $f_r(e^{i\theta})=F(re^{i\theta})$ where $F$ is a function as you describe, then your assumptions give $$ \int_{0}^{2\pi}|f_r(e^{i\theta})|d\theta \le M, \;\;\; 0 \le r < 1. $$ If you let $\rho_r(\theta)=\int_{0}^{\theta}F(re^{i\theta})d\theta$, then $\rho_r$ is a function of bounded variation with total variation $V(\rho_r)=\int_{0}^{2\pi}|F(re^{i\theta})|d\theta \le M$. The Helly Selection Theorem gives the existence of $\{ r_n \}$ tending upward to $1$ such that $\rho_{r_n}$ converges pointwise everywhere to some $\rho$ with $V(\rho)\le M$. This fact is easily used after integrating by parts: \begin{align} F(rse^{i\theta}) & = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}F(se^{i\theta'})d\theta' \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}\frac{d}{d\theta'}\int_{0}^{\theta'}F(se^{i\alpha})d\alpha \\ & = \left.\frac{1}{2\pi}\frac{1-r^2}{1+2r\cos(\theta-\theta')+r^2}\int_{0}^{\theta'}F(se^{i\alpha})d\alpha\right|_{\theta'=0}^{\theta'=2\pi} \\ & -\frac{1}{2\pi}\int_{0}^{2\pi}\left(\frac{\partial}{\partial\theta'}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}\right)\int_{0}^{\theta'}F(se^{i\alpha})d\alpha\, d\theta' \end{align} Setting $s=r_n$, letting $n\rightarrow\infty$, and integrating by parts using the Riemann-Stieltjes integral gives \begin{align} F(re^{i\theta})&=\left.\frac{1}{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}\rho(\theta')\right|_{\theta'=0}^{2\pi} \\ &-\frac{1}{2\pi}\int_{0}^{2\pi}\left(\frac{\partial}{\partial\theta'}\frac{1-r^2}{1-2r\cos(\theta'-\theta)+r^2}\right)\rho(\theta')d\theta' \\ & = \frac{1}{2\pi}\int_{0}^{2\pi}\frac{1-r^2}{1-2r\cos(\theta-\theta')+r^2}d\rho(\theta') \end{align} The function $\rho$ is of bounded variation $V(\rho) \le M$, which allows you convert the integral to a Lebesgue integral with respect to a complex measure $\rho$; but that's optional.