I believe there must be a mistake in the following argument but I am not sure where:
Suppose $f$ is harmonic in $D=\{z\in \mathbb{C}: |z|<1\}$ such that the functions $f_{r}(\theta)=f(re^{i\theta})$ are uniformly bounded in $L^{1}([-\pi,\pi])$-norm. Then $f(re^{i\theta})=\int_{-\pi}^{\pi}P_{r}(\theta -t)d\mu(t)$ for some finite complex Borel measure $\mu$ on the unit circle. (Here $P_{r}(t)$ is the Poisson kernel. This claim can be taken for granted since I know it to be true.) By Fatou's theorem, $d\mu=\frac{1}{2\pi}\tilde{f}d\theta+d\mu_{s}$ where $\tilde{f}(\theta)=\lim_{r\rightarrow 1}f(re^{i\theta})$ $a.e.$ and $d\mu_{s}$ and $d\theta$ (i.e. Lebesgue measure) are mutually singular. Hence $f(re^{i\theta})=\int_{-\pi}^{\pi}P_{r}(\theta -t)\tilde{f}(t)dt$, i.e., $f$ is the Poisson integral of $\tilde{f}\in L^{1}$ which means that $||f_{r}-\tilde{f}||_{1}\rightarrow 0$ as $r\rightarrow 1$. But does this make sense given that we started merely with the assumption that the $||f_{r}||_{1}$ are bounded as $r\rightarrow 1$?