F. & M. Riesz theorem Can someone explain me in which sense F. & M. Riesz theorem (https://en.wikipedia.org/wiki/F._and_M._Riesz_theorem) is important/interesting?
 A: Suppose $f(z)$ is a complex harmonic function on the unit disk. Then
$$
         \sup_{0 < r < 1}\int_{0}^{2\pi}|f(re^{i\theta})|d\theta < \infty
$$
holds iff there is a finite complex Borel measure $\mu$ on the unit circle $\mathbb{T}$ such that
$$
      f(z) = \int_{\mathbb{T}}\Re\left(\frac{w+z}{w-z}\right)d\mu(w),\;\;|z|< 1.
$$
The measure $\mu$ may have a singular continuous component. The odd part is that if $f$ is holomorphic in the unit disk with the same boundedness condition, then  the representing measure cannot have a singular continuous part. An interesting consequence of this is that every $f \in H^{1}$ (defined as holomorphic functions with this boundedness condition) is the product of two functions $f_1,f_2 \in H^{2}$, where $g \in H^{2}$ means that $g$ is holomorphic in the unit disk with
$$
     \sup_{0 < r < 1}\int_{0}^{2\pi}|g(re^{i\theta})|^{2}d\theta < \infty.
$$
Harmonic functions are very different in this regard. So $f \in H^{1}$ has an $L^{1}$ boundary function $f_1$, and $\|f_r-f_1\|_{L^{1}}\rightarrow 0$ as $r\uparrow 1$, where $f_r(e^{i\theta})=f(re^{i\theta})$. This is definitely not true for harmonic functions.
