F. and M. Riesz theorem

Remarks

Proof idea: the Poisson integral of $\mu$ is $F(z) = \frac{1}{2\pi} \sum_{n \ge 0} \hat\mu(n) z^n$, a holomorphic function. Its integrals $\int \abs{F(r e^{it})} \, dt$ are bounded by the total variation of $\mu$, so $F \in H^1$. A function in $H^1$ has boundary values $F^*$ with $F(r e^{it}) \to F^*(e^{it})$ in $L^1$, a fact that uses the inner-outer factorization. Hence the measures $F(r e^{it}) \, dt$ converge both to $\mu$ and to $F^* \, dt$, and $\mu = F^* \, dt$.

Example: if $\mu$ is a point mass at $t = 0$, then $\hat\mu(n) = 1$ for all $n$, so the hypothesis fails, as it must since a point mass is not absolutely continuous.

A consequence: if $\Omega$ is a domain bounded by a rectifiable Jordan curve, then harmonic measure on $\partial\Omega$ and arc length have the same null sets.

Going deeper

For a student

The theorem identifies the analytic measures on the circle with the boundary values of $H^1$ functions. It is a key step in the theory of Hardy spaces and in the study of boundary behaviour of conformal maps. Its consequence for rectifiable Jordan domains links the harmonic measure with arc length.