The theorem of F. and M. Riesz
1 Totally ordered groups
Suppose that $G$ is a locally compact abelian group and that $P\subset G$ is a semigroup (satisfies $P+P\subset P$) that is closed and satisfies $P\cap (P)=\{0\}$ and $P\cup (P)=G$. We define a total order on $G$ by $x\le y$ when $yx\in P$. We verify that this is indeed a total order. (We remark that nowhere in this do we show the significance of $P$ being closed; but in this note we shall be speaking about discrete abelian groups where any set is closed.)
If $x\le y$ and $y\le z$, then $yx\in P$ and $zy\in P$ and hence $zx=(zy)+(yx)\in P+P\subset P$, showing that $x\le z$, so $\le $ is transitive. If $x\le y$ and $y\le x$ then $yx\in P$ and $xy\in P$, the latter of which is equivalent to $yx=(xy)\in P$, hence $yx\in P\cap (P)$, and then $P\cap (P)=\{0\}$ implies that $yx=0$, i.e. $x=y$, so $\le $ is antisymmetric. If $x,y\in P$ then $yx$ is either $0$, in which case $x=y$, or it is contained in one and only one of $P$ and $P$, and then respectively $$ or $$, showing that $\le $ is total.
Moreover, the total order $\le $ induced by the semigroup $P$ is compatible with the group operation in $G$: if $x\le y$ and $z\in G$, then $(y+z)(x+z)=yx\in P$, showing that $x+z\le y+z$.
We say that $G$ with the total order induced by $P$ is a totally ordered group. We shall use the following lemma in the next section.^{1}^{1} 1 Walter Rudin, Fourier Analysis on Groups, p. 194, Theorem 8.1.2.
Lemma 1.
Suppose that $\mathrm{\Gamma}$ is a discrete abelian group. $\mathrm{\Gamma}$ can be totally ordered if and only if $\gamma \mathrm{\in}\mathrm{\Gamma}$ having finite order implies that $\gamma \mathrm{=}\mathrm{0}$.
2 Functions of analytic type
If $G$ is a compact abelian group, then $G$ is connected if and only if $\gamma \in \widehat{G}$ having finite order implies that $\gamma =0$.^{2}^{2} 2 Walter Rudin, Fourier Analysis on Groups, p. 47, Theorem 2.5.6. Combined with Lemma 1, we get that a compact abelian group is connected if and only if its dual group can be ordered.
Suppose in the rest of this section that $G$ is a connected compact abelian group, and let $\le $ be a total order on $\widehat{G}$ induced by some semigroup. We say that a function $f\in {L}^{1}(G)$ is of analytic type if $$ implies that $\widehat{f}(\gamma )=0$, and we say that a measure $\mu \in M(G)$ is of analytic type if $$ implies that $\widehat{\mu}(\gamma )=0$. (We denote by $M(G)$ the set of regular complex Borel measures on $G$.) For $1\le p\le \mathrm{\infty}$, we denote by ${H}^{p}(G)$ those elements of ${L}^{p}(G)$ that are of analytic type. We emphasize that the notion of a function or measure being of analytic type depends on the total order $\le $ on $\widehat{G}$.
We remind ourselves that when $\mathcal{M}$ is a $\sigma $algebra on a set $X$ and $\mu $ is a measure on $\mathcal{M}$, if $A\in \mathcal{M}$ and $\mu (E)=\mu (A\cap E)$ for all $E\in \mathcal{M}$ then we say that $\mu $ is concentrated on $A$. Measures $\lambda ,\mu $ on $\mathcal{M}$ are said to be mutually singular if they are concentrated on disjoint sets.
Let $m$ be the Haar measure on $G$ such that $m(G)=1$, and suppose that $\sigma $ is a positive element of $M(G)$. The Lebesgue decomposition tells us that there is a unique pair of finite Borel measures ${\sigma}_{s}$ and ${\sigma}_{a}$ on $G$ such that (i) $\sigma ={\sigma}_{s}+{\sigma}_{a}$, (ii) ${\sigma}_{a}$ is absolutely continuous with respect to $m$, and (iii) ${\sigma}_{s}$ and $m$ are mutually singular. Then the RadonNikodym theorem tells us that there is a unique nonnegative $w\in {L}^{1}(m)$ such that $d{\sigma}_{a}=wdm$. Thus,
$$d\sigma =d{\sigma}_{s}+wdm.$$ 
We define $\mathrm{\Omega}$ to be the set of all trigonometric polynomials $Q$ on $G$ such that $\widehat{Q}(\gamma )=0$ for $\gamma \le 0$. We also define $K=\{1+Q:Q\in \mathrm{\Omega}\}$. $K\subset {L}^{2}(\sigma )$, and we denote by $\overline{K}$ its closure in the Hilbert space ${L}^{2}(\sigma )$.
Lemma 2.
$\overline{K}$ is a convex set.
Proof.
Let $f,g\in K$ be distinct and let $0\le t\le 1$. There are ${P}_{n},{Q}_{n}\in \mathrm{\Omega}$ such that $1+{P}_{n}\to f$ and $1+{Q}_{n}\to g$, and
$$(1t)f+tg=\underset{n\to \mathrm{\infty}}{lim}((1t)(1+{P}_{n})+t(1+{Q}_{n}))=\underset{n\to \mathrm{\infty}}{lim}(1+(1t){P}_{n}+t{Q}_{n}).$$ 
For each $n$, $(1t){P}_{n}+t{Q}_{n}\in \mathrm{\Omega}$, so we have written $(1t)f+tg$ as a limit of elements of $K$, showing that $(1t)f+tg\in \overline{K}$ and hence that $\overline{K}$ is convex. ∎
As $\overline{K}$ is a closed convex set in the Hilbert space ${L}^{2}(\sigma )$, there is a unique $\varphi \in \overline{K}$ such that $d(0,\overline{K})=\parallel 0\varphi \parallel $ (namely, that attains the infimum of the distance of elements of $\overline{K}$ to the origin), which we can write as
$$\parallel \varphi \parallel =\underset{Q\in \mathrm{\Omega}}{inf}\parallel 1+Q\parallel .$$ 
$\varphi $ is the unique element of $\overline{K}$ such that
$$\u27e8\varphi ,\psi \varphi \u27e9=0,\psi \in \overline{K}.$$ 
The following lemma establishes properties of $\varphi $.^{3}^{3} 3 Walter Rudin, Fourier Analysis on Groups, p. 199, Lemma 8.2.2.
Lemma 3.

1.
$\varphi =0$ almost everywhere with respect to ${\sigma}_{s}$.

2.
$\varphi w\in {L}^{2}(m)$ and ${\varphi }^{2}w={\parallel \varphi \parallel}^{2}$ almost everywhere with respect to $m$.

3.
If $\parallel \varphi \parallel >0$ and $h=\frac{1}{\varphi}$, then $h\in {H}^{2}(m)$ and $\widehat{h}(0)=1$.
Proof.
We write $c=\parallel \varphi \parallel $. Let $1+{Q}_{n}\in K$ such that $1+{Q}_{n}\to \varphi $. If $g\in {L}^{2}(\sigma )$ and $\varphi +g\in \overline{K}$, then $\u27e8\varphi ,(\varphi +g)\varphi \u27e9=0$, i.e. $\u27e8\varphi ,g\u27e9=0$. Let $\gamma >0$. On the one hand, $\gamma \in \mathrm{\Omega}$ so $\varphi +\gamma ={lim}_{n\to \mathrm{\infty}}1+({Q}_{n}+\gamma )\in \overline{K}$, hence $\u27e8\varphi ,\gamma \u27e9=0$ and so $\u27e8\gamma ,\varphi \u27e9=0$. On the other hand, define $g=\varphi \gamma $, which satisfies
$$\varphi +g=\varphi (1+\gamma )=\underset{n\to \mathrm{\infty}}{lim}(1+{Q}_{n})(1+\gamma )=\underset{n\to \mathrm{\infty}}{lim}1+\gamma +{Q}_{n}+{Q}_{n}\gamma ,$$ 
and because $\gamma >0$, each term of $\gamma +{Q}_{n}+{Q}_{n}\gamma $ belongs to $\mathrm{\Omega}$, showing that $\varphi +g\in \overline{K}$, from which we get $\u27e8\varphi ,g\u27e9=0$ and so $\u27e8g,\varphi \u27e9=0$. We have proved that
$${\int}_{G}\u27e8x,\gamma \u27e9\overline{\varphi (x)}\mathit{d}\sigma (x)=0,\gamma >0,$$  (1) 
and
$${\int}_{G}\u27e8x,\gamma \u27e9{\varphi (x)}^{2}\mathit{d}\sigma (x)=0,\gamma >0.$$  (2) 
Taking the complex conjugate of (2) gives
$$ 
Defining $d\lambda ={\varphi }^{2}d\sigma $ we have $\lambda \in M(G)$. The above and (2) give
$$\widehat{\lambda}(\gamma )=0,\gamma \ne 0.$$ 
As well,
$$\widehat{\lambda}(0)={\int}_{G}{\varphi }^{2}\mathit{d}\sigma ={c}^{2}.$$ 
Because $\lambda \in M(G)$ and $\widehat{\lambda}\in {L}^{1}(\widehat{G})$, there is some $f\in {L}^{1}(G)$ such that $d\lambda =fdm$, defined by
$$f(x)={\int}_{\widehat{G}}\widehat{\lambda}(\gamma )\u27e8x,\gamma \u27e9\mathit{d}{m}_{\widehat{G}}(\gamma ),\gamma \in \widehat{G},$$ 
where ${m}_{\widehat{G}}$ is the Haar measure on $\widehat{G}$ that assigns measure $1$ to each singleton.^{4}^{4} 4 Walter Rudin, Fourier Analysis on Groups, p. 30. That is, $d\lambda =fdm$ where $f(x)={c}^{2}{m}_{\widehat{G}}(\{0\})={c}^{2}$, hence $d\lambda ={c}^{2}dm$. Combined with $d\lambda ={\varphi }^{2}d\sigma $ we get
$${\varphi }^{2}d\sigma ={c}^{2}dm.$$ 
Therefore ${\varphi }^{2}d\sigma $ is absolutely continuous with respect to $m$, and because ${\varphi }^{2}d\sigma ={\varphi }^{2}d{\sigma}_{s}+{\varphi }^{2}wdm$, it follows that ${\varphi }^{2}d{\sigma}_{s}=0$, that is, that $\varphi (x)=0$ for ${\sigma}_{s}$almost all $x\in G$, proving the first claim. Furthermore, ${\varphi }^{2}d\sigma ={\varphi }^{2}wdm$ and using ${\varphi }^{2}d\sigma ={c}^{2}dm$ we get ${\varphi (x)}^{2}w(x)={c}^{2}$ for $m$almost all $x\in G$. Because $w\in {L}^{1}(m)$ and ${\varphi w}^{2}={c}^{2}w$, we get $\varphi w\in {L}^{2}(m)$, proving the second claim.
So far we have not supposed that $c>0$. If indeed $c>0$, then ${h}^{2}={\varphi }^{2}={c}^{2}w$, giving $h\in {L}^{2}(m)$. For $\gamma \in \widehat{G}$,
${\int}_{G}}h(x)\u27e8x,\gamma \u27e9\mathit{d}m(x)$  $=$  ${\int}_{G}}{\varphi (x)}^{2}\overline{\varphi (x)}\u27e8x,\gamma \u27e9\mathit{d}m(x)$  
$=$  ${c}^{2}{\displaystyle {\int}_{G}}\u27e8x,\gamma \u27e9\overline{\varphi (x)}w(x)\mathit{d}m(x)$  
$=$  ${c}^{2}{\displaystyle {\int}_{G}}\u27e8x,\gamma \u27e9\overline{\varphi (x)}\mathit{d}\sigma (x).$ 
This and (1) yield
$${\int}_{G}h(x)\u27e8x,\gamma \u27e9\mathit{d}m(x)=0,\gamma >0,$$ 
in other words,
$$ 
namely, $h$ is of analytic type, i.e. $h\in {H}^{2}(m)$. Moreover, for each $n\in \mathbb{N}$ we check that ${Q}_{n}+\varphi \in \overline{K}$ and hence that $\u27e81+{Q}_{n},\varphi \u27e9=\u27e81,\varphi \u27e9$, giving
$${c}^{2}\widehat{h}(0)={\int}_{G}\overline{\varphi}\mathit{d}\sigma ={\int}_{G}(1+{Q}_{n})\overline{\varphi}\mathit{d}\sigma .$$ 
This is true for all $n\in \mathbb{N}$, so we obtain
$${c}^{2}\widehat{h}(0)={\int}_{G}{\varphi }^{2}\mathit{d}\sigma ={\parallel \varphi \parallel}^{2}={c}^{2},$$ 
i.e. $\widehat{h}(0)=1$, proving the third claim. ∎
The above lemma is used to prove the following theorem.^{5}^{5} 5 Walter Rudin, Fourier Analysis on Groups, p. 200, Theorem 8.2.3. The proof of this theorem in Rudin is not long, but I don’t understand the first step in his proof so I have not attempted to write it out.
Theorem 4.
Suppose that $G$ is a connected compact abelian group and that $\mu \mathrm{\in}M\mathit{}\mathrm{(}G\mathrm{)}$ is of analytic type. If the Lebesgue decomposition of $\mu $ is
$$d\mu =d{\mu}_{s}+fdm,$$ 
where ${\mu}_{s}$ and $m$ are mutually singular and $f\mathrm{\in}{L}^{\mathrm{1}}\mathit{}\mathrm{(}m\mathrm{)}$, then ${\mu}_{s}\mathrm{\in}M\mathit{}\mathrm{(}G\mathrm{)}$ is of analytic type and $f$ is of analytic type, and ${\widehat{\mu}}_{s}\mathit{}\mathrm{(}\mathrm{0}\mathrm{)}\mathrm{=}\mathrm{0}$.
3 The theorem of F. and M. Riesz
We are now equipped to prove the theorem of F. and M. Riesz.^{6}^{6} 6 Walter Rudin, Fourier Analysis on Groups, p. 201, §8.2.4.
Theorem 5 (F. and M. Riesz).
If $\mu \mathrm{\in}M\mathit{}\mathrm{(}\mathrm{T}\mathrm{)}$ and $\widehat{\mu}\mathit{}\mathrm{(}n\mathrm{)}\mathrm{=}\mathrm{0}$ for every negative integer $n$, then $\mu $ is absolutely continuous with respect to Haar measure.
Proof.
Write $d\mu =d{\mu}_{s}+fdm$, where ${\mu}_{s}$ and $m$ are mutually singular and $f\in {L}^{1}(m)$. Theorem 4 tells us that ${\mu}_{s}$ is of analytic type, i.e. ${\widehat{\mu}}_{s}(n)=0$ for $$, and that ${\widehat{\mu}}_{s}(0)=0$. Therefore, if ${\mu}_{s}\ne 0$ then there is a minimal positive integer ${n}_{0}$ for which ${\widehat{\mu}}_{s}({n}_{0})\ne 0$. Defining $\widehat{\lambda}(n)={\widehat{\mu}}_{s}({n}_{0}+n)$, we get that $\lambda \in M(\mathbb{T})$ and that $\lambda $ and $m$ are mutually singular. But $\widehat{\lambda}(n)={\widehat{\mu}}_{s}({n}_{0}+n)=0$ for $$, so $\lambda $ is of analytic type, and therefore Theorem 4 says that ${\widehat{\mu}}_{s}({n}_{0})=\widehat{\lambda}(0)=0$ (because $\lambda $ and $m$ are mutually singular), a contradiction. Hence ${\widehat{\mu}}_{s}(n)=0$ for all $n\in \mathbb{Z}$, which implies that ${\mu}_{s}=0$. But this means that $\mu $ is absolutely continuous with respect to $m$, completing the proof. ∎