I search the reference for the proof of the following theorem: Let G be a locally compact group. If group G is discrete, then the measure algebra $M(G) = L^{1}(G)$.
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A discrete group is always locally compact. – tomasz Jan 04 '24 at 23:49
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1Anyway, isn't this obvious? I'm not sure what your definition of the measure algebra is, but since compact sets are finite, for every Radon measure the measure of a set will be the sum of point measures. – tomasz Jan 04 '24 at 23:53
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@tomasz This is not true; a discrete group can be uncountable, hence not $\sigma$-compact (as the Haar measure is the counting measure). – Sha Vuklia Jan 06 '24 at 22:08
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@tomasz my bad, I read $\sigma$-compact instead of locally compact – Sha Vuklia Jan 07 '24 at 15:20
1 Answers
I'm not aware of a full reference, but I've patched together two references along with an argument by myself:
Let $m$ denote the Haar measure on $G$. We consider $L^1(G)\subset M(G)$ by sending $\phi$ to $\phi m$ (see Example VII.1.11 in Conway's Functional Analysis, or this post). In Conway's example the Radon-Nikodym theorem is invoked ($G$ is assumed to be $\sigma$-compact) to conclude that $fm=gm$ implies that $f=g$. However, it is not necessary to invoke Radon-Nikodym (see Proposition 9.16 in Yeh's Real Analysis).
For the other inclusion, let $\mu\in M(G)$. Since $\mu$ is finite, so is $\vert\mu\vert$ (by Proposition C.4 in Conway's Functional Analysis), which implies that the subset of $G$ consisting of elements $x$ such that $\vert\mu\vert(\{x\})=\vert\mu(\{x\})\vert\neq 0$ is countable (see this post). This will justify the summations below.
Let $$ \phi:G\to\mathbb F,x\mapsto \mu(\{x\}). $$ Since $G$ is discrete, $\phi$ is continuous. Note that $$ \int \vert\phi\vert dm=\sum_{x\in G}\vert\mu(\{x\})\vert=\vert\mu\vert(G)\vert<\infty, $$ and hence $\phi\in L^1(G)$. Note that for any $f\in C_c(G)$, $$ (\phi m)(f)=\int f\phi dm=\sum_{x\in G} f(x)\phi(x)=\sum_{x\in G} f(x)\mu(\{x\})=\int f d\mu, $$ hence $\phi m=\mu$ (having invoked the Riesz representation theorem implicitly).

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2To the downvoter: what wasn't good about the post? I'm happy to receive feedback to improve the post. – Sha Vuklia Jan 05 '24 at 18:14