# Sufficient condition for Radon measure be Haar measure

I am reading Dinakar Ramakrishnan&Robert J.Valenza Fourier Analysis on Number Fields.

I want to understand following proposition.

Prop 1-7 (ii) Let $G$ be a locally compact group with nonzero Radon measure $\mu$.

The measure $\mu$ is a left Haar measure on G if and only if $$\int_{G} L_s f d\mu = \int _G f d\mu$$ for all $f \in \mathcal{C}_c^{+}$ and $s \in G$

And following is the proof of this prop.

Convrsely, from the positive linear functional $\int_G \cdot d\mu$ on $\mathcal{C_c}(G)$ we can, by the Riesz representation theorem, explicitly recover the Radon measure $\mu$ of any open subset $U \subset G$ as follows: $$\mu (U) = sup \{\int_G f d\mu \mid f \in \mathcal{C_c}(G), \ \|f\|_u \leq 1, \ and \ supp(f) \subset U \}$$

I am wondering why the last equality holds.

• Notice how $Fourier Analysis \ on \ Number Fields$ has weird spacing? It is better to do it with stars instead of dollar-signs: Fourier Analysis on Number Fields. – GEdgar Apr 29 '18 at 17:00
Let $K$ be compact subset of $U$. Since $G$ is locally compact there exists $f \in C_c(G)$ such that $0 \leq f \leq 1$, $f=1$ on $K$ with support inside $U$. Note that $\mu (K) \leq \int f d\mu$ which does not exceed the RHS. Since $\mu$ is Radon we can take supremum over all $K$ to get LHS $\leq$ RHS. The reverse inequality is obvious.