How does Riesz representation theorem give us the left invariance of the Radon measure from a left invariant positive linear functional on $C_c(G)$?

Question

I am trying to understand the end of the (canonical) proof of the existence of Haar measure on a locally compact group by Riesz representation theorem.

Let $G$ be a locally compact group and $C_c(G)$ denote the set of all compactly supported function on $G$. Suppose $I$ is a left-invariant positive linear functional on $C_c(G)$ (Here, by left-invariant we mean $I(L_x f)=I(f)$ for any $x\in G$, where $L_x f(y)=f(xy)$ ).

By Riesz representation theorem, there exists a unique Radon measure $\mu$ on $G$ such that

$$I(f)=\int_Gfd\mu, \forall f\in C_c(G).$$

From here I can use the Urysohn's lemma, monotone convergence theorem, etc. to conclude that $$\mu(xU)=\mu(U)$$

for any $x\in G$ and open subsets $U$ of $G$. But I wonder how to prove that

$$\mu(xE)=\mu(E)$$

for any $x\in G$ and Borel subset $E$ of $G$ (so that $\mu$ is a left invariant Haar measure on $G$).

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Note that the latter doesn't follow from the former immediately.

Answer 1

Use the uniqueness part of the theorem. For each $x\in G$, the measure $\mu_x(E)=\mu(xE)$ satisfies $$\int f d\mu_x=\int L_{x^{-1}}f d\mu=I(L_{x^{-1}}f)=I(f)$$ for all $f\in C_c(G)$. By the uniqueness of $\mu$, this implies $\mu=\mu_x$.

Or more directly, the fact that $\mu$ (and hence also $\mu_x$) is a Radon measure implies that it is uniquely determined by its values on open sets, by outer regularity.