I've been reading about the difference between Radon and regular measures and am confused on a small point. Let $X$ be a locally compact Hausdorff space, and $\mu$ a Borel measure on $X$ (that is, a measure on the sigma algebra of Borel sets of $X$). Suppose that $\mu$ is finite on compact sets, every Borel set can be approximated from above by open sets, and every open set can be approximated from below by compact sets. Then $\mu$ is called a Radon measure.
If, moreover, every Borel set with finite measure can be approximated from below by compact sets, then $\mu$ is called a regular measure.
Every regular measure is of course a Radon measure, but not conversely.
The Riesz Representation Theorem states that given any positive linear functional $T: C_c(X,\mathbb C) \rightarrow \mathbb C$, there exists a unique regular Borel measure $\mu$ on $X$ such that
$$\int\limits_X f(x)d\mu(x) = T(f). \tag{$f \in C_c(X,\mathbb C)$}$$
Now, assume $X$ is a locally compact Hausdorff topological group, with left Haar measure $\lambda$. Then $\lambda$ is always a Radon measure, but not necessarily a regular measure. Define a positive linear functional $T: C_c(X,\mathbb C) \rightarrow \mathbb C$ by
$$T(f) = \int\limits_X f(x)d\lambda(x).$$
By the Riesz representation theorem, there is a unique regular Borel measure $\mu$ such that
$$\int\limits_X f(x)d\mu(x) = \int\limits_X f(x)d\lambda(x). \tag{$f \in C_c(X,\mathbb C)$}$$
This implies $\mu = \lambda$, and therefore that $\lambda$ is a regular Borel measure.
This argument seems to show that every Haar measure is regular. But I know there are Haar measures which are not regular. What is going on here?