Is a real closed, bounded interval a locally compact Hausdorff space? Does this hold? I've been confused by the statement of the Riesz-Markov-Kakutani representation theorem; that is, the formulation is as follows: 

Let $X$ be a locally compact Hausdorff space. For any positive
  linear functional  $\psi$  on $C_c(X)$, there is a unique regular Borel
  measure $\mu$ on $X$ such that
$$\psi(f) = \int_X f(x) \, d \mu(x) \quad$$  for all $f$ in $C_c(X)$.

whereas Riesz proved simply that 

Every continuous linear functional $A[f]$ over the space $C([0, 1])$ of
  continuous functions in the interval $[0,1]$ can be represented in the
  form
$$A[f] = \int_0^1 f(x)\,d\alpha(x),$$ where $α(x)$ is a function of bounded
  variation on the interval [0, 1], and the integral is a
  Riemann–Stieltjes integral.

Reading this led me to believe that the former, more general formulation would hold with any closed bounded interval in $\mathbb{R}$.
Any help would be appreciated.
 A: Yes.
Clearly, $\mathbb R$ is Hausdorff when endowed with the Euclidean topology (it's a metric space after all), and any topological subspace of a Hausdorff space is Hausdorff with respect to the relative topology. To see this, let $(X,\tau)$ be a Hausdorff topological space and $Y\subseteq X$ a non-empty subset (note: $Y$ need not be open). Endow $Y$ with the relative topology: $$\tau_Y\equiv\{U\cap Y\,|\,U\in\tau\}.$$
Now, pick any distinct pair $u\in Y$ and $v\in Y$. Since $X$ is Hausdorff, there exist disjoint open subsets $U$ and $V$ of $X$ such that $u\in U$ and $v\in V$. Also, $u\in U\cap Y$ and $v\in V\cap Y$, and $U\cap Y$ and $V\cap Y$ are disjoint. They are also open with respect to the relative topology by definition. It follows that $(Y,\tau_Y)$ is Hausdorff.
Therefore, $[a,b]$ is a Hausdorff space whenever $a\in\mathbb R$, $b\in\mathbb R$, and $a<b$ (when endowed with the relatively topology).
Moreover, $[a,b]$ is compact (it is closed and bounded—c.f. the Heine–Borel theorem). Now, every compact space is locally compact. Why? Let $(X,\tau)$ be a compact topological space and take any $x\in X$. Does $x$ have a compact neighborhood? Just take the whole set $X$!
