Positive functionals on C_c(X) are integration against a Radon measure

Let $0\le f\in C_c(X)$ and choose $N$ with $f\le N\epsilon$. For $1\le n\le N$ set $K_n=\{f\ge n\epsilon \}$ and $f_n=\min \{\epsilon ,(f-(n-1)\epsilon {)}_{+}\}\mathrm{.}$ Then $f={\sum }_{n=1}^N{f}_n$, $\epsilon {\mathbf{1}}_{K_n}\le f_n\le \epsilon {\mathbf{1}}_{K_{n-1}}$, where $K_0=\mathrm{supp}f$.

The upper compact estimate $0\le h\le {\mathbf{1}}_K\Rightarrow \mathrm{\Lambda }(h)\le \mu (K)$ follows directly from [L1] by comparing $h$ with functions majorizing ${\mathbf{1}}_K$; the lower estimate ${\mathbf{1}}_K\le h\Rightarrow \mu (K)\le \mathrm{\Lambda }(h)$ is [L1]. Applying these to $f_n\mathrm{/}\epsilon$ gives $\epsilon \mu (K_n)\le \mathrm{\Lambda }(f_n)\le \epsilon \mu (K_{n-1})\mathrm{.}$ Since both indicators are measurable simple functions, [L2] gives the same inequalities for $\int f_{n}d\mu$.

Each $f_n$ is supported on the compact set $K_0$ and is bounded, hence integrable because $\mu (K_0)<\mathrm{\infty }$. Summing step 1.2 using [L3] gives $\mid \mathrm{\Lambda }(f)-\int fd\mu \mid \le \epsilon \mu (K_0)\mathrm{.}$ Letting $\epsilon \downarrow 0$ proves equality for nonnegative $f$. Applying it to $f^{+}$ and $f^{-}$ and using [L3] proves the result for every real $f\in C_c(X)$.