Example

— Let $X$ be a discrete space, $\mu$ a positive measure on $X$, and set $\alpha(x) = \mu(\varphi_{\{x\}})$ for all $x \in X$. Every numerical function $f$ defined on $X$ is then continuous; for such a function $f \geq 0$, $\mu^*(f) = \sum_{x \in X} \alpha(x)f(x)$, where we agree to set $\alpha(x)f(x) = 0$ whenever $\alpha(x) = 0$ and $f(x) = +\infty$. For,

where $M$ runs over the set of all finite subsets of $X$. If there exists an $x_0 \in X$ such that $f(x_0) = +\infty$ and $\alpha(x_0) > 0$, then $\sum_{x \in M} \alpha(x)f(x) = +\infty$ whenever $x_0 \in M$, and, on the other hand, $f \geq n \cdot \varphi_{\{x_0\}}$ for every integer $n > 0$, therefore $\mu^*(f) \geq n \alpha(x_0)$ and so $\mu^*(f) = +\infty$. If, on the contrary, $\alpha(x) = 0$ at all the points where $f(x) = +\infty$, then the function $g$ equal to $f$ at the points $x \in M$ where $\alpha(x) > 0$ and to 0 elsewhere belongs to $\mathscr{H}_+$, and so, by virtue of the conventions made, $\mu(g) = \sum_{x \in M} \alpha(x)f(x)$, which again proves the relation $\mu^*(f) = \sum_{x \in X} \alpha(x)f(x)$.