Proposition 5

In order that every lower semi-continuous real-valued function $f$ (finite or not) on a topological space $X$ should be the upper envelope of the continuous real-valued functions on $X$ (finite or not) which are $\leq f$, it is necessary and sufficient that $X$ be uniformizable.

The condition is necessary. Let $x_0$ be any point of $X$ and let $V$ be any open neighbourhood of $x_0$; then the characteristic function $\varphi_V$ of the set $V$ is lower semi-continuous (Chapter IV, § 6, no. 2, Corollary to Proposition 1); by hypothesis, there is therefore a continuous real-valued function $g$ on $X$ such that $g \leq \varphi_V$ and $g(x_0) = a > 0$. The continuous function $\inf \left( 1, \frac{1}{a} g^+ \right)$ takes its values in $[0, 1]$, is equal to 0 in $C_V$, and equal to 1 at $x_0$. Hence (Theorem 2) $X$ is uniformizable.

The condition is sufficient. Consider first the case in which $f$ takes its values in $[-1, +1]$. We have to show that, for each $x_0 \in X$ and each number $a < f(x_0)$, there is a continuous real-valued function $g$ on $X$ such that $g \leq f$ and $g(x_0) \geq a$. If $a \leq -1$, we may take $g$ to be the constant $-1$. If $-1 < a < f(x_0)$, there is a neighbourhood $V$ of $x_0$ such that $f(x) \geq a$ for all $x \in V$. Since $X$ is uniformizable, there is a continuous real-valued function $h$ on $X$, with values in $[0, 1]$, such that $h(x_0) = 0$ and $h(x) = 1$ for all $x \in C_V$. We may then take $g(x) = a - (a + 1)h(x)$, and we have a continuous function satisfying the stated conditions. Note that this function takes its values in $[-1, +1]$.

The general case follows by transfer of structure; for there is a strictly increasing homeomorphism of $[-1, +1]$ onto $\overline{\mathbf{R}}$ (Chapter IV, § 4, no. 2, Proposition 2), and the definition of a semi-continuous function involves only the order structure and the topology of $\overline{\mathbf{R}}$.