Theorem 2

Let $X$ be a Baire space and let $(f_\alpha)$ be a family of lower semi-continuous real-valued functions on $X$ such that, at every point $x$ of $X$, the upper envelope $\sup_\alpha f_\alpha(x)$ is finite. Then every non-empty open set in $X$ contains a non-empty open set on which the family $(f_\alpha)$ is uniformly bounded above.

This theorem may also be stated in the form that the set of points in the neighbourhood of which the family $(f_\alpha)$ is uniformly bounded above is a dense open set.

Let $f = \sup_\alpha f_\alpha$ be the upper envelope of the family $(f_\alpha)$. The function $f$ is lower semi-continuous (Chapter IV, § 6, no. 2, Theorem 4) and finite at every point of $X$. It is therefore enough to carry through the proof in the case where the family $(f_\alpha)$ consists of a single function $f$. Let $A_n$ be the set of points $x \in X$ such that $f(x) \leq n$; $A_n$ is closed (Chapter IV, § 6, no. 2, Proposition 1), and the assumptions imply that $X$ is the union of the sets $A_n$; hence at least one of the $A_n$ has an interior point, and therefore there is a non-empty open set on which $f$ is bounded above (by an integer $n$). If we apply this result to any non-empty open subset of $X$ (this subspace is also a Baire space by Proposition 3 of no. 3), we have the theorem.

In applications of this theorem it is generally the case that the $f_\alpha$ are continuous on $X$.