Proposition 13

— Let $\mu$ be a bounded measure on $X$. Let $\mathscr{B}$ be a filter base on $\mathscr{L}_F^p$ having the following properties: $1^\circ$ there exists a set $M \in \mathscr{B}$ such that the functions $f \in M$ are uniformly bounded on $X$; $2^\circ$ $\mathscr{B}$ converges uniformly on every compact subset of $X$ to a function $f_0$. Under these conditions, $f_0$ belongs to $\mathscr{L}_F^p$ and $\mathscr{B}$ converges in mean of order $p$ to $f_0$.

We note first of all that if $|f(x)| \leq a$ for every $x \in X$ and every function $f \in M$, then also $|f_0(x)| \leq a$ for every $x \in X$. This being so, for every $\varepsilon > 0$ there exist a compact set $K$ such that $|\mu|(CK) \leq \varepsilon^p$ and a set $N \in \mathscr{B}$ such that, for every function $f \in N$, $|f(x) - f_0(x)| \leq \varepsilon(|\mu|(K))^{-1/p}$ for all $x \in K$. Now, we may write

it follows from the foregoing that if $f \in M \cap N$ then $N_p((f - f_0)\varphi_K) \leq \varepsilon$ and $N_p((f - f_0)\varphi_{CK}) \leq 2a\varepsilon$, whence $N_p(f - f_0) \leq (2a + 1)\varepsilon$, which proves the proposition.