Lemma 3

Let Y be a locally compact subspace of X; for every measure $\mu$ on X, $|\mu_Y| = |\mu|_Y$.

Let f be a function in $\mathscr{K}_+(Y)$ and $\varepsilon$ any number $> 0$; by definition, there exists a function $g \in \mathscr{K}(Y; \mathbf{C})$ such that $|g| \leq f$ and $|\mu_Y|(f) \leq |\mu_Y(g)| + \varepsilon$. Denoting by $f'$ and $g'$ the functions obtained by extending f and g, respectively, to be 0 on $X - Y$, we have $\mu_Y(g) = \mu(g')$ and, since $|g'| \leq f'$,

whence $|\mu_Y|(f) \leq |\mu|_Y(f) + \varepsilon$ and, since $\varepsilon$ is arbitrary,

On the other hand, let K be the support of f and let U be a compact neighborhood of K in X such that $|\mu|(U - K) \leq \varepsilon$; by Urysohn's theorem, there exists a function $f_1 \in \mathscr{K}_+(X)$, extending f, with support contained in U and such that $\|f_1\| = \|f\|$. There exists a function $h_1 \in \mathscr{K}(X; \mathbf{C})$ such that $|h_1| \leq f_1$ and $|\mu|(f_1) \leq |\mu(h_1)| + \varepsilon$. If h is the restriction of $h_1$ to Y, then $h \in \mathscr{K}(Y; \mathbf{C})$, $|h| \leq f$ and $\mu(h_1) - \mu_Y(h) = \mu(h_1 \varphi_{U-K})$, therefore

moreover, we have similarly

From this we infer that

and since $\varepsilon$ is arbitrary, $|\mu|_Y(f) \leq |\mu_Y|(f)$, which completes the proof.