The following exercise provides an independent proof of Theorem 17.12 in \(\mathbb{R}^{n}\). Set \(d(x, A):=\inf _{a \in A}|x-a|\) and assume that all measures are finite on compact sets.

(i) (Urysohn's lemma) Let \(K \subset \mathbb{R}^{n}\) be a compact set and \(U_{k}:=K+B_{1 / k}(0)\). Show that \(U_{k}\) is open and that \(u_{k}(x):=d\left(x, U_{k}^{c}ight) /\left(d\left(x, U_{k}^{c}ight)+d(x, K)ight)\) are continuous, compactly supported functions such that \(u_{k} \downarrow \mathbb{1}_{K}\).

(ii) Use monotone convergence to show that \(\int u d \mu=\int u d u\) for \(u \in C_{c}^{+}\left(\mathbb{R}^{n}ight)\) implies that \(\mu(K)=u(K)\) on all compact sets \(K\).

(iii) Use the uniqueness of measures theorem to show that \(\mu=u\).

(iv) Replace \(\mathbb{R}^{n}\) by \((X, d)\) such that \(X\) is locally compact, i.e. each \(x \in X\) has a compact neighbourhood. Show that \(\bar{U}_{k}\) from (i) is compact and that the other steps go through without changes.

[ the compactness of \(K\) gives \(K+B_{1 / k}(0)=\bigcup_{i

Data from theorem 17.12

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Theorem 17.12 Let (X, d) be a o-compact metric space and p, v be measures which are finite on compact sets. Then the families C(X) and C.(X) are determining. If , v are finite measures with (X)=v(X), o-compactness is not needed. Proof Let , v be measures with fu du = fu dv for all u Cc (X); in particu- lar, 1K L () n L (v) for all compact sets K. From Theorem 17.8 we know that C. (X) is dense in L (u) L (v), i.e. there is a sequence (un)neN such that Un 1K both in L () and L (v). Therefore, assumption Ju (K)= 1kdu = lim = [1kd 11700 un dv= = [1 k dv=v(K). Since X is o-compact, there is an increasing sequence of compacts KX. This means that the assumptions of the uniqueness theorem of measures (Theorem 5.7) are satisfied, and we get u=v. If (X)=v(x)