61.3. TIGHT MEASURES 1993

Proof: By tightness, there exists an increasing sequence of compact sets, {Kn} suchthat

µ (Kn)> 1− 1n

for all µ ∈ Λ. Now letting µ ∈ Λ and φ ∈C (Kn) such that ||φ ||∞≤ 1, it follows∣∣∣∣∫Kn

φdµ

∣∣∣∣≤ µ (Kn)≤ 1

and so the restrictions of the measures of Λ to Kn are contained in the unit ball of C (Kn)′ .

Recall from the Riesz representation theorem, the dual space of C (Kn) is a space of com-plex Borel measures. Theorem 17.5.5 on Page 462 implies the unit ball of C (Kn)

′ is weak∗ sequentially compact. This follows from the observation that C (Kn) is separable whichis proved in Corollary 61.3.4 and leads to the fact that the unit ball in C (Kn)

′ is actuallymetrizable by Theorem 17.5.5 on Page 462. Therefore, there exists a subsequence of Λ,{µ1k} such that their restrictions to K1 converge weak ∗ to a measure, λ 1 ∈C (K1)

′. Thatis, for every φ ∈C (K1) ,

limk→∞

∫K1

φdµ1k =∫

K1

φdλ 1

By the same reasoning, there exists a further subsequence {µ2k} such that the restrictionsof these measures to K2 converge weak ∗ to a measure λ 2 ∈ C (K2)

′ etc. Continuing thisway,

µ11,µ12,µ13, · · · → Weak∗ in C (K1)′

µ21,µ22,µ23, · · · → Weak∗ in C (K2)′

µ31,µ32,µ33, · · · → Weak∗ in C (K3)′

...

Here the jth sequence is a subsequence of the ( j−1)th. Let λ n denote the measure inC (Kn)

′ to which the sequence {µnk}∞

k=1 converges weak∗. Let {µn} ≡ {µnn} , the diag-onal sequence. Thus this sequence is ultimately a subsequence of every one of the abovesequences and so µn converges weak∗ in C (Km)

′ to λ m for each m. Note that this is allhappening on different sets so there is no contradiction with something converging to twodifferent things.

Claim: For p > n, the restriction of λ p to the Borel sets of Kn equals λ n.Proof of claim: Let H be a compact subset of Kn. Then there are sets, Vl open in Kn

which are decreasing and whose intersection equals H. This follows because this is a metricspace. Then let H ≺ φ l ≺Vl . It follows

λ n (Vl) ≥∫

Kn

φ ldλ n = limk→∞

∫Kn

φ ldµk

= limk→∞

∫Kp

φ ldµk =∫

Kp

φ ldλ p ≥ λ p (H) .

Now considering the ends of this inequality, let l→ ∞ and pass to the limit to conclude

λ n (H)≥ λ p (H) .