Real and Complex Analysis (Rudin)

44 REAL AND COMPLEX ANALYSIS
PROOF We first show that
if Kl and K2 are disjoint compact sets. Choose E > O. By Urysohn's lemma,
there exists f E Cc(X) such that f(x) = 1 on K 1, f(x) = 0 on K 2, and
o 5, f 5, 1. By Step II there exists g such that
Kl u K2 -< g and Ag < J.I(K 1 U K 2) + E.
Note that Kl - O. Since Ei E IDlF , there are compact sets Hi c Ei with
(i = 1, 2, 3, ... ). (11)
Putting Kn = HI U ... u Hn and using induction on (10), we obtain
n
n
J.I(E) ~ J.I(Kn) = L J.I(Hi) > L J.I(Ei) - E. (12)
i= 1 i= 1
Since (12) holds for every n and every E > 0, the left side of (9) is not smaller
than the right side, and so (9) follows from Step I.
But if J.I(E) < 00 and E > 0, (9) shows that
N
J.I(E) 5, L J.I(Ei) + E (13)
i= 1
for some N. By (12), it follows that J.I(E) 5, J.I(K N ) + 2E, and this shows that E
satisfies (3); hence E E IDlF .
IIII
STEP V If E E IDlF and E > 0, there is a compact K and an open V such that
K c E c V and J.I(V - K) < E.
PROOF Our definitions show that there exist K c E and V ~ E so that
E
E
J.I(V) - 2: < J.I(E) < J.I(K) + 2:.
Since V - K is open, V - K E IDlF' by Step III. Hence Step IV implies that
J.I(K) + J.I(V - K) = J.l{V) < J.I(K) + E. IIII
STEP VI If A E IDlF and B E IDlF' then A - B, A u B, and A n B belong to IDlF •