Real and Complex Analysis (Rudin)

68 REAL AND COMPLEX ANALYSIS
Put
k
gk= Llfni+l-fnil,
i= 1
00
g = L Ifni+1 - fni I·
i= 1
(2)
Since (1) holds, the Minkowski inequality shows that IIgkllp < 1 for k = 1,
2, 3, .... Hence an application of Fatou's lemma to {gf} gives IIgllp:S; 1. In
particular, g(x) < 00 a.e., so that the series
00
fnl(x) + L (fni+l(x) - fnJx» (3)
i= 1
converges absolutely for almost every x e X. Denote the sum of (3) by f(x),
for those x at which (3) converges; put f(x) = 0 on the remaining set of
measure zero. Since
we see that
k-l
fnl + L (ffti+1 - fn,) =f .... , (4)
i= 1
f(x) = lim fn,(x) a.e. (5)
i-+ 00
Having found a function f which is the pointwise limit a.e. of UnJ, we
now have to prove that this f is the l!'-limit of {f..}. Choose € > O. There
exists an N such that IIfn - fm lip < € if n > Nand m> N. For every m > N,
Fatou's lemma shows therefore that
We conclude from (6) that f - fm e l!'(Jl), hence that f e l!'(Jl) [since f =
(f - fm) + f"J, and finally that IIf - fm IIp-+ 0 as m-+ 00. This completes the
prooffor the case 1 :s; p < 00.
In LOO(Jl) the proof is much easier. Suppose {in} is a Cauchy sequence in
LOO(Jl), let Ak and Bm. n be the sets where I fk(x)l > II fk II 00 and where
I fn(x) - fm(x) I > 11f.. - fm 1100' and let E be the union of these sets, for k, m,
n = 1, 2, 3, .... Then Jl(E) = 0, and on the complement of E the sequence Un}
converges uniformly to a bounded functionJ. Definef(x) = 0 for x e E. Then
fe LOO(Jl), and Ilfn - flloo-+ 0 as n-+ 00.
IIII
The preceding proof contains a result which is interesting enough to be
stated separately:
3.12 Theorem If 1 :s; p :s; 00 and if Un} is a Cauchy sequence in l!'(Jl), with
limit J, then Un} has a subsequence which converges pointwise almost everywhere
to f(x).
(6)