Real and Complex Analysis (Rudin)

134 REAL AND COMPLEX ANALYSIS
Prove this by completing the following outline.
X,c - x.1 dll. Then (!In, p) is a complete metric space (modulo sets of measure
Define peA, B) = I 1
0), and E -+ IE I. dll is continuous for each n. If £ > 0, there exist Eo, ~, N (Exercise 13, Chap. S) so
that
IL(/. -IN) dill < £ if ptE, Eo) N. (*)
If I4A) < ~, (*) holds with B = Eo - A and C = Eo u A in place of E. Thus (*) holds with A in place
of E and 2£ in place of £. Now apply (a) to {fl' ... ' IN}: There exists~' > 0 such that
Iff. dill < 3£ if I4A) < ~', n = 1,2,3, ....
11 Suppose II is a positive measure on X,I4X) < 00,1. e V(Il) for n = 1,2, 3, ...,f.(x)-+ lex) a.e., and
there exists p > 1 and C < 00 such that Ix 1 In I· dll < C for all n. Prove that
lim fl/-Inldll=O.
"-00 Jx
Hint: {f.} is uniformly integrable.
12 Let!lJl be the collection of all sets E in the unit interval [0, 1] such that either E or its complement
is at most countable. Let II be the counting measure on this a-algebra !lJI. If g(x) = x for 0 s: x s: 1,
show that g is not !In-measurable, although the mapping
1-+ L x/(x) = fIg dll
makes sense for every Ie VCIl) and defines a bounded linear functional on VCIl). Thus (IJ)* in
this situation.
13 Let L«> = L«>(rn), where rn is Lebesgue measure on I = [0, 1]. Show that there is a bounded linear
functional A that is 0 on C(I), and that therefore there is no g e V(rn) that satisfies At =
Illg drn for every Ie L«>. Thus (L«»*