 
Summary: 3 Spaces Lp
1. In this part we fix a measure space (, A, µ) (we may assume that µ is
complete), and consider the A measurable functions on it.
2. For f L1
(, µ) set
f 1 = f L1 = f L1(,µ) =
f dµ.
It follows from the above inequalities that for c C1
f + g 1 f 1 + g 1,
cf 1 = c f 1.
With more work we derive that
f 1 = 0 f = 0 µ  a.e..
Factorising by the subspace of functions equal 0 µ a.e., or redefining the
symbol = between to functions, we conclude that L1
(, µ) is a normed
vector space.
3. Definition of a normed space from B, Ch.2 (B2). Prove that spaces in
B2, Examples 1, parts 1, 2, 4, 5, 9 (cases l1
, l
