
“real: chapter_13” — 2011/5/22 — 23:35 — page 67 — #67
Lebesgue Measure and Integration 13-67
Theorem 13.6.30 Let (X , M, µ) be a measure space.
(i) Suppose f : X →[0, ∞] is measurable, E ∈
M and
E
fdµ = 0.
Then f = 0 a.e. on E.
(ii) Suppose f ∈ L
1
(µ) and
E
fdµ = 0 for every E ∈ M. Then
f = 0 a.e. on X .
(iii) Suppose f ∈ L
1
(µ) and
"
"
"
"
"
"
X
fdµ
"
"
"
"
"
"
=
X
|f |dµ.
Then there exists a constant α such that αf =|f |a.e. on X .
Proof (i) If A
n
={x ∈ E/f (x)>
1
n
} (n = 1, 2, 3, ...), then
1
n
µ(A
n
) ≤
A
n
fdµ ≤
E
fdµ = 0
so that µ(A
n
) = 0. Since {x ∈ E/f (x)>0}=
∞
n=1
A
n
, (i) follows.
(ii) Put f (x) = u(x) + iv(x). Let E ={x/u(x) ≥ 0}. The real part
of
E
fdµ is then
E
u
+
dµ. (Note ...