9.16. FADDEYEV’S LEMMA 231
Do∫
Ωdµ to both sides and use µ (Ω) = 1. Thus
φ
(∫Ω
f dµ
)≤∫
Ω
φ ( f )dµ +λ
(∫Ω
f dµ−∫
Ω
f dµ
)=∫
Ω
φ ( f )dµ.
There are no difficulties with measurability because φ is continuous. ■
Corollary 9.15.2 In the situation of Lemma 9.15.1 where µ(Ω) = 1, suppose f hasvalues in [0,∞) and is measurable. Also suppose φ is convex and increasing on [0,∞).Then φ(
∫Ω
f du)≤∫
Ωφ( f )dµ .
Proof: Let fn (ω) = f (ω) if f (ω) ≤ n and let fn (ω) = n if f (ω) ≥ n. Then bothfn,φ ( fn) are in L1 (Ω) . Therefore, the above holds and φ(
∫Ω
fndu) ≤∫
Ωφ( fn)dµ. Let
n→ ∞ and use the monotone convergence theorem. ■
9.16 Faddeyev’s LemmaThis next lemma is due to Faddeyev. I found it in [35].
Lemma 9.16.1 Let f ,g be nonnegative measurable nonnegative functions on a measurespace (Ω,µ). Then
∫f gdµ =
∫∞
0∫[g>t] f dµdt =
∫∞
0∫
∞
0 µ ([ f > s]∩ [g > t])dsdt.
Proof: First suppose g = aXE where E is measurable, a > 0. Now [g > t] = /0 ift ≥ a and it equals XE if t < a. Thus the right side equals
∫ a0∫
E f dµdt =∫ a
0∫
XE f dµ =∫aXE f dµ which equals the left side. Thus the first equation is true if g = aXE . Similar
reasoning shows that when you have g a nonnegative simple function, g = ∑ni=1 aiXEi
where we can arrange to have {ai} increasing, the first equation still holds. Now by themonotone convergence theorem, this yields the desired result for the first equation.
To get the second equal sign, note that∫∞
0
∫[g>t]
f dµdt =∫
∞
0
∫X[g>t] f dµdt =
∫∞
0
∫∞
0µ([
X[g>t] f > s])
dsdt
=∫
∞
0
∫∞
0µ ([ f > s]∩ [g > t])dsdt ■
9.17 Exercises1. Let Ω = N={1,2, · · ·}. Let F = P(N), the set of all subsets of N, and let µ(S) =
number of elements in S. Thus µ({1}) = 1 = µ({2}), µ({1,2}) = 2, etc. In thiscase, all functions are measurable. For a nonnegative function, f defined on N, show∫N f dµ = ∑
∞k=1 f (k) . What do the monotone convergence and dominated conver-
gence theorems say about this example?
2. For the measure space of Problem 1, give an example of a sequence of nonnegativemeasurable functions { fn} converging pointwise to a function f , such that inequalityis obtained in Fatou’s lemma.
3. If (Ω,F ,µ) is a measure space and f ≥ 0 is measurable, show that if g(ω) = f (ω)a.e. ω and g≥ 0, then
∫gdµ =
∫f dµ. Show that if f ,g ∈ L1 (Ω) and g(ω) = f (ω)
a.e. then∫
gdµ =∫
f dµ .