1.1 Section 4.3 Problem 1.
Suppose s1 and s2 are simple function with s1 = s2 a.e. Let E be the set of measure 0 where s1 6= s2. Observe that,
n
Z
s1dµ = Xaiµ(Ai)
i=1
n
= Xaiµ(Ai \ E) + aµ(E)
i=1
n
= Xaiµ(Ai \ E) + a · 0
i=1
n
= Xaiµ(Ai \ E)
i=1
and,
m
Z
s2dµ = Xbiµ(Bi)
i=1 m
= Xbiµ(Bi \ E) + bµ(E)
i=1 m
= Xbiµ(Bi \ E) + a · 0
i=1 m
= Xbiµ(Bi \ E)
i=1
But observe that ∪ni=1Ai \E Bi \E = X \E and, by definition, s1 = s2 everywhere on this set. Since we are integrating the same function on the same set, we must have that R s1dµ = R s2dµ.
Problem 3.
Suppose s is a nonnegative simple function. Suppose RX sdµ = 0. Then,
n
Z
sdµ = Xαiµ(Ei)
x i=1
= 0
Since s ≥ 0, we have that αi ≥ 0 for all i. Moreover, we have that µ(Ei) ≥ 0 for all i by properties of measure. Thus, in order for this sum to equal 0, we must have that αiµ(Ei) = 0 for all i. Note that if µ(Ei) = 0 for all i, then µ(X) = 0. Hence, even if s > 0 on all of X, it is vacuously true that s = 0 a.e. since the whole space is measure 0. So let us assume that at least one Ei has positive measure. If αiµ(Ei) > 0 for this i, then Pni=1 αiµ(Ei) > 0, a contradiction. Hence, for every set Ei of positive measure, we must have that s = 0. Thus, s = 0 a.e. in X.
Now let us assume s = 0 a.e. in X. Then we have that there is a set E with µ(E) = 0 such that s > 0 on E and s = 0 on X \ E. Since s = 0 a.e. and 0 is a simple function (with every coefficient αi set to 0), we can apply Exercise 1 and state that we must have,
Z Z
sdµ = 0dµ
X X n
= X0 · µ(Ai)
i=1
= 0
as required.
1.2 Section 4.4
Problem 2.
Let f be a nonnegative measurable function and let A be a measurable set. Suppose RA f dµ = 0. Then,
Z Z
f dµ = sup{ s dµ : s is simple and 0 ≤ s ≤ f} = 0
A A
. Since the supremum of this set is 0 and every nonnegative simple function has a nonnegative integral, then for any s in the above set, we must have that,
n
Z
s dµ = Xaiµ(Ai)
A i=1
= 0
By Section 4.3 Problem 3, we thus have that s = 0 a.e. in A. Since s was arbitrary, this holds for every s in the above set. Let us suppose that it is not the case that f = 0 a.e. That is, there is a set of positive measure where f > 0. Then there would be a simple function s with 0 ≤ s ≤ f such that s > 0 on this set as well and hence it would not be the case that s = 0 a.e., a contradiction. Hence, we must have that f = 0 a.e.
Now suppose that f = 0 a.e. Then there is a set E of measure 0 such that f = 0 everywhere on A\E. Fix a simple function s on A such that 0 ≤ s ≤ f. For this to hold, we must have that s = 0 on A \ E. We can have that 0 < s ≤ f on E, but note that µ(E) = 0.
Thus,
n
Z
s = Xai µ(Ai)
A i=1
n m
= Xai µ(Ai \ E) + Xei µ(Ei)
i=1 i=1 n
= Xai µ(Ai \ E)
i=1
= 0
where ∪Ei = E (note since Ei ⊂ E we have µ(Ei) ≤ µ(E) and since measure is nonneg-
ative and µ = 0, we get µ(Ei) = 0).
Since s was arbitrary on A with the property 0 ≤ s ≤ f, this must hold for every element of the set {RA s dµ : s is simple and 0 ≤ s ≤ f}. Thus, RA s = 0 for every element s in the set and so,
Z Z
f dµ = sup{ s dµ : s is simple and 0 ≤ s ≤ f}
A A
= 0
as required.
Problem 3.
Suppose f is a nonnegative measurable function and let A,B be measurable sets with A ⊂ B. We have that
Z Z
f = sup{ s dµ : s is simple and 0 ≤ s ≤ f}
A A
and
Z Z
f = sup{ s dµ : s is simple and 0 ≤ s ≤ f}
B B
Observe that for every s in the set {RB s dµ : s is simple and 0 ≤ s ≤ f}, we can write,
n
Z
s = Xbnµ(Bi)
B i=1
n m
= Xbiµ(Bi \ A) + Xaiµ(Ai)
i=1 i=1
Z Z
= s + s
B\A A
where ∪Ai = A. Since the integral of any nonnegative simple function is nonnegative, we must have from the above that RB s ≥ RA s. Observe that every element of {RA s dµ : s is simple and 0 ≤ s ≤ f} can be extended to a simple function on B by taking s = 0 on B \ A, so we have {RA s dµ : s is simple and 0 ≤ s ≤ f} ⊂ {RB s dµ : s is simple and 0 ≤ s ≤ f}. Moreover, as we have shown above, if RB\A > 0, then RB s > RA s. Thus, we must have that,
Z Z
sup{ s dµ : s is simple and 0 ≤ s ≤ f} ≤ sup{ s dµ : s is simple and 0 ≤ s ≤ f}
A B and so,
Z Z
f ≤ f
A B
as required.
Problem 4.
Let f be a nonnegative measurable function and {Aj} be a sequence of disjoint measurable sets. We have,
Z Z
f dµ = sup{ s dµ : s is simple and 0 ≤ s ≤ f}
tAj tAj
Fix s in the above set. Then,
n
Z
s = Xbi µ(Bi) tAj i=1
for some sets Bi such that ∪Bi = tAj. However, we could alternatively divide up tAj so that each subset in our sum corresponds to only one Aj. That is, we can write,
n
Z
s = XXaji µ(Aji) tAj j=1 i=1
n n
= Xa1i µ(A1i) + Xa2i µ(A2i) + ···
i=1 i=1
Z Z
= s + s + ···
A1 A2
Now note that RAi s is contained in the above set for any i because we can set RAj s = 0 for any j 6= i and maintain its status as a simple function. Hence,
Z Z
X
f dµ = f dµ tAi j Aj Problem 5.
Consider
1[0,1/2](x) n odd
fn(x) =
1[1/2,1](x) n even
Observe that liminfn→∞ fn(x) = 0. Thus,
Z Z
liminf fn(x)dµ = 0dµ
n→∞
= 0
Now note that R fndµ = for all n. Thus,
Z
liminf fndµ
n→∞ n→∞ 2
Thus, we have that R liminfn→∞ fn(x)dµ < liminfn→∞R fndµ as required.
