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
s1d = Xai(Ai)
i=1
n
= Xai(Ai E) + a(E)
i=1
n
= Xai(Ai E) + a 0
i=1
n
= Xai(Ai E)
i=1
and,
m
Z
s2d = Xbi(Bi)
i=1 m
= Xbi(Bi E) + b(E)
i=1 m
= Xbi(Bi E) + a 0
i=1 m
= Xbi(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 s1d = R s2d.
Problem 3.
Suppose s is a nonnegative simple function. Suppose RX sd = 0. Then,
n
Z
sd = Xi(Ei)
x i=1
= 0
Since s 0, we have that i 0 for all i. Moreover, we have that (Ei) 0 for all i by properties of measure. Thus, in order for this sum to equal 0, we must have that i(Ei) = 0 for all i. Note that if (Ei) = 0 for all i, then (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 i(Ei) > 0 for this i, then Pni=1 i(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 (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 i set to 0), we can apply Exercise 1 and state that we must have,
Z Z
sd = 0d
X X n
= X0 (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 d = 0. Then,
Z Z
f d = sup{ s d : 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 d = Xai(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 AE. 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 (E) = 0.
Thus,
n
Z
s = Xai (Ai)
A i=1
n m
= Xai (Ai E) + Xei (Ei)
i=1 i=1 n
= Xai (Ai E)
i=1
= 0
where Ei = E (note since Ei E we have (Ei) (E) and since measure is nonneg-
ative and = 0, we get (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 d : s is simple and 0 s f}. Thus, RA s = 0 for every element s in the set and so,
Z Z
f d = sup{ s d : 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 d : s is simple and 0 s f}
A A
and
Z Z
f = sup{ s d : s is simple and 0 s f}
B B
Observe that for every s in the set {RB s d : s is simple and 0 s f}, we can write,
n
Z
s = Xbn(Bi)
B i=1
n m
= Xbi(Bi A) + Xai(Ai)
i=1 i=1
Z Z
= s + s
BA 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 d : 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 d : s is simple and 0 s f} {RB s d : s is simple and 0 s f}. Moreover, as we have shown above, if RBA > 0, then RB s > RA s. Thus, we must have that,
Z Z
sup{ s d : s is simple and 0 s f} sup{ s d : 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 d = sup{ s d : s is simple and 0 s f}
tAj tAj
Fix s in the above set. Then,
n
Z
s = Xbi (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 (Aji) tAj j=1 i=1
n n
= Xa1i (A1i) + Xa2i (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 d = f d 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)d = 0d
n
= 0
Now note that R fnd = for all n. Thus,
Z
liminf fnd
n n 2
Thus, we have that R liminfn fn(x)d < liminfnR fnd as required.
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