Description
1.1 Section 4.6 Problem 2.
Observe that |f| is defined as,
f(x) if f(x) ≥ 0
|f| =
−f(x) if f(x) < 0
In addition, we have,
+ f(x) if f(x) ≥ 0
f =
0 if f(x) < 0
and,
f− = −f(x) if f(x) ≤ 0
0 if f(x) > 0
Now let us consider f+(x) + f−(x) for 3 cases: when f(x) > 0, when f(x) = 0, and when f(x) < 0. When f(x) > 0, we have that f+(x) = f(x) and f−(x) = 0. Hence, f+(x) + f−(x) = f(x) in this case, just as in the case of |f|. Now suppose f(x) = 0. Then f+(x) = 0 and f−(x) = 0, so f+(x) + f−(x) = 0. Again, |f(x)| = 0 when f(x) = 0, so they coincide in this case as well. Now suppose f(x) < 0. Then f+(x) = 0 and f−(x) = −f(x). Thus, f+(x) + f−(x) = −f(x). This is precisely the same as |f|. Hence, in all 3 possible cases for f(x), we have that f+ + f− coincides with |f|, and so f+ + f− = |f|.
Problem 3.
Suppose f is integrable. Then R f+dµ < ∞ and R f−dµ < ∞. Hence, R f+dµ+R f−dµ < ∞. By Lemma 4.6.2 Part 2, we have that,
Z Z Z
f+dµ + f−dµ = (f+ + f−)dµ
< ∞
By Exercise 2, we thus have that,
Z Z
+ −
(f + f )dµ = |f|dµ
< ∞
Now suppose |f| is integrable. Then R |f|+dµ < ∞ and R |f|−dµ < ∞. But observe that, since |f| ≥ 0 everywhere, then |f|− = 0 everywhere. Hence, we from this and our work in Exercise 2 that,
Z Z Z
|f|dµ = |f|+dµ − |f|−dµ
Z
= |f|+dµ
Z
= (f+ − f−)dµ
Then, by Lemma 4.6.2 Part 2, we have that f+ and f− are integrable and thus, R f+dµ < ∞ and R f−dµ < ∞. Hence,
Z Z
f+dµ − f−dµ < ∞
And thus, we have that R fdµ = R f+dµ − R f−dµ is integrable, as required.
Problem 4.
Suppose f is an integrable function and fix a ∈ R. We have that,
Z Z Z
fdµ = f+dµ − f−dµ
with R f+dµ < ∞ and R f−dµ < ∞. Observe that f+,f− are thus nonnegative integrable functions. Thus, applying Theorem 4.4.5, we have that, af+ and af− are integrable. Thus,
R af+dµ < ∞ and R af−dµ < ∞ and so,
Z Z
+ −
af dµ − af dµ < ∞
But the above is precisely the definition of R afdµ, and so we must have that af is integrable.
Problem 5.
Suppose that f ≤ g a.e. Then of course f+ ≤ g+ a.e. If this were not the case, then there would be a set of positive measure on which f+ > g+, which, by the definition of f+ and g+, would imply that there is a set of positive measure on which f > g, a contradiction.
In addition, we have that f− ≥ g− a.e. Suppose that this is not the case. Then there is a set of positive measure on which f− < g−. But this implies that f(x) > g(x) for x in this set (either f(x) ≥ 0 or f(x) is a negative number greater than g(x)). This is a contradiction, and so we must have f− ≥ g− a.e. Hence, we have that,
Z Z Z
+ −
fdµ = f dµ − f dµ
where f+ ≤ g+ a.e. and f− ≥ g− a.e. Now let us consider R f+dµ and R g+dµ. We have,
Z Z + +
f dµ = sup{ sdµ : s is simple and 0 ≤ s ≤ f }
Z Z + +
g dµ = sup{ sdµ : sg is simple and 0 ≤ s ≤ g }
We have that f+ ≤ g+ a.e. Let sg be the supremum of simple function in the above set, and the same for sf. Then we must have sf ≤ sg a.e. as well. Observe that the set X where sf > sg is measure 0, and so it does contribute at all to the value of the integral by Corollary 4.3.3. Thus, we can disregard X when calculating the integral and so apply Theorem 4.3.2(2) which states that,
Z Z
sfdµ ≤ sgdµ
Since these were the supremum of simple functions approximating f+ and g+, we have
that,
Z Z
+ +
f dµ ≤ g dµ
Similarly, we have that,
Z Z
− −
f dµ ≥ g dµ
These two inequalities and the fact that all of these integrals are nonnegative give us that,
Z Z Z Z Z Z
+ − + −
fdµ = f dµ − f dµ ≤ g dµ − g dµ = gdµ
as required.
Problem 6.
Let f be an integrable function and suppose that RA fdµ = 0 for all measurable sets A. Then we have,
Z Z
+ −
f dµ − f dµ = 0
A A
Since both of the above integrals are nonnegative, we must have that RA f+dµ = 0 = R f−dµ. Since f+ and f− are both nonnegative measurable functions, by Problem 4.4.2
A
(solved on the previous HW), we have that f+ = 0 a.e. and f− = 0 a.e. on A. Let X be the set where f+ > 0 and Y be the set where f− > 0 and let Z = X ∪ Y . Then µ(Z) ≤ µ(X) + µ(Y ) = 0 + 0 = 0. Note that Z is precisely the set where f 6= 0 (since when f = 0 we have f+ = f− = 0 and when f 6= 0, one of f+ and f− is greater than 0). Hence, the set of values where f 6= 0 on A has measure 0 and so f = 0 a.e. on A.
Problem 7.
Suppose f is a nonnegative integrable function and that {Ep}p>0 is a sequence of decreasing (Ep+1 ⊂ Ep) measurable sets. Furthermore, suppose limp→∞ µ(Ep) = 0. We want to show that Z
fdµ = 0
∩p>0Ep
We know that since limp→∞ µ(Ep) = 0 that for every > 0, there exists N ∈ N such that for all n > N, we have that |En| < . Observe that since we have a decreasing sequence of sets, for any finite subset {E0,E1,…,Ek}, we have that Ek and so µ∩ik=0Ei = µ(Ek). Thus, by Proposition 2.5.2(2), we have
µ(∩p>0Ep) = lim µ(Ep)
p→∞
= 0
Note that since µ(∩p>0Ep), any measurable subset of ∩p>0Ep must have measure 0 as well. Thus, for any simple function s defined on µ(∩p>0Ep), we have that s = 0. But this implies that R∩p>0Ep sdµ for all simple functions s by the formulation of the integral given by Corollary 4.3.3. But if the integral of any simple function on ∩p>0Ep is 0, then we have that the integral of non-negative simple function must be 0 as well because, for a nonnegative function g,
Z Z
gdµ = sup{ sdµ : s is simple and 0 ≤ s ≤ g}
∩p>0Ep ∩p>0Ep
= sup{0}
= 0
Since f+ and f− are nonnegative measurable functions defined on ∩p>0Ep, we have that
Z Z
+ −
f dµ = 0 = f dµ
∩p>0Ep ∩p>0Ep
And, hence,
Z Z Z
+ −
fdµ = f dµ − f dµ
∩p>0Ep ∩p>0Ep ∩p>0Ep = 0 − 0 = 0
as required.
Problem 9.
Let f : X → R∗ be a measurable function and f is integrable. Now suppose |f(x)| = ∞ on a set X with µ(X) > 0. Then f = ∞ or f = −∞ (or both) on a set of positive measure. Thus, we have that either f+ = ∞ or f− = ∞ (or both) on a set of positive measure. Thus, a maximal s approximating simple function on f+ or f− (or both) must attain ∞ on a set of positive measure. Hence, we have s = Pn aiµ(Ai) = a0µ(A0)+a1µ(A1)+···+∞µ(Ak)+ ··· + anµ(An) = ∞. Since ∞ is the max value attainable in R∗ it must be the supremum of any subset of R∗ containing it, and so,
Z Z + +
f dµ = sup{ sdµ : s is simple and 0 ≤ s ≤ f }
= ∞
or,
Z Z
− −
f dµ = sup{ sdµ : s is simple and 0 ≤ s ≤ f }
= ∞
or both. But note that f is only Lebesgue integrable if both R f+ < ∞ and R f− < ∞. Hence, we have that f is not integrable, a contradiction. Thus, we must have that the set where |f(x)| = ∞ has measure 0. That is, |f(x)| < ∞ a.e.
