[Solved] Dynamical Homework 10

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⁺d < and ^R fd < . Hence, ^R f⁺d+^R fd < . By Lemma 4.6.2 Part 2, we have that,

Z Z Z

f⁺d + fd = (f⁺ + f)d

<

By Exercise 2, we thus have that,

Z Z

+

(f + f )d = |f|d

<

Now suppose |f| is integrable. Then ^R |f|⁺d < and ^R |f|d < . 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|d = |f|⁺d |f|d

Z

= |f|⁺d

Z

= (f⁺ f)d

Then, by Lemma 4.6.2 Part 2, we have that f⁺ and f are integrable and thus, ^R f⁺d < and ^R fd < . Hence,

Z Z

f⁺d fd <

And thus, we have that ^R fd = ^R f⁺d ^R fd is integrable, as required.

Problem 4.

Suppose f is an integrable function and fix a R. We have that,

Z Z Z

fd = f⁺d fd

with ^R f⁺d < and ^R fd < . 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⁺d < and ^R afd < and so,

Z Z

+

af d af d <

But the above is precisely the definition of ^R afd, 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

+

fd = f d f d

where f⁺ g⁺ a.e. and f g a.e. Now let us consider ^R f⁺d and ^R g⁺d. We have,

Z Z

f d = sup{ sd : s is simple and 0 s f }

Z Z

g d = sup{ sd : s_g is simple and 0 s g }

We have that f⁺ g⁺ a.e. Let s_g be the supremum of simple function in the above set, and the same for s_f. Then we must have s_f s_g a.e. as well. Observe that the set X where s_f > s_g 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

s_fd s_gd

Since these were the supremum of simple functions approximating f⁺ and g⁺, we have

that,

Z Z

f d g d

Similarly, we have that,

Z Z

f d g d

These two inequalities and the fact that all of these integrals are nonnegative give us that,

Z Z Z Z Z Z

+ +

fd = f d f d g d g d = gd

as required.

Problem 6.

Let f be an integrable function and suppose that ^R_A fd = 0 for all measurable sets A. Then we have,

Z Z

+

f d f d = 0

A A

Since both of the above integrals are nonnegative, we must have that ^R_A f⁺d = 0 = ^R fd. 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 (Z) (X) + (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 {E_p}_p>0 is a sequence of decreasing (E_p+1 E_p) measurable sets. Furthermore, suppose lim_p (E_p) = 0. We want to show that Z

fd = 0

p>0Ep

We know that since lim_p (E_p) = 0 that for every > 0, there exists N N such that for all n > N, we have that |E_n| < . Observe that since we have a decreasing sequence of sets, for any finite subset {E₀,E₁,,E_k}, we have that E_k and so _i^k₌₀E_i = (E_k). Thus, by Proposition 2.5.2(2), we have

(_p>0E_p) = lim (E_p)

p

= 0

Note that since (_p>0E_p), any measurable subset of _p>0E_p must have measure 0 as well. Thus, for any simple function s defined on (_p>0E_p), we have that s = 0. But this implies that ^R_p>0Ep sd 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>0E_p 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

gd = sup{ sd : 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>0E_p, we have that

Z Z

+

f d = 0 = f d

p>0Ep p>0Ep

And, hence,

Z Z Z

+

fd = f d f d

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 (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 a_i(A_i) = a₀(A₀)+a₁(A₁)++(A_k)+ + a_n(A_n) = . 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 d = sup{ sd : s is simple and 0 s f }

=

or,

Z Z

f d = sup{ sd : 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.