Lemma 3.15. (Vitali type covering lemma. ) Let be a collection of open balls in , and let . If , there exist disjoint such that
Proof. The set is open, then for there exists a compact set such that
.
Since is compact, there exists finitely many balls such that . Let be the collection of the balls. Choose that has the largest radius. Let be the collection of balls from that is disjoint from . Choose the ball that has the largest radius. In general, suppose that are choosen. Let be the collection of balls in that are disjoint from the previous balls . Choose that has the largest radius. Since there are only balls, the process will stop. We assume that are the finally chosen balls. Given any ball , we claim that will intersect with some ball from . Otherwise, one more ball will be chosen. A contradiction. Hence
So .
Definition. A measurable function is called locally integrable with respect to the Lebesgue measure if for every bounded measurable set The space of such functions is denoted by
Definition. If , we define
.
Lemma 3.16. If , is jointly continuous in .
Proof. Given . Since , . By Corollary 3.6, given , there exists such that if ,
So given , if ,
.
So So
Hence as ,
Definition. If , the Hardy-Littlewood maximal function by
Remark. The function is measurable for
is open for any .
Theorem 3.17. There is a constant such that for any and ,
.
Proof. Let . For , there exists such that
.
Let . For any , by Vitali type lemma 3.15, there exists disjoint balls ,
Note that the right hand side does not depend on . Let , we prove that
This proves Theorem 3.17.
We shall use the notion of limit superior for real valued functions of a real variable.
So is equivalent to
Theorem 3.18. If , then .
Proof. Since we are investigating continuity of a funtion at a point, it is a local property. So we may assume that . Secondly, it is easy to see that the claim is true for a continuous function.
Now given , there exists a continuous function such that Consider
Hence
Given any ,
By the maximal theorem,
Let , we see that . We rationalize , we see that
Definition. For any , define the Lebesgue set of to be
.
Theorem. If , then
Proof. For any , . Then for ,
Let . Then
For any , there exists such that ,
So
Let ,
Thus . Since ,
Definition. A Borel measure is called “regular" if (i). for every compact . (ii). , for any . A signed measure or complex measure is called
“regular” if is regular.
Theorem 3.22. Let be a regular signed or complex Borel measure on , and be its Lebesgue-Radon-Nikodym representation. Then for ,
.
Proof. We make two observations. (i). If is regular, so are . (ii). We may assume that , because .
We know that are mutually singular. Let be a Borel set such that Let
. We will show that for all , which will complete the proof with the aid of the Lebesgue-Radon-Nikodym theorem.
By the regularity of , for any , there exists open set such that and For any , there exists such that . Let and , there exists and disjoint balls such that
Let . We see that . Since , . So
Theorem 3.23. Let be increasing, and let .
(a). The set of points at which is discontinuous is countable.
(b). and are differentiable , and
Proof. (a). Let . It is clear that is the set of points at which is discontinuous. Since is increasing, for ,
Also for , the interval contains a rational number. So is at most countable.
(b). We observe that is increasing and right continous. except perhaps where is discontinuous. Moreover
We know that is regular. So by Theorem 3.22, exists .
Let . We need to show that exists and equals zero Let be an enumeration of points at which , i.e., . Given any ,
Let . Then is a Borel measure that is finite on compact sets by . Then is a Lebesgue-Stieltjes measure.
We know that
. This goes to zero as for by Theorem 3.22. Indeed, let . Then since is at most countable; . So . Thus exists , and
Definition. If and , we define
is called the total variation of $F$.
Remark. It is easy to observe that for $a<b$,
Thus is an increasing function with values in .
Definition. (1). If , is said to be of bounded variation on . We denote the space of all such functions by .
(2). The supremum in is called the total variation of on . If it is finite, then we say that .
Examples. (1). If is bounded and increasing, then
(2). If and , then
(3). If is differentiable on and is bounded, then . Hint: using the mean value theorem.
(4). If , then for , but One can see it by taking .
(5). If for and , then for . One can see it by taking .
Lemma 3.26. If and is bounded, then is increasing.
Proof. If , then by ,
.
Theorem 3.27. (1). iff .
(2). If , then iff is the difference of two bounded increasing functions. For the direction, we take .
(3). If , then exist for all . This follows from (1) and (2).
(4). If , the set of points at which is discontinuous is countable. This follows from (1) and (2).
(5). If and , then exist and are equal a.e. This follows from (1), (2) and Theorem 3.23.
Definition. NBV=\{F\in BV:\, F \text{ is right continuous and } F(-\infty) =0\}.
Remark. If , then the function defined by is in NBV and . Indeed, ; so the difference of two increasing functions. is again the difference of two increasing functions. So . is obvious.
Theorem 3.28. If , then . If is also right continuous, so is .
Proof. Given and , then by the definition of , there exists such that
So by ,
. This proves that Since is increasing, for all . So .
For the second claim, we define and assume that . For any , since $F$ is right continuous and is defined, there exists such that for ,
For any such , there exists such that
, and so
Similarly on the interval , we apply the same reasoning to conclude that there exists so that
.
So on ,
On the other hand,
So that . Since is arbitrary, a contradiction. Therefore .
Theorem 3.29. If is a complex measure on and , then . Conversely if , there is a unique complex Borel measure such that ; moreover .
Proof. Suppose that is a complex measure. By decomposing into real and complex parts, and then considering the positive and negative parts of measures,
. Since is a complex measure, are finite positive measures. Let . Then are right continuous, increasing functions. Also and . By Theorem 3.27 (1) and (2),
Conversely, for , we write by Theorem 3.27 (1) and (2). Then by Theorem 3.27, each is bounded and increasing. Then . By Theorem 3.28, is right continuous and so $T_F$ is right continuous. So each is right continuous by Theorem 3.28. Also it is easy to see that each by Theorem 3.28 again. So by Theorem 1.16, for some finite Borel measure . Let .
The last step is to show that . It is contained in Exercise 28 and 21 in Folland’s book.
Theorem 3.30. If , then . Morevoer iff , and iff
Proof. Since , then $\mu_F$ in Theorem 3.29 is a complex Borel measure that is also regular. Then by the Radon-Nikodym theorem,
, where , where by Theorem 3.22. The rest conclusions follows easily from the Radon-Nikodym representation above.
Proposition 3.32. If , then is absolutely continuous iff .
Proof. If , then by Theorem 3.5, we see that the claim holds.
We need to prove that for , if , then . Since is absolutely continuous, then for any , there exists such that for any finite disjoint intervals , then if ,
Since , there exists a sequence of open sets such that , latex \mu_F$ is regular, we can find a sequence of open sets such that such that . That is to say, as . Therefore is a sequence of open sets that are decreasing and contains , and moreover . We abbreviate it by .
Each is an at most countable union of disjoint open intervals . For any ,
So Since , we see that
Since is arbitrary, .
Corollary 3.33. If , then the function is in NBV and is absolutely continuous, and Conversely if is absolutely continuous, then and .
Proof. Suppose that . Then , right continuous and by dominated convergence theorem. So That is absolutely continuous follows from Corollary 3.6. By Theorem 3.32, the deduced measure and also . by Theorem 3.30. On the other hand, gives rise to the same function . By the uniqueness in Theorem 3.29, $fdm$ and $F’dm$ are the same complex Borel measures. Hence
Conversely, if and is absolutely continuous, then by Theorem 3.32. Hence by Theorem 3.30, Considering the real and imaginary parts of complex valued functions, and positive and negative parts of real-valued functions, we see that because is a complex Borel measure.
Lemma 2.34 If is absolutely continuous on , then .
Proof. We know that is absolutely continuous. Let be given. There exists such that for disjoint intervals with , then
Let $N$ be the smallest integer that is larger than . We divide into consecutive segments with length that is at most . For any points , if necessary adding more endpoints of the previous consecutive segments , so that these points can be grouped into subgroups. On each subgroup, if the points are denoted by , So the total sum is majorized by This holds true for any partition of . So .
Theorem 3.35 (The fundamental theorem of calculus for Lebesgue integrals.) If and , the following are equivalent:
(a) is absolutely continuous on $[a,b]$
(b) for some .
(c) is differentiable on , , and .
Proof. By subtracting and extending to trivially, we may assume that by Lemma 3.34. So by Corollary 3.33, follows.
is trival.
Extending trivially to , i.e., . Then we invoke Corollary 3.33.