MEASURE THEORY

WOMP PROGRAM SEPTEMBER 16, 2004

1. Introduction: Failings of the Riemann Integral

There are two ways in which the Riemann integral fails to be adequate for the purposes of analysis. The first way is that the class of functions which you can integrate is rather small. But more seriously is the failure of the Riemann integral to behave well with limiting processes. Because in analysis it is in escapable to need to pass to limits, we really \underline{need} something better then the Riemann Integral.

Examples 1.

(1) Let C denote the Cantor set in [0, 1] and let f(x) = χ_{C}(x). Then f(x) is not Riemann integrable. But, when looked at the right way we can give a (seemingly) reasonable definition of ∫_{0}^{1} f(x) dx.

To do simply consider the following since 0 ≤ f(x) ≤ 1 on [0, 1] we better have that 0 ≤ ∫_{0}^{1} f(x) dx ≤ 1. But we can do better because f(x) = 0 on (1/3, 2/3) so the really when you think about it 0 ≤ ∫_{0}^{1} f(x) dx ≤ 1−1/3 by a similar token we can continue taking into account the intervals on which f(x) = 0 and we would get that 0 ≤ ∫_{0}^{1} f(x) dx ≤ 1 − 1/3 ∑_{k=0}^{∞}(2/3)^{k} = 0. Hence it seems reasonable to say that ∫_{0}^{1} f(x) dx = 0.

(2) Another shortcoming of the Riemann integral is that you can take a very nice sequence of integrable functions, whose limit fails to to be Riemann integrable. For example that

f_{n} = \begin{cases}1 & for x ∈ ℚ s.t. x = p/q with p, q ∈ ℤ and q ≤ n \\ 0 & otherwise \end{cases}

Then clearly f_{n}(x) → χ_{ℚ}(x), for each x. Notice further that for any n, ∫_{0}^{1} f_{n}(x) dx = 0. But I leave it as an exercise to show that χ_{ℚ} is not Riemann integrable.

The key idea to developing something better is to consider what is going on in the first Example. In order to integrate this function what we needed to understand was the size of the sets on which χ_{C} took its values. After examining the function we realized that χ_{C} took the value 0 on a set whose total length seemed to be 1. And we were integrating over a set whose length was 1, so there was “nothing” left over. It turns out that we can push this idea very far to integrate. But first we need to understand the “length”, “volume”, “size” or measure of sets.

2. Measurable spaces

Definition. Let X be a set and let M be a collection of subsets of X. We shall say M is a σ-algebra in X if

(1) ∅ is in ℳ
(2) If A is in ℳ then A^{c} = X \ A is in ℳ
(3) If A_{n} is in ℳ, for n = 1, 2, …, then A = ⋃_{i=1}^{∞} A_{n} is in ℳ

Further the sets in ℳ are called the measurable sets of X, and X (together with ℳ is called a measure space.

Remark. Take X and ℳ as above. Notice that we know X ∈ ℳ, finite unions are in ℳ (why?). Countable intersections of elements of ℳ are in ℳ, because ⋂_{i=1}^{∞} A_{i} = (⋃_{i=1}^{∞} A_{i}^{c})^{c}. The collection ℳ is also closed under finite intersections and set differences.

So our goal today will be talking about some basic properties of measure spaces and functions on them and apply this knowledge specifically to X = ℝ. Before we start constructing examples of measure spaces lets point out one useful facts that makes it much easier to find examples.

Example 1. Given any set X, let ℳ be the collection of all subsets of X. Then ℳ is a σ-algebra in X.

Exercise 1. If ℱ is any collection of sets of X show that there is a smallest σalgebra containing ℱ. This σ-algebra is called the σ-algebra generated by ℱ and is denoted by σ(ℱ).

Example 2. If we consider X = ℝ and take ℱ to be the collection of all intervals in ℝ, then σ(ℱ) is called the Borel sets of ℝ. Notice it contains all open sets (and hence all closed sets). All countable unions and intersections of open and closed sets, etc.

Remark. It is frequently useful to allow functions to take values of ±∞. For this purpose we introduce a topological space ℝ̄, which is as a set simply ℝ∪{∞}∪{−∞}, and has open sets given by sets of the form: (a, b), [−∞, b), (a, ∞]. (By this you know what I mean, and that is what is meant). This topological space is usually called the extended real numbers.

Definition. Let X be a measure space and let f : X → ℝ̄. We say f is a measurable function if f^{−1}(V) is a measurable set for any open set V ⊂ ℝ̄.

A quick question, why is this the right kind of function to think about?

Here is my favorite type of measurable function.

Exercise 2. Let E be a measurable set show that

χ_{E} = \begin{cases}1 & for x ∈ E \\ 0 & for x ∉ E \end{cases}

The first thing we will need about measurable functions is the following lemmas.

Lemma. Let X be a measure space. Let f : X → ℝ̄ be a measurable function and g : ℝ̄ → ℝ̄ be a continuous function then h = f ∘ g is a measurable function.

Proof. exercise □

Lemma. Let f and g be real measurable functions on a measure space X and let Φ : ℝ^{2} → ℝ be a continuous map. Then Φ(f(x), g(x)) is a measurable function.

Proof. Notice if we let F (x) = (f(x), g(x)) then we are trying to show that F ∘ Φ is a measurable mapping. By the previous lemma it is enough to show that F is measurable function. In order to do this first consider a open rectangle, R = I × J, in ℝ^{2} with sides parallel to the axes. Then F^{−1}(R) = f^{−1}(I) ∩ g^{−1}(J). Hence F^{−1}(R) is a measurable set by the measurability of f and g. Since any open set V = ⋃)_{i=1}^{∞}R_{i}, we have that

F^{−1}(V) = ⋃_{i=1}^{∞} F^{−1}(R_{i}).

But this is measurable because each F^{−1}(R_{i}) is measurable. □

What was the point of this lemma well...

Exercise 3. Show that if f and g are measurable functions then:

(1) |f| is a measurable function;

(2) f + g is a measurable function;

(3) fg is a measurable function.

Exercise 4. Let X be a measure space and suppose that f : X → ℝ̄ is a function such that f^{−1}((α, ∞]) is measurable for all α. Then show that f is a measurable function.

I had said that one of the most important operations for analysis was that of taking limits. Let’s see the class of measurable functions does under these operations.

Theorem. If f_{n} : X → ℝ̄ are measurable for n = 1, 2, 3, … then the following functions are measurable:

(1) g(x) = sup_{n≥1} f_{n}(x);

(2) g(x) = inf_{n≥1} f_{n}(x);

(3) g(x) = lim sup_{n→∞} f_{n}(x);

(4) g(x) = lim inf_{n→∞} f_{n}(x);

(5) g(x) = lim_{n→∞} f_{n}(x);

(6) g(x) = max{f_{1}(x), f_{2}(x)}

(7) g(x) = min{f_{1}(x), f_{2}(x)} Proof. For obvious reasons we will prove only 1 and 2 and leave the rest to the reader.

For 1 notice that g^{−1}((α, ∞]) = ⋃_{n=1}^{∞} f_{n}^{−1}((α, ∞]). Hence it is measurable. To see that 3 holds notice (by definition) we have that

lim sup_{n→∞} f_{n}(x) = inf_{n≥1} ( sup_{m≥n} f_{m}(x)) .

So the measurability follows from 1 and 2. □

Now let’s introduce a very important class of measurable functions.

Definition. A measurable function s : X → ℝ (not ℝ̄!) is said to be a simple function if its range consists of only finitely many values.

Notice if we define A_{i} = {x ∈ X | s(x) = α_{i}}, then we can write s(x) = ∑_{i=1}^{N} α_{i}χ_{A_{i}}. Also there are no restrictions on the A_{i} other than measurability, hence simple functions are quite different in character the step functions.

Why are simple functions so nice? well, they are simple. Also, if there is one theme in analysis it is, prove your theorems for a “nice” class of functions that its easy to work with then try to extend this to all functions your interested in by some limiting process. Simple functions turn out to allow to do just this much of the time.

Theorem. Let f : X → [0, ∞] be a measurable function. Then there exists simple functions s_{i} such that

(1) 0 ≤ s_{1} ≤ s_{2} ≤ ⋯ ≤ f, and
(2) s_{n}(x) → f(x) as n → ∞ for every x ∈ X.

Proof. Let A_{n}k = {x ∈ X | k2^{−n} ≥ f(x) < (k + 1)2^{−n}} and let B_{n} = {x ∈ X | f(x) ≥ n}. Then we take

s_{n} = \begin{cases}n & for x ∈ B_{n} \\ k2^{−n} & for x ∈ A_{n}k \end{cases} .

Now fix x and suppose that s_{n}(x) = k2^{−n}. In this case k2^{−n} ≤ f_{n}(x) < (k +1)2^{−n}, therefore (2k)2^{−n−1} ≤ f(x) < (2k + 2)2^{−n−1}. Hence s_{n+1}(x) = 2k or 2k + 1, but either way s_{n}(x) ≤ s_{n+1}(x). To see that lim_{n→∞} s_{n}(x) = f(x) simply notice that (f(x) − 2^{−n}) < s_{n}(x) ≤ f(x). □

3. Measures

Definition. A positive measure is a function μ defined on a σ-algebra, ℳ to [0, ∞] which is countably additive. That is, if A_{i}, i = 1, 2, 3, … are \underline{disjoint} then

μ (⋃_{i∈ℕ} A_{i}) = ∑_{i=1}^{∞} μ(A_{i}).

Assume throughout that there exists sets of finite measure.

Exercise 5. Let μ be a positive measure on a σ-algebra ℳ Show that

(1) μ(∅) = 0;

(2) if A_{1}, A_{2}, …, A_{n} are disjoint sets in ℳ then μ(A_{1} ∪ … ∪ A_{n}) = μ(A_{1}) + … μ(A_{n});

(3) if A ⊂ B are measurable sets then μ(A) ≤ μ(B);

(4) if A_{1} ⊂ A_{2} ⊂ A_{3} ⊂ ⋯ are measurable sets and A = ⋃_{n=1}^{∞} A_{n}, then μ(A) = lim_{n→∞} μ(A_{n}).

(5) Suppose μ(A_{1}) < ∞ and A_{1} ⊃ A_{2} ⊃ A_{3} ⊃ ⋯ are measurable sets. If A = ⋂_{n=1}^{∞} A_{n} then lim_{n→∞} μ(A_{n}) = μ(A).

Here are two interesting examples of measures.

Example 3.

(1) Let X ⊆ ℤ, and let ℳ consist of all subsets of X. Then we can define a measure μ on X by letting μ(A) be the number of elements of A if A is finite and μ(A) = ∞ if A has infinitely many elements.

(2) Let X and ℳ be any measure space. And fix x_{0} ∈ X. We define a measure μ on X by letting μ(A) = 1 if x_{0} ∈ A and 0 if x_{0} ∉ A. This usually called the “delta function” at x_{0} or “a unit point mass at x_{0}”. And usually denoted by δ_{x_{0}}

4. The big example, thank you Mr. Lebesgue.

Here we will \underline{outline} how one constructs Lebesgue measure. First the obvious, how does one measure rectangles in R^{n}.

Definition. Let R = I_{1} ×I_{2} ×⋯×I_{n}, where I_{k} are intervals in R (open, close, half open, what ever floats your boat.) Let a_{k} and b_{k} be the left and right hand endpoints of I_{k}. Then we define the measure of R, denoted by m(R) = ∏_{k=1}^{n}(b_{i} − a_{i}).

Just what you were expecting I hope. We will say that two rectangles are non overlapping if there intersection is along a face (if you prefer, their intersection has empty interior.)

Definition. If E = ⋃_{i=1}^{∞} R_{i} where R_{i} are non-overlapping rectangles then we define m(E) = ∑_{i=1}^{∞} R_{i}

Really need to check if this is consistent if E is a rectangle. You may if you wish.

Theorem. Every open set can be written as a union of non-overlapping closed rectangles.

This last theorem is included for culture. It’s meant to suggest I could already define the measure of arbitrary open sets if I pleased. But instead I am going to introduce a set function defined for all sets that over approximates the measure of a set.

Definition. Let A be an arbitrary set. Let

|A|_{e} = inf ∑_{k} m(R_{k})

Where the infimum is taken over all coverings of A by finite or countably many rectangles R_{k}

Now there are things to check like | · |_{e} and m agree whenever both are defined. But I will take this for granted.

Exercise 6. Show that A ⊆ B then |A|_{e} ≤ |B|_{e}.

Theorem. Let A = ⋃_{n} A_{n}. Then |A|_{e} ≤ ∑_{n} |A_{n}|_{e}

Proof. Let S_{n} = {I_{n1}, I_{n2}, …} where I_{n}k satisfy A_{n} ⊂ ⋃_{k} I_{nk} and

|A_{n}|_{e} ≤ ∑_{k} m(I_{nk}) + ϵ/2^{n}

□

Definition. We will say that a set, A is measurable if, given ϵ > 0, there is an open set G ⊇ A so that |G − A|_{e} < ϵ

That is we can approximate A by an open set so well that the over approximation to our error is less then ϵ.

Theorem. Every open set is measurable.

Theorem. If |Z|_{e} = 0 then Z is measurable.

Theorem. The countable union of measurable sets is measurable.

Proof. Suppose A = ⋃_{n} A_{n} where each A_{n} is measurable. Then for each n we take G_{n} so that |G_{n} − A_{n}|_{e} < ϵ/2^{n+1}. Let G = ⋃_{n} G_{n}, then G − A = ⋃_{n} G_{n} − ⋃_{n} A_{n} ⊆ ⋃_{n} G_{n} − A_{n}. Hence |G − A|_{e} ≤ ∑_{n} |G_{n} − A_{n}| ≤ ∑_{n} ϵ/2^{n+1} = ϵ. Therefore, A is measurable. □

Lemma. If F_{1} and F_{2} are bounded disjoint closed sets then |F_{1}+F_{2}|_{e} = |F_{1}|e+|F_{2}|_{2}

Theorem. Every closed set is measurable.

Theorem. The complement of a measurable set is measurable.

Proof. Suppose A is a measurable set. Then for each n there is an G_{n} so that |G_{n} − A|_{e} ≤ 1/n If G_{n}^{c} = F_{n} we have F_{n} ⊆ A^{c}. A^{c} − ⋃_{n} F_{n} ⊆ F_{N} for any N. Hence |A^{c} − ⋃_{n} F_{n}|_{e} ≤ |A^{c} − F_{N}|_{e} ≤ 1/N since A^{c} − F_{N} = G_{N} − A. Since this holds for every N we have that |A^{c} − ⋃_{n} F_{n}|_{e} = 0. This means if we right A^{c} = (A^{c} − ⋃ F_{n}) + ⋃_{n} F_{n} we see that A^{c} is the union of two measurable sets and hence measurable. To be more explicit | ⋃_{n} F_{n}| is measurable because it is the union of closed sets and closed sets are measurable. A^{c} − ⋃_{n} F_{n} is measurable because it has outer measure 0. □

This now tells us that the measurable sets form a σ-algebra. From here we just need to see that outer measure is a measure on this σ-algebra.

Theorem. The countable intersection of measurable sets is measurable

Proof. Let A = ⋂_{n} A_{n}, then A^{c} = ⋃_{n} A_{n}^{c} and hence is the union of measurable sets, and therefore is measurable. But A = (A^{c})^{c} so A is measurable. □

Theorem. If A_{1}, A_{2}, … are a countable sequence of measurable sets which are mutually disjoint and if A = ⋃_{n} A_{n}, then |A| = ∑_{n} |A_{n}|.

Proof. We have immediately that |A| ≤ ∑_{n} |A_{n}|. To show the reverse inequality we consider two cases.

Case 1: All Each A_{n} is bounded. Take a closed F_{n} ⊆ A_{n} so that |A_{n} − F_{n}| ≤ ϵ/2^{n}. Now

A ⊇ ⋃_{n=1}^{N} A_{n} ⊇ ⋃_{n=1}^{n} F_{n} so |A| ≥ | ⋃_{n=1}^{N} F_{n}|.

But since F_{n} are closed, bounded, and mutually disjoint then | ⋃_{n=1}^{n} F_{n}| = ∑_{n=1}^{N} |F_{n}|. Hence letting N → ∞, |A| ≥ ∑_{n=1}^{∞} |A_{n}|. We may write A_{n} = F_{n} + (A_{n} − F_{n}), so that |A_{n}| ≤ |F_{n}| + ϵ/2^{n} or |F_{n}| ≥ |A_{n}| − ϵ/2^{n}. Therefore

|A| ≥ ∑_{n=1}^{∞} |A_{n}| − ϵ ∑_{n=1}^{∞} 1/2^{n} = ∑_{n=1}^{∞} |A_{n}| − ϵ

. And the theorem follows from letting ϵ → 0.

Case 2: A_{n} may be unbounded. Left as an exercise to the reader. Idea, simply partition the whole space into bounded disjoint sets, and intersect. □

5. Finally we integrate

Now that we have all this machinery behind us let’s define the integral. A quick note on arithmetic on ℝ̄, the following conventions make the definitions much simpler (for a < ∞) ∞ + a = a + ∞ = ∞ (and you can decide how to handle −∞ and

subtraction) and if a > 0 then a · ∞ = \begin{cases}∞ & a ≠ 0 \\ 0 & a = 0 \end{cases}. Again I leave other cases to

you.

Definition. If s = ∑_{n=1}^{N} a_{n}χ_{En} is a simple function then we define the integral of s with respect to Lebesgue measure by ∫_{ℝ} s dx = ∑_{n=1}^{N} a_{n}|E_{n}|.

Much like you’d expect I hope. Now for positive functions.

Definition. Let f be a non-negative measurable function (i.e. f : ℝ → [0, ∞]). Then we define the integral of f to be

∫_{ℝ} f(x) dx = sup ∫_{ℝ} s dx

where the supremum is taken over all non-negative simple functions s ≤ f.

Exercise 7. Show that if f is a simple function then this definition gives the same integral as the previous definition.

Notice with this definition of integral nothing stops ∫_{ℝ} f(x) dx from being infinite, and that’s fine. But we reserve the expression f(x)isintegrable to mean that we know how to integrate f \underline{and} the integral of f is finite.

Definition. Lastly if f(x) is an arbitrary measurable function then let

f^{+} = \begin{cases}f(x) & for f(x) ≥ 0 \\ 0 & otherwise \end{cases}, and let f^{−}(x) = f^{+}(x)−f(x). Then if either ∫_{ℝ} f^{+}(x) dx <

∞ or ∫_{ℝ} f^{−}(x) dx < ∞ then we define

∫_{ℝ} f(x) dx = ∫_{ℝ} f^{+}(x) dx − ∫_{ℝ} f^{−}(x) dx

6. References

I recommend.... W. Rudin; \underline{Real and complex Analysis}. Kolmogorov, Fomin; \underline{Introductory Real Analysis} A. Zygmund, R. Wheeden; \underline{Measure and Integration}

Never read the actual this last book, but the section on the construction of Lebesgue comes from the notes on which the book is based.