7.5: Measure Theory Review

The beginning of this course does not require the Lebesgue integral, however, knowing it may make some of the earlier results easier to understand and the exercises easier to do. In some of the later chapters, Lebesgue integral does become necessary in several places. To make the first reading of the entire book easier for a student who has not had a course on measure theory yet, we present the basic ideas of the Lebesgue integral and list the results that make it so useful. We avoid getting into the details of the definition and simply state the useful results without proof. A reader who is interested can consult, for example, [R1].

Given a set \(X\), we designate a collection \(\mathcal{M}\) of subsets of \(X\), called the measurable sets. On these sets we define a measure, that is, a function \(\mu \colon \mathcal{M} \to \mathbb{R}\), such that \(\mu \geq 0\), \(\mu(\emptyset) = 0\), and \(\mu\) is \(\sigma\)-additive, that is, the measure of a union of countably many disjoint sets is the sum of the measures. If \(X\) is the euclidean space \(\mathbb{R}^n\), there always exists a measure called the Lebesgue measure that will agree with the standard \(n\)-dimensional volume on simple sets such as rectangles. A complication is that not all subsets of \(\mathbb{R}^n\) can then be measurable. We say that \((X,\mathcal{M},\mu)\) is a measure space.

A function \(f \colon X \to \mathbb{R}\) is called measurable if its sublevel sets are measurable. Since one generally has to work hard to produce a nonmeasurable function in the usual measure spaces we consider, the reader may be forgiven for assuming every function that appears in this book is measurable. A function is simple if its support is of finite measure and it only has finitely many values, in which case the integral is defined as \[\int_X f\, d \mu \overset{\text{def}}{=} \sum_{y \in f(X)} y \, \mu\bigl(f^{-1}(y)\bigr) .\] That is, on the set where \(f(x)=y\) we define the integral as the value of the function times the measure of the set and then we add these up. If the function is actually a step function and the measure was the Lebesgue measure, this is the same as would be done for the Riemann integral. The integral of any nonnegative \(f\) is defined by \[\int_X f\, d \mu \overset{\text{def}}{=} \sup_{\substack{\varphi \leq f \\ \varphi \text{ is simple}}} \int_X \varphi \, d\mu .\] The integral of any real-valued measurable function is then defined by writing \(f = f_+ - f_-\) for nonnegative functions \(f_+\) and \(f_-\), as long as the integrals of at least one of these is not infinite. Similarly the integral of complex-valued functions is defined by writing \(f = u + i \, v\). The most common class of functions to deal with is then the \(L^1\)-integrable functions, which are the functions such that \[\int_X |f| \, d\mu < \infty .\] Note that for the purposes of integration we often allow nonnegative functions to take on the value \(\infty\) at some points. In general, we allow our functions to be be undefined on a set of measure zero if we are integrating them since changing a function on a measure zero set does not change the integral.

A couple of things to notice about the definition is the following. First, because step functions are simple functions with respect to the Lebesgue measure, the integration is a generalization of the Riemann integral on the real line and on \(\mathbb{R}^n\) in general in the sense that the two integrals agree when they are both defined.

Second, many more functions (all measurable functions in fact) can be limits of simple functions, and the integral is defined as a limit of such integrals, one would, rightly, expect that limits can easily pass under the integral and we no longer need to worry about integrability of the limit.

Besides integration, one often forgotten feature of this setup is that it also applies to series. For a countable set \(X\) such as \(\mathbb{N}\) we can define the so-called counting measure, where every set \(S \subset X\) is measurable and \(\mu(S)\) is simply the number of elements in \(S\). If \(z_n = f(n)\) is a function defined on \(X\), then we write \[\sum_{n \in X} z_n = \int_X f \, d\mu .\] So the following theorems also apply to series, where being \(L^1\) simply means that the series is absolutely summable.

We say that something happens almost everywhere if the set where it does not happen is of measure zero. Similarly we may say that this something happens for almost every \(x \in X\). Note that if a sequence of measurable functions converges to a function \(f\) almost everywhere, then this function can be assumed to be measurable (it is equal almost everywhere to a measurable function). We have the following three theorems, which despite appearances are actually just equivalent to each other, but each statement is useful in different situations.

Theorem \(\PageIndex{1}\)

Fatou's Lemma

Let \((X,\mathcal{M},\mu)\) be a measure space and \(\{ f_n \}\) a sequence of nonnegative measurable functions that converges almost everywhere to a function \(f\). Then \[\int_X f \, d\mu \leq \liminf_{n\to \infty} \int_X f_n \, d\mu .\]

Theorem \(\PageIndex{2}\)

Monotone Convergence Theorem

Let \((X,\mathcal{M},\mu)\) be a measure space and \(\{ f_n \}\) a sequence of nonnegative measurable functions that converges pointwise to a function \(f\). Then \[\int_X f \, d\mu = \lim_{n\to \infty} \int_X f_n \, d\mu .\]

Theorem \(\PageIndex{3}\)

Dominated Convergence Theorem

Let \((X,\mathcal{M},\mu)\) be a measure space and \(\{ f_n \}\) a sequence of measurable complex-valued functions that converges almost everywhere to a function \(f\). Suppose \(g\) is a nonnegative \(L^1\)-integrable function such that \(|f(x)| \leq g(x)\) for almost every \(x\). Then \[\int_X f \, d\mu = \lim_{n\to \infty} \int_X f_n \, d\mu .\]

A common application of the dominated convergence theorem is differentiating under the integral.

Theorem \(\PageIndex{4}\)

Differentiation under the Integral

Let \(U \subset \mathbb{R}\) be open and \((X,\mathcal{M},\mu)\) be a measure space. Suppose \(f \colon U \times X \to \mathbb{C}\) is a function such that for each \(t \in U\), \(x \mapsto f(t,x)\) is \(L^1\)-integrable, for almost every \(x\), \(\frac{\partial f}{\partial t}\) exists on all of \(U\), and there exists an \(L^1\)-integrable \(g \colon X \to [0,\infty]\) such that \(\left|\frac{\partial f}{\partial t}(t,x)\right| \leq g(x)\) for all \(t \in U\) and almost every \(x \in X\). Then for all \(t \in U\), \[\frac{d}{dt} \int_X f(t,x) \, d\mu(x) = \int_X \frac{\partial f}{\partial t}(t,x) \, d\mu(x) .\]

A measure space is \(\sigma\)-finite if it is a countable union of sets of finite measure. For example, the euclidean space with Lebesgue measure is \(\sigma\)-finite because it is a union of balls, which are of finite measure. We often want to write an integral over a product space as an iterated integral, such as writing an integral over a subset of \(\mathbb{R}^n\) using \(n\) one-dimensional integrals. If \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) are product spaces, we can define a product measure space by requiring that \(\mu \times \nu (A \times B) = \mu(A)\nu(B)\) (we again skip details). First, for nonnegative functions we obtain the following simple theorem where no integrability needs to be checked, and we are allowing things to be infinite if needed.

Theorem \(\PageIndex{5}\)

Tonelli

Suppose \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) are \(\sigma\)-finite measure spaces and \(f \colon X \times Y \to \mathbb{R}\) is a nonnegative measurable function. Then:

1. For almost every \(x \in X\), \(y \mapsto f(x,y)\) is measurable, and for almost every \(y \in Y\), \(x \mapsto f(x,y)\) is measurable.
2. The functions \(y \mapsto \int_X f(x,y) \, d\mu(x)\) and \(x \mapsto \int_Y f(x,y) \, d\nu(y)\) are measurable.
3. \[\begin{align}\begin{aligned} \int_{X \times Y} f(x,y) d(\mu\times \nu) & = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y) \\ & = \int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) . \end{aligned}\end{align}\]

In general there is the Fubini theorem. A measure is complete, if every subset of a measure zero set is also measurable. A measure can be completed by simply throwing those sets in, but it is a minor technicality that the product of two measure spaces is not in general complete and must be completed. This is an issue for measurability of the functions involved, but the functions that one usually considers in applications are easily shown measurable in all of these measure spaces and their completions.

Theorem \(\PageIndex{6}\)

Fubini

Suppose \((X,\mathcal{M},\mu)\) and \((Y,\mathcal{N},\nu)\) are complete measure spaces and \(f \colon X \times Y \to \mathbb{R}\) is \(L^1\)-integrable. Then:

1. For almost every \(x \in X\), \(y \mapsto f(x,y)\) is \(L^1\)-integrable, and for almost every \(y \in Y\), \(x \mapsto f(x,y)\) is \(L^1\)-integrable.
2. The functions \(y \mapsto \int_X f(x,y) \, d\mu(x)\) and \(x \mapsto \int_Y f(x,y) \, d\nu(y)\) is \(L^1\)-integrable.
3. \[\begin{align}\begin{aligned} \int_{X \times Y} f(x,y) d(\mu\times \nu) & = \int_Y \left( \int_X f(x,y) \, d\mu(x) \right) d\nu(y) \\ & = \int_X \left( \int_Y f(x,y) \, d\nu(y) \right) d\mu(x) . \end{aligned}\end{align}\]

Tonelli theorem is often applied in tandem with the Fubini theorem. Tonelli establishes integrability and Fubini is used to write the integral we need as iterated integral, or swap the order of integrations.

Note that the Tonelli and Fubini theorems are very useful in simplifying the development of the power series by using the counting measure as we mentioned above. They are also useful for swapping a series summation and integration such as \[\int_0^1 \sum_{n=1}^\infty a_n(x) \, dx = \sum_{n=1}^\infty \int_0^1 a_n(x) \, dx .\]