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The material of this chapter is concrete: here, for the first time, we fix \(X=\mathbb {R}^d\) and apply the abstract CarathΓ©odory theorem (the chapter on the CarathΓ©odory theorem) to a concrete premeasure β the volume of rectangles β in order to obtain the concrete Lebesgue measure \(\lambda\).
Lebesgue measure: applying the extension theorem to construct volume in \(\mathbb {R}^{d}\)
Let us apply the CarathΓ©odory theorem on the extension of a premeasure to an important special case β the rigorous construction of the notion of \(d\)-dimensional volume (length for \(d=1\), area for \(d=2\)) in \(\mathbb {R}^{d}\).
Definition 1 Let \(\mathcal{S} = \{ (a, b]: \; -\infty < a \leq b < +\infty \}\) be the semiring of half-open intervals on \(\mathbb {R}\). Define \(\lambda : \mathcal{S} \to [0, +\infty ]\) by \(\lambda ((a, b]) := b-a\). We have already shown that this function is a premeasure (a countably additive mapping on a semiring). Then \(\lambda^*\) is called the outer Lebesgue measure on \(\mathbb {R}\), and \(\mathcal{M}(\lambda^*)\) is the collection of Lebesgue-measurable sets.
The Lebesgue measure is the restriction of the outer Lebesgue measure to \(\mathcal{M}(\lambda^*)\). Strictly speaking, it should be denoted \(\lambda^*|_{\mathcal{M}(\lambda^*)}\), but for simplicity we will often just write \(\lambda\), since \(\lambda^*\) agrees on the semiring of half-open intervals with the measure \(\lambda\) originally introduced, by the CarathΓ©odory theorem. The Borel measure (or Lebesgue-Borel measure) is the restriction of the outer Lebesgue measure to \(\sigma (\mathcal{S}) = \mathcal{B}(\mathbb {R})\). The Borel measure is also frequently called the Lebesgue measure and denoted \(\lambda\) when it is clear from context that the outer measure \(\lambda^*\) is being considered precisely on Borel sets. By the CarathΓ©odory theorem, \((\mathbb {R}, \mathcal{M}(\lambda^*), \lambda )\) is a complete measure space.
Lebesgue measure on \(\mathbb {R}^{d}, d \geq 2\) is introduced analogously. It is denoted \(\lambda^d\) or simply \(\lambda\). Half-open intervals in \(\mathbb {R}^{d}: (\overrightarrow {a}, \overrightarrow {b}] = (a_1, b_1] \times \dots \times (a_d, b_d]\), \(\overrightarrow {a},\overrightarrow {b} \in \mathbb {R}^{d}\). Most of the statements and problems on Lebesgue measure in \(\mathbb {R}\) presented below can be generalized to the corresponding statements in \(\mathbb {R}^{d}\). For simplicity, however, we will typically state them in \(\mathbb {R}\).
Lemma 1 (Corollary of the premeasure extension theorem) Every Borel set is Lebesgue measurable.
The inclusion \(\mathcal{B}(\mathbb {R}) \subset \mathcal{M}(\lambda^*)\) is strict, if only because \(\left|\mathcal{B}(\mathbb {R}))\right| = \left|\mathbb {R}\right| < \left|2^{\mathbb {R}}\right| = \left|\mathcal{M}(\lambda^*)\right|\) . We will explicitly construct a non-Borel, Lebesgue-measurable set in one of the upcoming seminars. The inclusion \(\mathcal{M}(\lambda^*) \subset 2^{\mathbb {R}}\) is also strict, if we accept the axiom of choice. You should already be familiar with an example of a non-Lebesgue-measurable subset of the real line β this is the Vitali set.
Let \(A\) be the set of points on \([0,1]\) whose decimal expansion does not contain the digit 1. Prove that \(A \in \mathcal{M}(\lambda^*)\). Find \(\lambda (A)\).
Properties of Lebesgue measure
Theorem 1 (Regularity of Lebesgue measure) The Lebesgue measure of measurable sets can be approximated from above by measures of open sets and from below by measures of compact sets. That is, if \(A \in \mathcal{M}(\lambda_n^*) \subset 2^{\mathbb {R}^{n}}\), then \[
\begin{align} \lambda _n(A) & = \inf \left\{ \lambda _n(U): \; U \text{ -- open and } A \subset U\right\} \\ \lambda _n(A) & = \sup \left\{ \lambda _n(K): \; K \text{ -- compact and } K \subset A\right\} \end{align}
\]
Prove the regularity of Lebesgue measure.
Theorem 2 (Translation invariance) If \(a \in \mathbb {R}^{n}, A \subset \mathbb {R}^{n}\), then \(\lambda_n^*(A) = \lambda_n^*(A + a)\) and \(A \in \mathcal{M}(\lambda_n^*) \iff A + a \in \mathcal{M}(\lambda_n^*)\). It also holds that \(A \in \mathcal{B}(\mathbb {R}^{n}) \iff A + a \in \mathcal{B}(\mathbb {R}^{n})\)
Show that for any number \(\alpha \in (0,1)\) there exists a closed, nowhere dense set \(C \subset [0,1]\) such that \(\lambda (C) = \alpha\).
This problem disproves the intuitively plausible claim that every subset of \(\mathbb {R}\) of positive Lebesgue measure contains an interval of positive length. However, see Theorem Theorem 3.
Modify the construction scheme for the Cantor set, at step \(n\) removing \(2^{n-1}\) intervals not of fixed length \(\frac{1}{3^n}\), but of an arbitrary length \(p^n, 0 < p \leq \frac{1}{3}\)
The following statement is useful in some problems.
Theorem 3 Let \(A \subset \mathbb {R}^{n}\). Denote \(\operatorname {diff}(A) := \left\{ x-y: x \in A, y \in A\right\}\). Then if \(A \in \mathcal{M}(\lambda^*)\) and \(\lambda (A) > 0\), \(\operatorname {diff}(A)\) contains some neighborhood of zero.
In other words, for any measurable set of positive measure \(A \subset \mathbb {R}^{n}\) there exists \(\varepsilon > 0\) such that for any vector \(z \in U_\varepsilon (0)\) of length \(\left\| z\right\| < \varepsilon\), in \(A\) there exist elements \(x,y \in A\) such that \(x - y = z\).
Prove Theorem Theorem 3.
Use the regularity of Lebesgue measure. Note that the existence of an \(\varepsilon\)-neighborhood of the zero vector in \(\operatorname {diff}(A)\) is equivalent to \(A\) and \(A+x\) having nonempty intersection for any \(x\) such that \(\left\| x\right\| < \varepsilon\).
We consider the Lebesgue measure \(\lambda\) on the space
\(\mathbb {R}^{n}\), \(n \geq 2\);
\(\mathbb {R}\). Is the converse of the theorem on \(\operatorname {diff}\) true, i.e., is it true that if \(A \in \mathcal{M}(\lambda^*)\) and \(\operatorname {diff}(A)\) contains some neighborhood of zero, then \(\lambda (A) > 0\)?
Let \((X, \mathcal{F}, \mu )\) be a space with a \(\sigma\)-finite measure. Prove that \(X\) cannot be partitioned into an uncountable family of sets of positive measure. That is, there is no family \(\left\{ B_{\lambda }\right\}_{\lambda \in \Lambda } \subset \mathcal{F}\) such that \(\mu (B_{\lambda }) > 0\) for all \(\lambda \in \Lambda\), \(B_{\lambda_1} \cap B_{\lambda_2} = \varnothing\) for \(\lambda_1 \neq \lambda_2\), and \[
X = \bigsqcup _{\lambda \in \Lambda }B_{\lambda }
\]
Let \(\lambda\) be Lebesgue measure on \(\mathbb {R}^{n}\). Prove that the Lebesgue measure of any hyperplane of dimension less than \(n\) equals \(0\).
Lebesgue-Stieltjes measure
Definition 2 Consider Definition Definition 1. Suppose that instead of the original premeasure on the semiring \((a, b] \mapsto b - a\) we take the mapping \((a, b] \mapsto F(b) - F(a)\) for some nondecreasing, right-continuous function \(F: \mathbb {R} \to \mathbb {R}\). Denote it by \(\mu_F\). We have already shown that \(\mu_F\) is also a premeasure on \(\mathcal{S}\) (see the proof of countable subadditivity of \(\lambda_F\)). Hence the CarathΓ©odory extension theorem can be applied to \(\mu_{F}\) on the semiring of half-open intervals.
The restriction of \(\mu_F^*\) to \(\sigma (\mathcal{S}) = \mathcal{B}(\mathbb {R})\) is called the Lebesgue-Stieltjes measure. On half-open intervals \((a, b]\) its value agrees with \(\mu_F\). Therefore, for simplicity, we will denote \(\mu_F^*\) by \(\mu_F\). \((\mathbb {R}, \mathcal{B}(\mathbb {R}), \mu_F)\) is a measurable space equipped with a measure.
Suppose the right-continuous, nondecreasing function \(F: \mathbb {R} \to \mathbb {R}\) is piecewise constant and has finitely many points of discontinuity. That is, there exist points \(-\infty < x_{1} < x_2 < \ldots < x_{n} < \infty\) such that \(f\) is discontinuous only at these points and is constant on the intervals \((-\infty , x_{1})\), \((x_1, x_2)\), \((x_2, x_3)\), \(\ldots\), \((x_{n}, +\infty )\).
Find the class of sets that are CarathΓ©odory measurable with respect to the Lebesgue-Stieltjes measure \(\mu_{F}\).
Choose a function \(F: \mathbb {R} \to [0,1]\) such that every Borel subset of \((-\infty , 0]\) has zero Lebesgue-Stieltjes measure, while \((a,b]\) has measure \(e^{-a} - e^{-b}\) for \(0 \leq a < b\). In other words, \(\mu_{F}(A) = 0\) for \(A \in \mathscr {B}\left((-\infty , 0]\right)\), and \[
\mu _{F}\left((a,b]\right) = e^{-a} - e^{-b}, \qquad \forall a,b: 0 \leq a < b
\]