9  Lebesgue Measure

9.1 Nonmeasurable Sets

It is not clear whether all subsets of \(\mathbb R\) are measurable for the Lebesgue measure \(\lambda\) (as defined in Theorem 7.1). The proof that there are nonmeasurable sets for \(\lambda\) will need the axiom of choice.

Let us look at some nonmeasurable sets for other measures.

Example 9.1  

  • Let \(X = \{0,1\}\), and \(\mathcal S = \{\varnothing, X\}\) be the trivial \(\sigma\)-algebra. Let \(\mu(X) = 1\). Then \(\mu^*(\{0\})=\mu^*(\{1\})=1\), so \(\{0\}\) and \(\{1\}\) are not measurable for \(\mu^*\).
  • Another less trivial example: Let \(X\) be any uncountable set, \(\mathcal S := \{A \subset X: A \text{ or } X \setminus A \text{ is countable}\}\). Let \(m(A)=0\) for \(A\) countable and \(m(A) = 1\) if \(A\) has countable complement. Then \(\mathcal S\) is a \(\sigma\)-algebra and \(m\) is a measure on it. Any uncountable set \(C\) with uncountable complement in \(X\) is nonmeasurable since \[ m^*(C) + m^*(X\setminus C)= 1 + 1 = 2 \ne m^*(X) = m(X)=1 \]

A set \(E \subset \mathbb R\) is called Lebesgue measurable iff \(E \in \mathcal M(\lambda^*)\). We know from Lemma 6.2 and Lemma 6.4 that \(\lambda^*\) restricted to \(\mathcal M(\lambda^*)\) is a measure which extends \(\lambda\). This measure is also called \(\lambda\), or Lebesgue measure. \(\mathcal M(\lambda^*)\) is the largest domain on which \(\lambda\) is generally defined. From Proposition 8.1 and Theorem 8.2, we also know that for every Lebesgue measurable set \(E\) therer is a Borel set \(B\) with \(\lambda(E\Delta B)=0\).

It is going to be proved that not all sets are Lebesgue measurable.

One might ask whether there is a Lebesgue measurable set \(A\) of positive measure whose comlement also has positive measure and which is “spread evenly” over the line in the sense that, for some constant \(r \in (0,1)\), for all intervals \(J\) with \(0 < \lambda(J) < \infty\), we have the ratio \(\lambda(A \cap J) / \lambda(J) = r\). It turns out there is no such set \(A\). For any set of positive measure whose complement also has positive measure, some of the ratios must be close to \(0\) and others close to \(1\).

Proposition 9.1 Let \(E\) be any Lebesgue measurable set in \(\mathbb R\) with \(\lambda(E)>0\). Then for any \(\varepsilon > 0\) there is a finite, nontrivial interval \(J = [a,b]\) such that \(\lambda(E\cap J)>(1-\varepsilon)\lambda(J)\)

We may assume \(\lambda(E) < \infty\) and \(\varepsilon < 1\).

Take finite intervals \(J_n := (a_n, b_n]\) such that \(E \subset \bigcup_n J_n\) and \(\sum_{n\ge 1}\lambda(J_n)\le \lambda(E) / (1-\varepsilon)\).

Then \(\lambda(E) \le \sum_{n\ge 1}\lambda(J_n\cap E)\) and for some \(n\), \((1-\varepsilon)\lambda(J_n)\le \lambda(J_n\cap E)\).

A set \(E\) can have positive Lebesgue measure, and in fact, its complement can have measure \(0\) without \(E\) including any interval of positive length (e.g. the set of irrational numbers). But if we take all differences between elements of a set of positive measure, the next fact shows we will get a set including a nontrivial interval around \(0\).

Proposition 9.2 (Steinhaus) If \(E\) is any Lebesgue measurable set with \(\lambda(E) > 0\), then for some \(\varepsilon > 0\), \[ E - E := \{x-y: x,y \in E\}\supset [-\varepsilon, \varepsilon] \]

By Proposition 9.1, we can take an interval \(J\) such that \(E \cap J\) occupies a significant portion of \(J\), with \(\lambda(E\cap J) > 3\lambda(J)/4\).

Let \(\varepsilon := \lambda(J) / 2\). For any set \(C\subset \mathbb R\) and \(x \in \mathbb R\), let \(C + x := \{y+x: y\in C\}\).

Then, if \(|x|\le\varepsilon\),

\[ \begin{align*} &(E\cap J)\cup ((E\cap J)+x)\subset J \cup(J + x), &\text{ and }\\ &\lambda(J \cup (J+x))\le 3\lambda(J)/2, \end{align*} \] as it is evident that \((E\cap J)\subset J\) and also its translation. By our choice of \(\varepsilon\), then the measure of \(J \cup (J+x)\) cannot exceed that much measure. On the other hand, we have

\[ \begin{align*} &\lambda((E\cap J)+x) = \lambda(E\cap J) > 3 \lambda(J)/4, &\text{ while } \\ &\lambda((E\cap J)\cup ((E\cap J)+x)) \le 3\lambda(J) / 2 & \end{align*} \] which forces an overlap:

\[ \begin{align*} &((E\cap J) + x)\cap (E \cap J) \ne \varnothing, &\text{ and } \\ &x \in (E\cap J) - (E\cap J)\subset E - E. \end{align*} \]

We will use the structure of measurable sets in showing that nonmeasurable sets exist; specifically, a set \(E \subset [0,1]\) with outer measure \(1\), so \(E\) is “thick” in the whole interval, but such that its complement is equally thick.

Theorem 9.1 Assuming the axiom of choice, there exists a set \(E \subset \mathbb R\) which is not Lebesgue measurable. In fact, there is a set \(E \subset I := [0, 1]\) with \(\lambda^*(E) = \lambda^*(I\setminus E) = 1\).

Recall that \(\mathbb Z\) is the set of all integers. Let \(\alpha\) be a fixed irrational number, say \(\alpha = \sqrt{2}\).

Let \(G\) be the following additive subgroup of \(\mathbb R\): \[ G := \mathbb Z + \mathbb Z \alpha := \{m + n \alpha: m, n \in \mathbb Z\} \\ \]

We show that \(G\) is dense in \(\mathbb R\).

  • Suppose \(c := \inf \{g : g \in G, g > 0\}\).
  • If \(c = 0\), let \(0 < g_n < 1/n\), \(g_n \in G\). Then \(G\supset \{m g_n: m \in \mathbb Z, n = 1,2,...\}\), a dense set.
  • If \(c > 0\) and \(g_n\downarrow c\), \(g_n \in G\), \(g_n > c\), then \(g_n - g_{n+1} > 0\), belong to \(G\) and converge to 0, so \(c = 0\), a contradiction.
  • So \(c \in G\) and \(G = \{mc: m \in \mathbb Z\}\), a contradiction since \(\alpha\) is irrational.

Let \(H\) be the subgroup: \[ H := \{2m+n\alpha: m,n \in \mathbb Z\} \] Likewise, \(H\) and \(H+1\) are dense.

The cosets \(G+y\), where \(y \in \mathbb R\) are disjoint or identical. These cosets form a partition of \(\mathbb R\). By the axiom of choice, let \(C\) be a set containing exactly one element of each coset. Then \(\mathbb R = G + C\).

Let \(X := C+H\). Note that, \(G = H \cup (H+1)\). Then \(\mathbb R \setminus X = C + H + 1\).

Now \((X - X)\cap (H+1)=\varnothing\), since \(X-X=\{(c_1 - c_2) + (h_1 - h_2)\}\) and any difference \(c_1 - c_2\) cannot belong to \(G\) unless \(c_1 = c_2\). Because \(H+1\) is dense, by Proposition 9.2, \(X\) does not include any measurable set with positive Lebesgue measure.

Let \(E := X \cap I\). Then \(\lambda^*(I \setminus E) = 1\). Likewise, \((\mathbb R \setminus X) - (\mathbb R \setminus X) = (C+H+1)-(C+H+1) = (C+H)-(C+H)\) is disjoint from \(H+1\), so \(\lambda^*(E) = 1\).

Hence, Lebesgue measure is not defined on all subsets of \(I\).