Probability with Measure

1.4 The Borel \(\sigma \)-field and Lebesgue Measure

In this section we take \(S\) to be the real number line \(\R \). We want to describe a measure \(\lambda \) that captures the notion of length as we discussed at the beginning of this chapter. So we should have \(\lambda ((a,b)) = b-a\). The first question is - which \(\sigma \)-field should we use? We have already argued that the power set \({\cal P}(\R )\) is too big. Our \(\sigma \)-field should contain open intervals, and also unions, intersections and complements of these.

Note that \({\cal B}(\R )\) also contains isolated points \(\{a\}\) where \(a \in \R \). To see this first observe that \((a, \infty ) \in {\cal B}(\R )\) and also \((-\infty , a) \in {\cal B}(\R )\). Now by S(iii),\((-\infty , a] = (a, \infty )^{c} \in {\cal B}(\R )\) and \([a, \infty ) = (-\infty , a)^{c} \in {\cal B}(\R )\). Finally as \(\sigma \)-fields are closed under intersections, \(\{a\} = [a, \infty ) \cap (-\infty , a] \in {\cal B}(\R )\). You can show that \({\cal B}(\R )\) also contains all closed intervals (see Problem ??).

We make two observations:

• 1. \({\cal B}(\R )\) is defined quite indirectly and there is no “formula” that can be used to give the most general element in it. However it is very hard to find a subset of \(\R \) that isn’t in \({\cal B}(\R )\) – we will give an example of one in Section 1.5.

• 2. \({\cal B}(S)\) makes sense on any set \(S\) for which there are subsets that can be called “open” in a sensible way. In particular this works for metric spaces. The most general type of \(S\) for which you can form \({\cal B}(S)\) is a topological space.

The measure that precisely captures the notion of length is called Lebesgue measure in honour of the French mathematician Henri Lebesgue (1875-1941), who founded the modern theory of integration. We will denote it by \(\lambda \). First we need a definition.

Let \(A \in {\cal B}(\R )\) be arbitrary. A covering of A is a finite or countable collection of open intervals \(\{(a_{n}, b_{n}), \nN \}\) so that

\[ A \subseteq \bigcup _{n=1}^{\infty }(a_{n}, b_{n}).\]

It would take a long time to prove that \(\lambda \) really is a measure, and it wouldn’t help us understand \(\lambda \) any better if we did it, so we’ll omit that from the course. For the proof, see the standard text books e.g. Cohn, Schilling or Tao.

Let’s check that the definition (1.4) agrees with our intuitive ideas about length.

• 1. If \(A = (a,b)\) then \(\lambda ((a,b)) = b-a\) as expected, since \((a,b)\) is a covering of itself and any other cover will have greater length.

• 2. If \(A = \{a\}\) then choose any \(\epsilon > 0\). Then \((a - \epsilon /2, a + \epsilon /2)\) is a cover of \(a\) and so \(\lambda (\{a\}) \leq (a + \epsilon /2) - (a - \epsilon /2) = \epsilon \). But \(\epsilon \) is arbitrary and so we conclude that \(\lambda (\{a\}) = 0\).

  From (1) and (2), and using M(ii), we deduce that for \(a < b,\)

  \[\lambda ([a, b)) = \lambda (\{a\} \cup (a, b)) = \lambda (\{a\}) + \lambda ((a, b)) = b-a.\]

• 3. If \(A = [0, \infty )\) write \(A = \bigcup _{n=1}^{\infty }[n-1, n)\). Then by M(ii), \(\lambda ([0, \infty )) = \infty \). By a similar argument, \(\lambda ((-\infty , 0)) = \infty \) and so \(\lambda (\R ) = \lambda ((-\infty , 0)) + \lambda ([0, \infty )) = \infty \).

• 4. If \(A\in \mc {B}(\R )\), and for some \(x\in \R \) we define \(A_x=\{x+a\-a\in A\}\), then \(\lambda (A)=\lambda (A_x)\).

  In words, if we take a set \(A\) and translate it (by \(x\)), we do not change its measure. This is easily seen from (1.4), because any cover of \(A\) can be translated by \(x\) to be a cover of \(A_x\).

In simple practical examples on Lebesgue measure, it is generally best not to try to use (1.4), but to just apply the properties (1) to (4) above:

e.g. to find \(\lambda ((-3, 10) - (-1, 4))\), use (1.2) to obtain

\[\begin {aligned} \lambda ((-3, 10) - (-1, 4)) & = \lambda ((-3, 10)) - \lambda ((-1, 4)))\\ & = (10 - (-3)) - (4 - (-1)) = 13 - 5 = 8. \end {aligned}\]

If \(I\) is a closed interval (or in fact any Borel set) in \(\R \) we can similarly define \({\cal B}(I)\), the Borel \(\sigma \)-field of \(I\), to be the smallest \(\sigma \)-field containing all open intervals in \(I\). Then Lebesgue measure on \((I, {\cal B}(I))\) is obtained by restricting the sets \(A\) in (1.4) to be in \({\cal B}(I)\).

Sets of measure zero play an important role in measure theory. Here are some interesting examples of quite “large” sets that have Lebesgue measure zero

• 1. Countable Subsets of \(\R \) have Lebesgue Measure Zero

  Let \(A \subset \R \) be countable. Write \(A = \{a_{1}, a_{2}, \ldots \} = \bigcup _{n=1}^{\infty }\{a_{n}\}\). Since \(A\) is an infinite union of point sets, it is in \({\cal B}(\R )\). Then

  \[\lambda (A) = \lambda \left (\bigcup _{n=1}^{\infty }\{a_{n}\}\right ) = \sum _{n=1}^{\infty }\lambda (\{a_{n}\}) = 0.\]

  It follows that

  \[ \lambda (\mathbb {N}) = \lambda (\mathbb {Z}) = \lambda (\mathbb {Q}) = 0.\]

  The last of these is particularly intriguing as it tells us that the only contribution to length of sets of real numbers comes from the irrationals.

• 2. The Cantor Set has Lebesgue Measure Zero

  Recall the construction of the Cantor set \(C = \bigcap _{n=1}^{\infty }C_{n}\) given earlier in this chapter. Recall also that the \(C_n\) are decreasing, that is \(C_{n+1}\sw C_n\), and hence also \(C\sw C_n\) for all \(n\).

  Since \(C_{n}\) is a union of intervals, \(C_{n} \in {\cal B}(\R )\) for all \(\nN \). Hence \(C \in {\cal B}(\R )\). We easily see that \(\lambda (C_{1}) = 1 - \frac {1}{3}\) and \(\lambda (C_{2}) = 1 - \frac {1}{3} - \frac {2}{9}\). Iterating, we deduce that \(\lambda (C_{n}) = 1 - \sum _{r=1}^{n}\frac {2^{r-1}}{3^{r}}\) and since \(\lambda (C)\leq \lambda (C_n)\) we thus have

  \[\lambda (C)\leq \lambda (C_n)=1 - \ds \sum _{r=1}^{n}\frac {2^{r-1}}{3^{r}}.\]

  Letting \(n\to \infty \), and using that limits preserve weak inequalities, we obtain \(\lambda (C)\leq 0\). But by definition of a measure we have \(\lambda (C)\geq 0\). Hence \(\lambda (C)=0\).