10  Measurable sets

Measure theory tries to assign measure to subsets of some set \(X\). This can be lengths of intervals of \(\mathbb{R}\), areas in \(\mathbb{R}^2\), et.c. A little thought shows that we need to consider quite general subsets. For example, I expect to be able to measure the area of a circle, but also the circle with a single point, or a countable number of points, punched out. Moreover, when integrating functions, we compute areas under a curve. Changing a function at a single point in \(\mathbb{R}\), or a whole surface in \(\mathbb{R}^3\), does not change the value of the integral of the function. We need to have a more rigorous understanding of measure.

Consider the real line \(\mathbb{R}\). Our intuition tells us that an interval \([a,b]\subset\mathbb{R}\) has length \(\mu([a,b])=b-a\). The symbol \(\mu\) means “measure”. Similarly, the length of \(\mathbb{R}\) must be infinite, \(\mu(\mathbb{R}) = +\infty\). We also want the measure of a union of two disjoint intervals to be the sum of the lengths, \(\mu([a,b]\cup [c,d]) = \mu([a,b]) + \mu([c,d])\), with \(a < b\leq c < d\). And the measure of a single point should be zero, \(\mu(\{x\}) = 0\). The measure of finitely many points should also be zero, but what about infinitely many points? If the set is contably infinite, we expect the measure to be zero, but the interval is an uncountably infinite set, with positive measure.

This leads to the notion of a \(\sigma\)-algebra; an all-important notion in measure theory:

Definition 10.1 (\(\sigma\)-algebra) Let \(X\) be a set. A \(\sigma\)-algebra \(\boldsymbol{X}\) is a family of subsets of \(X\) such that:

  1. \(\emptyset, X \in \boldsymbol{X}\) empty set and \(X\)

  2. \(A \in \boldsymbol{X}\) if and only if \(A^\complement = X \setminus A \in \boldsymbol{X}\) complements

  3. \(A_i \in \boldsymbol{X}\) for \(i \in I \subset\mathbb{N}\) implies that \(\bigcup_{i \in I} A_i \subset \boldsymbol{X}\) countable unions

The pair \((X,\boldsymbol{X})\) is called a measurable space, and \(A \in \mathcal{X}\) is called a measurable (sub)set.

Reading between the lines, so to speak, we dont’t expect to be able to measure all subsets of \(X\). Indeed, it turns out it is not possible in general to define a measure on all subsets of, say, \(\mathbb{R}\). There will be non-measurable sets, to which we cannot assign a meaningful measure.

Example 10.1 (Example) The smallest possible \(\sigma\)-algebra: \[\boldsymbol{X} = \{\emptyset, X\}\] The largestpossible \(\sigma\)-algebra: \[\boldsymbol{X} = 2^X = \{ \text{all subsets of $X$} \}\] Interesting cases lie between these.

The following lemma is useful, because it shows that one can always find a smallest \(\sigma\)-algebra fulfilling some condition:

Lemma 10.1 (Intersections of \(\sigma\)-algebras) Let \(I\) be a set, and let \(\{\boldsymbol{X}_i \mid i \in I\}\) be a family of \(\sigma\)-algebras. Then the intersection of all the \(\sigma\)-algebras is again a \(\sigma\)-algebra: \[\boldsymbol{X} = \left\{ A \mid A \in \boldsymbol{X}_i \; \text{for all $i\in I$} \right\}\]

Turning to Euclidean space, or any other metric space, the open sets form subsets that we wisth to make measurable.

Definition 10.2 (Borel \(\sigma\)-algebra) Let \(X\) be a metric space. The Borel \(\sigma\)-algebra \(\boldsymbol{B}(X)\) is the smallest \(\sigma\)-algebra that contain all open subsets of \(X\). (It is enough to require that it contains all \(\epsilon\)-balls.)

Such a smallest \(\sigma\)-algebra exists by the above lemma. We say that the Borel \(\sigma\)-algebra is generated by the open sets.

Which subsets are included in the Borel \(\sigma\)-algebra?

Example 10.2 (Example) Consider \(\mathbb{R}\), and its Borel \(\sigma\)-algebra. It contains all subsets consisting of a single point, since \(\{x\} = (]-\infty,x[ \cup ]x,+\infty[)^\complement\). But then we can take countable unions of such points, e.g., the rational numbers \(\mathbb{Q}\), which is also measurable. But then we can take the complement, to get the irrational numbers \(\mathbb{R}\setminus\mathbb{Q}\), also measurable.

We can also consider the Cantor set: Start with an interval, and remove the middle third. Remove the middle third of those two again, and continue. It is not hard to see that this produces a measurable set, and in fact a fractal.

So the Borel \(\sigma\)-algebra \(\boldsymbol{B}(\mathbb{R})\) contains quite complicated sets.