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Lebesgue Measure

Disclosure: these are my notes on the presentation of Lebesgue measures, as given in these lecture notes. I have a vague, but not complete idea of how it lines up with a more general presentation, e.g. the one given here by Brent Nelson.

Pre-requisite notes:

0.1 Lebesgue measure on \((0,1]\)

The most relevant context is Caratheodory's Extension Theorem. We will be using this theorem to specify the uniform probability measure on \([0,1]\) (this is the Lebesgue measure). What follows is an outline of how that goes (see Extension Theorem for a more general outline):

  1. Define an algebra \(\mathcal{F}_0\) on which we will define our measure
  2. Define our measure \(\mathbb{P}_0\)
  3. Get extension \(\mathbb{P}\) measure on \(\mathcal{F} = \sigma(\mathcal{F}_0)\) from the Extension Theorem.

Here is a more detailed outline. First, we decide to work with \((0,1]\) instead (this will make sense when we define our algebra \(\mathcal{F}_0\)). Let \(C\) be all intervals \([a,b]\) in \((0,1]\). Then, the \(\sigma\) -field \(\sigma(C)\) generated by \(C\) is called the Borel \(\sigma\) -algebra \(\mathcal{B}\).

Next, we define an algebra \(\mathcal{F}_0\) which we will define our measure on. \(\mathcal{F}_0\) consists of:

  • the empty set \(\emptyset\)
  • sets that are a finite union of intervals of the form \((a,b]\). So every \(A\in \mathcal{F}_0\) is of the form: \[ A = (a_1,b_1] \cup \cdots \cup (a_n, b_n] \] where \(0 \leq a_1 < b_1 \leq a_2 < b_2 \leq ... \leq a_n < b_n \leq 1\) and \(n\in \mathbb{N}\)

The notes give the following lemma:

0.1.1 lemma 1 \(\sigma(\mathcal{F}_0) = \sigma(C) = \mathcal{B}\)

This ensures that the algebra we are dealing with actually generates the \(\sigma\) -algebra that we are interested in.

0.1.2 lemma 2.1 \(\mathcal{F}_0\) is an algebra

0.1.3 lemma 2.2 \(\mathcal{F}_0\) is not a \(\sigma\) -algebra

0.1.4 defining our measure \(\mathcal{P}_0\)

For every \(A_0 \in \mathcal{F}_0\) of the form: \[ A = (a_1, b_1] \cup \cdots \cup (a_n, b_n] \] we define: \[ \mathbb{P}_0 = \sum_{i=1}^n (b_i - a_i) \]

0.1.5 Lemma 3 \(\mathbb{P}_0\) is \(\sigma\) -additive (or countably additive) on \(\mathcal{F}_0\)

(proof in the notes) This is one of the pieces we need to apply the Extension theorem.

0.1.6 Apply Extension Theorem

Finally, we apply Caratheodory's Extension Theorem to arrive at \(\mathbb{P}\), called the Lebesgue measure on \((0,1]\)

1 Lebesgue measure on \(\mathbb{R}\)

1.1 definition: Borel measurable

First, we define a \(\sigma\) -algebra on \(\mathbb{R}\) (all of these will be called also the Borel measure \(\mathcal{B}\)). This can be done in 3 (equivalent) ways:

  1. Let \(C\) be the collection of all intervals of the form \([a,b]\) and let \(\mathcal{B} = \sigma(C)\) be the \(\sigma\) -algebra that \(C\) generates.
  2. Let \(D\) be the collection of all intervals of the form \((-\infty, b]\) and let \(\mathcal{B} = \sigma(D)\)
  3. For any \(n\), we define the Borel \(\sigma\) -algebra of \((n, n+1]\) as the \(\sigma\) -algebra generated by the set of all intervals of the form \([a,b] \subset (n, n+1]\) (this is how we defined the Borel \(\sigma\) -algebra of (0,1] above). Then, we say that \(A\) is a Borel subset of \(\mathbb{R}\) if \(A \cap (n, n+1]\) is a Borel subset of \((n, n+1]\) for every \(n\). This definition is important for the \(\mu\) that we are about to specify since it guarantees that \(A \cap (n, n+1]\) is Borel measurable.

Useful fact: every singleton is Borel measurable.

1.2 Lebesgue measure

Let \(\mathbb{P}_n\) be the Lebesgue measure on \((n, n+1]\) (remember from the above section that this is defined on the Borel \(\sigma\) -algebra of \((n,n+1]\)). So, given a set \(A \subset \mathbb{R}\) we define: \[ \mu(A) = \sum_{-\infty}^{\infty} \mathbb{P}_n \left(A \cap (n, n+1]\right) \]

It turns out that \(\mu\) is a measure on \((\mathbb{R}, \mathcal{B})\). It is also called the Lebesgue measure. Question: how do we know that \(A \cap (n, n+1]\) is Lebesgue-measurable? Remember that \(A\) is not just any subset of \(\mathbb{R}\), it is a Borel measurable subset of \(\mathbb{R}\). And (3) from the previous section tells us that all such intersections are also Borel subsets.

1.3 technical correction: some notes on completion

Note that when most people refer to the Lebesgue measure, they are referring to a measure defined on the completion of \(\mathcal{B}\). Consider a measure space \((\Omega, \mathcal{F}, \mu)\). Then, if some set \(N\in \mathcal{F}\) has \(\mu(N)=0\) it is called a null-set. Then, \((\Omega, \mathcal{F}, \mu)\) is said to be complete if for every null-set \(N\), all subsets of \(N\) are measurable and have measure 0.

For a measure space, \((\Omega, \mathcal{F}, \mu)\) we can define the completion \((\Omega, \bar{\mathcal{F}}, \bar{\mu})\) where:

  • \(\bar{\mathcal{F}}\) is the smallest \(\sigma\) -algebra that contains \((\mathcal{N} \cup \mathcal{F})\) where \(\mathcal{N}\) is the set of all null-sets in \(\mathcal{F}\).

Pay attention: when we add in all the subsets of a null-set \(N\) into our algebra, we aren't placing any restrictions on what those subsets can look like. That is, we're not looking for "nice" subsets. We're taking all possible subsets of \(N\).

Extending the Lebesgue measure as discussed above yields a new measure, which is also called the Lebesgue measure.

1.4 alternative definition

This is the definition given on wikipedia and in these notes. Let \(\Omega = \mathbb{R}\), with metric \(d(x,y) = x-y\). Let \(F:\mathbb{R}\rightarrow\mathbb{R}\) be an increasing, continuous function. Then, you can define a measure \(\mu_{F}\): \[ \mu_F(S) := \inf\left\{\sum_{j=1}^{\infty} [F(b_j) - F(a_j)] : \bigcup_{j=1}^{\infty} (a_j,b_j] \supset S\right\} \] In words: consider all possible coverings of \(S\) formed by a countable union of disjoing half-open intervals. Then, the measure of \(S\) is the infimum of the measures of these coverings.

Then, let \(F(x) = x\). The completion of \((\mathbb{R}, \mathcal{B}, \mu)\) yields the measure space \((\mathbb{R}, \mathcal{L}, m)\), for which \(m\) is called the Lebesgue measure.

I think I can square this definition with the other definition given above, which also involves a cover involving countably many intervals.

2 FAQ

  • Is every singleton \(\{x\}\) Borel measurable? yes

3 see also

Created: 2021-09-14 Tue 21:44