What different ways are there to construct the Lebesgue measure?

Question

I have recently learned of different approaches (which I've included below) to constructing the Lebesgue measure, and I'm somewhat startled by how much each approach can illuminate the theory as a whole.

Are there still others approaches to defining the Lebesgue measure? If so, what are their benefits and disadvantages? Where may I read about them?

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$$\textbf{First Approach}$$ Starting with the premeasure $\mu$ defined only on boxes, and extending it into a measure through the follwing theorem:

Carathéodory's Extension Lemma: let $\mu_0:\Sigma_0\to [0,\infty]$ be a pre-measure on the algebra $\Sigma_0$ of $X$. Then $\mu_0$ can be extended to a measure $$\mu:\Sigma\to[0,\infty]$$ where $\Sigma:=\sigma(\Sigma_0)$ and $\mu|_{\Sigma_0}=\mu_0$. Furhermore, if $\mu_0$ is finite, then the extension $\mu$ is unique.

This approach is used in Williams' Probability with Martingales (PWM). It is worth mentioning that -excluding the proof of the theorem above- this is the simplest construction I know of. However, both of the proofs I know of the theorem above (one found at the end of PWT, and the other one found here) first use the pre-measure to construct an outer measure $\mu^*$ and then restrict the outer measure to Lebesgue measurable sets.

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$$\textbf{Second Approach}$$ Parting from the outer Lebesgue measure $\mu^*$ and restricting it through the following theorem:

Carathéodory's Restriction Lemma: let $\mu^*:2^X\to [0,\infty]$ be an outer measure on the power set $2^X$ of $X$. Then $\mu^*$ can be restricted to a measure $$\mu:\Sigma\to[0,\infty]$$ where $\Sigma := \Big\{ C \in 2^X : C \text{ is Caratheodory measurable} \Big\}$ and $\mu := \mu^*|_{\Sigma}$.

(I've never seen the 'Restriction Lemma' named as such, but I find the name appropriate). The approach may be found in Tao's An Introduction to Measure Theory (IMT), as well as Hunter's Measure Theory (MT). This approach is, I believe, the most common one, though it left me puzzled as to why the restriction was made to Carathéodory measurable sets specifically; some intuition may be found in this post or by studying equivalent (and more geometrically intuitive) definitions of Lebesgue measurable sets (as in Exercise 1.2.7 in IMT).

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$$\textbf{Third Approach}$$ Parting again from the Lebesgue outer measure $\mu^*$, one defines Lebesgue measurable sets as those which can be 'approximated from above' by open sets arbitrarily well:

Definition: a set $E\subseteq \mathbb{R}^d$ is Lebesgue measurable iff for every $\varepsilon>0$ there is some open set $U\subseteq \mathbb{R}^d$ containing $E$ such that $\mu^*(U\setminus E)<\varepsilon$.

Then the Lebesgue measure is characterized as the restriction of $\mu^*$ to Lebesgue measurable sets, and a bit of work shows that the collection of Lebesgue measurable sets is a $\sigma$-algebra, as well as that $\mu$ is a measure.

The approach is not as generalizable, although it is arguably more geometrically intuitive (specially when similarities and dissimlarities with the Jordan measure are made). It is also found in IMT.

Answer 1

I'm confused. All the approaches you write construct the Lebesgue measure in the same way: define the Lebesgue outer measure (using the same formula basically, infimum over total premeasures of covers), and restrict. The only difference is perhaps how one defines the measurable sets (the Caratheodory condition, the outer-inner measure condition, or the open-closed approximation condition --- an old article of mine discussing all three definitions of measurable sets may interest you, but beware there are some wrong proofs; though the intuition/motivation/interconnectedness of concepts should still be decent), but Lebesgue measure itself still is constructed in exactly the same way.

For genuinely different approaches, I already mentioned the functional analytic/category theoretic philosophy in my answer to one of your previous questions --- see also this MO thread https://mathoverflow.net/questions/38439/integrals-from-a-non-analytic-point-of-view?noredirect=1&lq=1, but another approach is to define the Lebesgue integral first (radical, right?) via taking the metric completion$\dagger\dagger\dagger$ of the vector space of continuous functions on $[0,1]$ with the metric induced by the norm $\|f\|_1 := \int_0^1 |f| d t$ where the integral denotes a Riemann integral: https://math.stackexchange.com/a/1259106/405572. The resulting normed vector space turns out to be isomorphic to $L^1([0,1])$, so one can define the Lebesgue measure of a set $E$ by $\|1_E\|_1$, as long as the characteristic function $1_E$ exists in the completion (i.e. for all $\epsilon>0$ there exists a continuous function $g$ s.t. $\|1_E-g\|_1<\epsilon$).

However I think this approach of defining Lebesgue measure or $L^1([0,1])$ is completely backward, since "points" in metric completion of continuous functions under the Riemann $\|\bullet\|_1$-norm are $\|\bullet\|_1$-Cauchy sequences of continuous function; it's not clear that they refer to actual functions $[0,1]\to \mathbb R$ (can't even take pointwise limits of such Cauchy-sequences of continuous functions unless you know already know about the theorem about convergence of a Cauchy-sequence of measurable functions $f_n:[0,1] \to \mathbb R$ in the $\|\bullet\|_1$-norm implies that a subsequence $f_{n_k}$ converges pointwise almost everywhere to some function $f:[0,1]\to \mathbb R$... just overall seems to do things upside down).

$\dagger \dagger \dagger$ (think about the construction of the real numbers as the metric-completion of the metrized vector space $\mathbb Q$, i.e. all real numbers are Cauchy-sequences of rational numbers identified modulo some equivalence relation)

Ultimately, the Caratheodory approach is the most effective generally-applicable tool at extending/constructing/defining a measure, and a collection of measurable sets to go with that measure, in the following sense: the other approaches to define measurable sets (outer-inner and open-closed approximation) are less general, since they only really apply if the total space $X$ is bounded and/or $X$ has some topological niceties (need a rich enough class of open/compact subsets of $X$, for instance).

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EDIT 12/26/22: I remembered another approach defining $L([0,1])$ and Lebesgue integration first: https://mathoverflow.net/a/167693/112504. The idea is that $L([0,1])$ (plus a little extra data, “describing the notion of mean/average”) is the (necessarily unique) initial object in some category (of Banach spaces plus a little extra data), and Lebesgue integration is the unique map (in the category, i.e. a norm-decreasing map between Banach spaces preserving the extra data) to $\mathbb R$. Though from this abstract description of $L([0,1])$, it seems difficult to isolate the indicator functions of measurable sets to define Lebesgue measure.

Answer 2

Here is a probabilistic way to construct Lebesgue's measure on $[0,1]$ first, and then extend to $\mathbb{R}$ by translation.

Let $(X_n:n\in\mathbb{B})$ be an i.i.d. sequence of Bernoulli $0-1$ random variables with $\mathbb{P}[X_1=0]=1/2=\mathbb{P}[X_1=1]$. A probability space with such a sequence exists, for example consider the space $(\{0,1\},\mathcal{P}(\{0,1\}),\nu)$ where $\nu(\{0\})=\nu(\{1\})=1/2$. Let $\Omega=\{0,1\}^{\mathbb{N}}$ equipped with the product $\sigma$-algebra $\mathcal{F}:=\mathcal{P}(\{0,1\})^{\otimes\mathbb{N}}$. By Ionescu-Tulcea's theorem there is a unique $\mu:\mathcal{F}\rightarrow [0, 1]$ such that $\mu(A_1\times\ldots \times A_n \times E_{n+1} \times \ldots)=\nu(A_1)\cdot\ldots\cdot\nu(A_n)$ for all $n\in\mathbb{N}$ and $A_j\subset E_j= \{0,1\}$.

The function $\phi:\Omega\rightarrow[0,1]$ given by $$\phi(\omega)=\sum^\infty_{n=1}2^{-n}X_n(\omega)$$ is measurable, and it induces a measure $m=\mu\circ \phi^{-1}$ on $[0,1]$. It is not difficult to show that the induced measure $m$ is the Lebesgue measure restricted to $[0,1]$. More precisely, recall that every $x\in[0,1]$ has a unique binary expansion $x=\sum_{n\geq1}d_n(x)/2^n$ where $d_n(x)\in\{0,1\}$, and $\sum_{n\geq1}d_n(x)=\infty$ for $x>0$. Notice that for each $n\in\mathbb{N}$, the $n$--th bit map $x\mapsto d_n(x)$ defines a measurable function from $([0,1],\mathscr{B}([0,1]))$ to $(\{0,1\},\mathscr{P}(\{0,1\}))$. Therefore, the map $\beta:[0,1]\rightarrow\{0,1\}^{\mathbb{N}}$ given by $x\mapsto(d_n(x))$ is measurable.

Lemma A: Suppose $\theta\sim U[0,1]$, and for each $n\in\mathbb{N}$ let $X_n=d_n\circ\theta$. Then, $(X_n:n\in\mathbb{N})$ is an i.i.d. Bernoulli sequence with rate $p=\tfrac12$. Conversely, if $(X_n)$ is an i.i.d. Bernoulli sequence with rate $p=\tfrac12$, then $\theta=\sum_{n\geq1}2^{-n}X_n\sim U[0,1]$.

The key to this is to notice that for any $N\in\mathbb{N}$ and $k_1,\ldots,k_N\in\{0,1\}$, \begin{align} \bigcap^N_{j=1}\{x\in(0,1]:d_j(x)=k_j\}&=(\sum^N_{j=1}\tfrac{k_j}{2^j}, \sum^N_{j=1}\tfrac{k_j}{2^j}+\tfrac{1}{2^N}]\\ \{x\in(0,1]: d_N(x)=0\}&=\bigcup^{2^{N-1}-1}_{j=0}(\tfrac{2j}{2^N},\tfrac{2j+1}{2^N}]\\ \{x\in(0,1]:d_N(x)=1\}&=\bigcup^{2^{N-1}-1}_{j=0} (\tfrac{2j+1}{2^N},\tfrac{2(j+1)}{2^N}] \end{align}

I leave the details to the OP.

Answer 3

A freshman's dream is to define the Lebesgue measure of a set $A\subseteq[0,1]$ by counting the elements of the set and dividing by the total number of points in $[0,1]$, similarly to the case of a finite sample space. While such a naive approach is impossible, a modification thereof can be used to define the Lebesgue measure using nonstandard analysis (NSA), providing a genuinely different approach to the Lebesgue measure. Here one needs a few preliminaries concerning NSA.

In NSA, every set $A\subseteq[0,1]$ (and of course more general sets) has a natural extension to a set $A^\ast$ contained in the hyperreal interval $[0,1]^\ast$. The natural numbers $\mathbb N$ have a natural extension $\mathbb N^\ast$ where the complement $\mathbb N^\ast\setminus\mathbb N$ consists of infinite hyperintegers. Similarly, the set of subsets $\mathbb P= \mathcal P([0,1])$ has a natural extension $\mathbb P^\ast$. Elements of $\mathbb P^\ast$ are called internal subsets of $[0,1]^\ast$. Internal sets admitting an internal bijection with an infinite hyperinteger $H$ are called hyperfinite sets. For each element $x\in[0,1]^\ast$, there is a unique real number $r\in[0,1]$ infinitely close to $x$, i.e., such that the difference $x-r$ is infinitesimal. The corresponding association $st:[0,1]^\ast\to[0,1]$, $x\mapsto r$ is called the standard part. Define the hyperfinite grid $G_H\subseteq[0,1]^\ast$ by $G_H=\big\{0,\frac{1}{H}, \frac{2}{H},\ldots,\frac{H-1}{H},1\big\}$.

Now we can define the Lebesgue measure of $A\subseteq[0,1]$ by "counting" as follows. Consider the inverse image $B=st^{-1}(A^\ast\cap G_H)$. We would like to define the Lebesgue measure of $A$ via the ratio $\frac{|B|}{H}$ (i.e., "counting" the number of points in $B$ and dividing by the total number $H$ of points in the grid). This doesn't quite work because $B$ may not be an internal set. However, a slight modification works: we consider all internal sets $C$ containing $B$, and define the measure as the infinimum of $\frac{|C|}{H}$ over all such internal sets. The result coincides with the Lebesgue measure of $A$ for all Lebesgue-measurable sets. Further details can be found in

Goldblatt, Robert Lectures on the hyperreals. An introduction to nonstandard analysis. Graduate Texts in Mathematics, 188. Springer-Verlag, New York, 1998.

Answer 4

There seems to be a modern style of teaching measure theory without mentioning inner measure. Let $m^o$ denote outer measure. Let $c(E)=\{F: F=\overline F\subseteq E\}.$ The inner measure of $E$ is $m^i(E)=\sup \{m^o(F):F\in c(E)\}.$ Now if $m^o(E)<\infty,$ define $E$ to be measurable iff $m^i(E)=m^o(E).$ And in general, any $E$ is measurable iff $E$ is the union of a countable family of measurable sets, each of which has finite outer measure. It follows that any $E$ is measurable iff there exists a $G_{\delta}$ set $A$ and an $F_{\sigma}$ set $B$ with $B\subseteq E\subseteq A$ and $m(A)=m(B).$