## 1.3. Constructing Measures

≪ 1.2. Measures and Their Properties | Table of ContentsExplicitly describing measures is very, very difficult. I don’t like beating dead horses, but the Vitali set shows that even something as intuitive as “the length of an interval” is very hard to extend to a full measure. What even would the length of the Cantor set be? (This is actually doable, and I’ll leave it as an exercise.)

Sort of like how \(\sigma\)-algebras are easily described by generating sets, we can use easier-to-describe “almost-measures” to then create a measure.

**Definition 7. Algebra of Subsets**

Let \(X\) be a set. A nonempty \(\mathcal A\subset\mathcal P(X)\) is an algebra of subsets on \(X\) if it is closed under finite unions and complements.

These are more appropriately called boolean algebras, and looking at our discussion of \(\sigma\)-algebras, it’s not hard to see that these are also closed under finite intersections. Furthermore, we know that \(\varnothing, X\) are in every algebra of subsets.

**Definition 8. Premeasures**

Let \(X\) be a set and \(\mathcal A\) be an algebra of subsets of \(X\). A function \(\mu_0:\mathcal A\to [0,\infty]\) is called a premeasure if

- \(\mu_0(\varnothing)=0\).
- If \(\lbrace A_n\rbrace\) is a disjoint sequence in \(\mathcal A\) such that \(\bigcup_{n=0}^\infty A_n\in\mathcal A\), then \(\mu_0\left(\bigcup_{n=0}^\infty A_n\right)=\sum_{n=0}^\infty\mu_0\left(A_n\right)\).

Algebras are premeasures are significantly easier to construct and define because we no longer need to worry about arbitrary countable unions: we only need to care about their consistency when a countable union lands back inside the algebra. Premeasures retain the properties that measures have, provided that the unions and intersections fall in the premeasure’s domain.

**Definition 1. Outer Measures**

Let \(X\) be a set. A function \(\mu^\star:\mathcal P(X)\to [0,\infty]\) is called an outer measure if:

- \(\mu^\star(\varnothing)=0\).
- If \(\lbrace A_n\rbrace\) is a disjoint sequence in \(\mathcal P(X)\), then \(\mu^\star\left(\bigcup_{n=0}^\infty A_n\right)\leq\sum_{n=0}^\infty\mu^\star\left(A_n\right)\).

This second requirement is called disjoint sub-additivity. Just like with regular measures, this property ensures monotonicity: if \(E\subset F\) are two subsets of \(X\), then \(\mu^\star(E)\leq\mu^\star(F)\).

An important thing to note is that since every algebra of sets contains \(X\), this outer measure is actually defined on \(\mathcal P(X)\).

The big picture is that we can extend premeasures to outer measures, which are defined on \(\mathcal P(X)\) but may not have the right additivity properties everywhere. Then, we refine the domain of the outer measure to some \(\sigma\)-algebra \(\mathcal M\) so that the outer measure becomes a measure. We’ll develop this idea in full generality while also building up the Lebesgue measure — the “lengths” of (measurable) subsets of \(\R\).

Let’s consider the algebra of sets on \(\R\) given by \[\mathcal I=\left\lbrace\bigcup_{j=1}^n \left[a_j, b_j\right) : a_j < b_j \leq a_{j+1}\right\rbrace. \] In words, this is the algebra of sets whose elements are exactly the subsets of \(\R\) that are expressible as finite unions of half-intervals. To abuse notation, we’ll allow \(a_j=-\infty\) and \(b_j=\infty\), understanding that \([-\infty, b)=\lbrace x\in\R: x< b\rbrace \).

It can be easily verified that \(\mathcal I\) is an algebra. In particular, when \(n=0\), the above union is the empty set by convention. We can then define a premeasure on \(\mathcal I\) by
\[\lambda\left(\bigcup_{j=1}^n \left[a_j, b_j\right)\right) = \sum_{j=1}^n\left(b_j-a_j\right). \]
It can also be pretty swiftly verified that this is indeed a premeasure. The one difficulty is that elements of \(\mathcal I\) do not necessarily have a unique representation as a finite union of half-intervals: for instance, \([0, 2)=[0, 1)\cup[1, 2)\). Nevertheless, the sum on the right *is* well-defined, and I’ll leave this as a very tedious exercise for the reader.

**Theorem 2. **

Let \(X\) be a set, \(\mathcal A\) be an algebra of sets on \(X\), and \(\mu_0\) a premeasure on \(\mathcal A\). Then, the function \[\mu^\star(B):=\inf\left\lbrace \sum_{n=0}^\infty\mu_0\left(A_n\right) : A_n\in\mathcal A\ \forall n,\ B\subseteq\bigcup_{n=0}^\infty A_n\right\rbrace\] is an outer measure on \(X\), and \(\mu^\star(A)=\mu_0(A)\) for all \(A\in\mathcal A\).

**Proof:** Clearly, \(\mu^\star(\varnothing)=0\), so we only need to check sub-additivity. Let \(\left\lbrace B_n\right\rbrace\) be a disjoint sequence of subsets of \(X\), and let \(\epsilon>0\). For each \(n\), let \(\left\lbrace A_{n, j}\right\rbrace_{j=0}^\infty\) be any sequence in \(\mathcal A\) such that \(B_n\subseteq\bigcup_{j=0}^\infty A_{n, j}\) and \(\sum_{j=0}^\infty\mu_0\left(A_{n,j}\right)<\mu^\star\left(B_n\right)+\epsilon 2^{-n}\). Then, if \(B=\bigcup_{n=0}^\infty B_n\), we have that \(B\subseteq\bigcup_{n=0}^\infty\bigcup_{j=0}^\infty A_{n, j}\), which is still a countable union. Then,
\[\mu^\star(B)\leq\sum_{n,j}\mu_0\left(A_{n, j}\right)<\sum_{n=0}^\infty\mu^\star\left(B_n\right)+\sum_{n=0}^\infty\epsilon 2^{-n}=\sum_{n=0}^\infty\mu^\star\left(B_n\right)+\epsilon.\]
Sending \(\epsilon\to 0\) gives sub-additivity. Finally, it’s obvious that \(\mu^\star(A)\leq\mu_0(A)\) for all \(A\in\mathcal A\), so we only need to show the reverse inequality. Suppose \(A\subseteq\bigcup_{n=0}^\infty A_n\), and suppose without loss of generality that the \(A_n\) are disjoint: we can otherwise consider \(B_0=A_0\) and \(B_n=A_n\setminus\bigcup_{j=0}^{n-1}A_j\). Then, \(A=\bigcup_{n=0}^\infty A\cap A_n\). Since \(A\cap A_n\subseteq A_n\) for all \(n\), disjoint additivity and monotonicity gives
\[\mu_0(A)=\sum_{n=0}^\infty\mu_0\left(A\cap A_n\right)\leq\sum_{n=0}^\infty\mu_0\left(A_n\right).\]
This is true for all such sequences \(A_n\), so we conclude that \(\mu_0(A)\leq\mu^\star(A)\). \(\square\)

There is one subtle detail that must be noted. We implicitly renumbered the sequence \(A_{n, j}\) (as \(A_m\), say), but we cannot yet justify that \(\sum_{m=0}^\infty\mu_0\left(A_m\right)=\sum_{n, j}\mu_0\left(A_{n, j}\right)\). However, it is true that the sum over \(m\) is less than the iterated sum over \(n\) and \(j\), so the result still works out.

Let’s go back to the premeasure \(\lambda\) on \(\mathcal I\). Let \(\theta\) be the outer measure induced by \(\lambda\) in accordance with the above. Notice that if \(\left\lbrace I_n\right\rbrace\) is a sequence of half-intervals in \(\mathcal I\), we can express \(\bigcup_{n=0}^\infty I_n\) as a countable union of intervals: that is, \(\bigcup_{n=0}^\infty I_n=\bigcup_{j=0}^\infty \left[a_j,b_j\right)\). Thus, we may write
\[\theta(B)=\inf\left\lbrace \sum_{j=0}^\infty\left(b_j-a_j\right) : B\subseteq\bigcup_{j=0}^\infty\left[a_j, b_j\right)\right\rbrace .\]
\(\theta\) is called the *Lebesgue outer measure*, and it gives an intuitive illustration of how outer measures work. \(\theta\) in particular uses simple building blocks that are easy to measure — half-intervals — to produce overestimates of how “long” any arbitrary subset of \(\R\) should be. It then says that the length of a subset of \(\R\) is the smallest possible overestimate.

The last and significantly more difficult step is to restrict \(\mu^\star\) to an appropriate domain. The way we do that is with Carathéodory’s method, which describes a condition for subsets of \(X\) to be “measurable” by \(\mu^\star\).

**Theorem 3. Carathéodory's Method**

Let \(X\) be a set and \(\mu^\star\) be an outer measure on \(X\). The set \[\mathcal M:=\left\lbrace A\subset X : \mu^\star(E)=\mu^\star(E\cap A)+\mu^\star(E\setminus A)\quad\forall E\subset X\right\rbrace\] is a \(\sigma\)-algebra on \(X\), and the restriction \(\mu=\left.\mu^\star\right|_{\mathcal M}\) is a measure.

In essence, Carathéodory’s method gives a condition on “compatibility” with \(\mu^\star\), and it says that the restriction to the compatible sets is a measure. I find it helpful to examine a simple example. Take any set \(X\) with more than \(1\) element (hopefully many more) and consider the algebra \(\mathcal A=\lbrace\varnothing, X\rbrace\). Define the premeasure \(\mu_0(\varnothing)=0\) and \(\mu_0(X)=1\). Then, if we extend this to an outer measure \(\mu^\star\) on \(\mathcal P(X)\), we see that \(\mu^\star(A)=1\) for any nonempty subset \(A\subseteq X\).

We now see that no nonempty *proper* subset of \(X\) is “compatible” with \(\mu^\star\). Indeed, if \(A\subsetneq X\), \(\mu^\star(X)=1\) but \(\mu^\star(X\cap A)+\mu^\star(X\setminus A)=2\): neither of the two arguments is empty. In words, comparing the measurements of \(X\) with the measurements of \(A\) and \(A^\complement\) show that the latter two picked up some extra “phantom measure”, meaning that one or the other was strictly overestimated by \(\mu^\star\).

This example was somewhat contrived, and the proof of the theorem in its full generality takes significant effort.

**Proof:** It’s obvious that \(\varnothing, X\in\mathcal M\), i.e. that \(\mathcal M\) is nonempty. Let \(A\in\mathcal M\). Then, \(E\cap A=E\setminus A^\complement\) and \(E\setminus A=E\cap A^\complement\), so
\[\mu^\star(E)=\mu^\star(E\cap A)+\mu^\star(E\setminus A)=\mu^\star\left(E\setminus A^\complement\right)+\mu^\star\left(E\cap A^\complement\right).\]
Hence, \(A^\complement\in\mathcal M\) and \(\mathcal M\) is closed under complements. Next, we’ll show that \(\mathcal M\) is closed under finite unions. For this, it suffices to show that if \(A, B\in\mathcal M\), then \(A\cup B\in\mathcal M\); closure under finite unions follows inductively.

We wish to show that for any \(E\subseteq X\), \(\mu^\star(E)=\mu^\star(E\cap(A\cup B))+\mu^\star(E\setminus(A\cup B))\). First, we’ll rewrite \(E\setminus(A\cup B)\) as \((E\setminus A)\setminus B\) for insidious reasons that will become clear later. Second, we’ll use the fact that \(A\in\mathcal M\) to write \[\begin{align*} \mu^\star(E\cap(A\cup B)) &=\mu^\star(E\cap(A\cup B)\cap A)+\mu^\star(E\cap (A\cup B)\setminus A) \\ &=\mu^\star(E\cap A)+\mu^\star((E\setminus A)\cup B). \end{align*}\] The second equality was done by performing some clever set-theoretic manipulations. Putting this back together gives us

\[\begin{align*} &\mu^\star(E\cap(A\cup B))+\mu^\star(E\setminus(A\cup B)) \\ =&\mu^\star(E\cap(A\cup B)\cap A)+\mu^\star(E\cap(A\cup B)\setminus A)+\mu^\star((E\setminus A)\setminus B) \\ =&\mu^\star(E\cap A)+\mu^\star((E\setminus A)\cap B)+\mu^\star((E\setminus A)\setminus B). \\ \end{align*}\]

We can now collect the last two terms because \(B\in\mathcal M\), and we can use \(A\in\mathcal M\) one last time to get: \[\mu^\star(E\cap(A\cup B))+\mu^\star(E\setminus(A\cup B))=\mu^\star(E\cap A)+\mu^\star(E\setminus A)=\mu^\star(E).\] Thus, we conclude that \(A\cup B\in\mathcal M\), and so \(\mathcal M\) is closed under finite unions. This argument is best served with a diagram, and I encourage the reader to provide their own.

It now remains to show that \(\mathcal M\) is closed under countable unions. Let \(\left\lbrace A_n\right\rbrace\) be a sequence in \(\mathcal M\), and let \(A=\bigcup_{n=0}^\infty A_n\). Since \(\mu^\star\) is an outer measure, we immediately have that \(\mu^\star(A)\leq\mu^\star(E\cap A)+\mu^\star(E\setminus A\) for any \(E\subseteq X\); it therefore suffices to show the reverse inequality.

First, notice that since \(E\setminus A\subseteq E\setminus\bigcup_{j=0}^nA_j\) for all \(n\), we have that \(\mu^\star(E\setminus A)\leq\mu^\star\left(E\setminus\bigcup_{j=0}^nA_j\right)\) for all \(n\). The quantity on the right is a nonincreasing sequence, so \[\mu^\star(E\setminus A)\leq\inf_n\mu^\star\left(E\setminus\bigcup_{j=0}^nA_j\right)=\lim_{n\to\infty}\mu^\star\left(E\setminus\bigcup_{j=0}^nA_j\right).\] Next, let \(B_0=A_0\) and \(B_n=A_n\setminus\bigcup_{j=0}^{n-1}A_j\) for \(n\geq 1\). Since we already showed that \(\mathcal M\) is closed under countable unions, we know that \(B_n\in\mathcal M\) for all \(n\). We now claim that \(\sum_{j=0}^n\mu^\star\left(E\cap B_j\right)=\mu^\star\left(E\cap\bigcup_{j=0}^nB_j\right)=\mu^\star\left(E\cap\bigcup_{j=0}^nA_j\right)\). The second equality is clear, and the first equality becomes clear after sitting in the dark for an hour and wondering why the proof in your own notes was wrong.

Consider for a moment that \[\begin{align*} \mu^\star\left(E\cap\left(A_0\cup A_1\right)\right) &=\mu^\star\left(E\cap\left(A_0\cup A_1\right)\cap A_0\right)+\mu^\star\left(E\cap\left(A_0\cup A_1\right)\setminus A_0\right) \\ &=\mu^\star\left(E\cap A_0\right)+\mu^\star\left(E\cap\left(A_1\setminus A_0\right)\right) \\ &=\mu^\star\left(E\cap B_0\right)+\mu^\star\left(E\cap B_1\right). \end{align*}\] With a bit of imagination, this can be easily extended inductively. So, we get that \[\mu^\star(E\cap A)\leq\sum_{j=0}^\infty\mu^\star\left(E\cap B_j\right)=\lim_{n\to\infty}\sum_{j=0}^n\mu^\star\left(E\cap B_j\right)=\lim_{n\to\infty}\mu^\star\left(E\cap\bigcup_{j=0}^n A_j\right).\] Putting this all together and remembering that \(\bigcup_{j=0}^n A_j\in\mathcal M\), we have: \[\begin{align*} \mu^\star(E\cap A)+\mu^\star(E\setminus A) &\leq\lim_{n\to\infty}\mu^\star\left(E\cap\bigcup_{j=0}^nA_j\right)+\lim_{n\to\infty}\mu^\star\left(E\setminus\bigcup_{j=0}^nA_j\right) \\ &=\lim_{n\to\infty}\left(\mu^\star\left(E\cap\bigcup_{j=0}^nA_j\right)+\mu^\star\left(E\setminus\bigcup_{j=0}^nA_j\right)\right) \\ &=\lim_{n\to\infty}\mu^\star(E)=\mu^\star(E). \end{align*}\] We thus conclude that \(\mathcal M\) is a \(\sigma\)-algebra. To show disjoint additivity of \(\mu=\left.\mu^\star\right|_{\mathcal M}\), one can adapt the argument used with the \(B_n\)’s above to an arbitrary disjoint sequence and an appropriate choice of \(E\); I’ll leave this as a relatively simple exercise for the reader. Hence, \((X,\mathcal M,\mu)\) is a measure space, as desired. \(\square\)

So far, we have shown a two-step process for constructing measures from premeasures, but we’ve only discussed the two steps in isolation. In particular, if \(\mu_0\) is a premeasure on an algebra \(\mathcal A\) over \(X\), \(\mu^\star\) is the outer measure induced by \(\mu_0\), and \(X,\mathcal M,\mu\) is the measure space obtained with Carathéodory’s method, is there any relationship between \(\mathcal M\) and \(\mathcal A\)?

**Proposition 4. **

Let \(\mu_0\) be a premeasure on an algebra \(\mathcal A\) over \(X\); let \(\mu^\star\) be the induced outer measure. Then, for all \(A\in\mathcal A\) and \(E\subseteq X\), \[\mu^\star(E)=\mu^\star(E\cap A)+\mu^\star(E\setminus A).\]

**Proof:** Let \(\epsilon>0\). Pick a sequence \(\left\lbrace B_n\right\rbrace\) in \(\mathcal A\) such that \(\sum_{n=0}^\infty\mu^\star\left(B_n\right)\leq\mu^\star(E)+\epsilon\) and \(E\subseteq\bigcup_{n=0}^\infty B_n\). Then, since each \(B_n\) is in \(\mathcal A\) and \(\mu^\star\) coincides with the premeasure \(\mu_0\) on \(\mathcal A\), we may use disjoint additivity to rewrite
\[\sum_{n=0}^\infty\mu^\star(B_n)=\sum_{n=0}^\infty\left(\mu^\star\left(B_n\cap A\right)+\mu^\star\left(B_n\setminus A\right)\right).\]
Since \(E\subseteq\bigcup_{n=0}^\infty\), we can collapse these sums and use monotonicity to write

\[\begin{align*} \mu^\star(E)+\epsilon &\geq\mu^\star\left(\bigcup_{n=0}^\infty B_n\cap A\right)+\mu^\star\left(\bigcup_{n=0}^\infty B_n\setminus A\right) \\ &\geq\mu^\star(E\cap A)+\mu^\star(E\setminus A). \end{align*}\]

Since this is true for any \(\epsilon\), we conclude that \(\mu^\star(E)\geq\mu^\star(E\cap A)+\mu^\star(E\setminus A)\). The reverse inequality is clear from disjoint subadditivity, giving the claim. \(\square\)

This lines up with our intuition: the outer measure agrees with the premeasure, and so the domain from Carathéodory’s method should contain the \(\sigma\)-algebra generated by \(\mathcal A\). Note that it doesn’t have to be exactly this \(\sigma\)-algebra, and it’s often quite a bit larger.

This pipeline only gives one way to construct measures out of premeasures, and it’s worth considering if other measures that extend the same premeasure exist.

**Theorem 5. **

Let \(\mathcal A\) be an algebra on a set \(X\), and let \(\mathcal M\) be the \(\sigma\)-algebra generated by \(\mathcal A\). Let \(\mu_0\) be a premeasure on \(\mathcal A\), and let \(\mu^\star\) be the outer measure induced by \(\mu_0\). Let \(\mu=\left.\mu^\star\right\rvert_{\mathcal M}\), which is a measure such that \(\left.\mu\right\rvert_{\mathcal A}=\mu_0\). If \(\nu\) is another such measure on \(\mathcal M\) that extends \(\mu_0\), then \(\nu\leq\mu\) everywhere, and \(\nu=\mu\) whenever \(\mu\) is finite. In particular, \(\mu\) is unique if it is \(\sigma\)-finite.

**Proof:** Let \(A\in\mathcal M\), \(\epsilon>0\), and let \(\left\lbrace A_n\right\rbrace\) be a disjoint sequence in \(\mathcal A\) such that \(A\subseteq\bigcup_{n=0}^\infty A_n\) and \(\mu\left(A\right)+\epsilon\geq\sum_{n=0}^\infty\mu\left(A_n\right)\). Since \(\mu\) and \(\nu\) both agree with \(\mu_0\) on \(\mathcal A\), we get that \(\mu\left(A\right)+\epsilon\geq\sum_{n=0}^\infty\nu\left(A_n\right)\geq\nu\left(A\right)\), where the right inequality comes from monotonicity of \(\nu\). Since this is true for all \(\epsilon>0\), it follows that \(\nu\left(A\right)\leq\mu\left(A\right)\).

In particular, when \(\mu\left(A\right)<\infty\), we may subtract things and write that \[\begin{align*} \mu(A) &=\mu\left(\bigcup_{n=0}^\infty A_n\right)-\mu\left(\bigcup_{n=0}^\infty A_n\setminus A\right) \\ &\leq\sum_{n=0}^\infty\mu\left(A_n\right)-\nu\left(\bigcup_{n=0}^\infty A_n\setminus A\right) \\ &=\sum_{n=0}^\infty \nu\left(A_n\right)-\nu\left(\bigcup_{n=0}^\infty A_n\setminus A\right) \\ &=\nu\left(\bigcup_{n=0}^\infty A_n\right)-\nu\left(\bigcup_{n=0}^\infty A_n\setminus A\right) \\ &=\nu(A) \end{align*}\] For the first inequality, we used the fact that \(\nu\leq\mu\) everywhere. Note that we assumed that \(\left\lbrace A_n\right\rbrace\) was a disjoint sequence so that we could swap the unions and sums. Since \(\mu(A)\leq\nu(A)\) whenever \(\mu(A)\) is finite, we conclude that \(\mu=\nu\) when \(\mu<\infty\).

Finally, if \(\mu\) is \(\sigma\)-finite, we can express \(X\) as a union of disjoint sets \(\left\lbrace A_n\right\rbrace\) such that \(\mu\left(A_n\right)<\infty\) for all \(n\). Then, for any \(A\in\mathcal M\), we have \[\mu(A)=\mu\left(\bigcup_{n=0}^\infty A\cap A_n\right)=\sum_{n=0}^\infty\mu\left(A\cap A_n\right)=\sum_{n=0}^\infty\nu\left(A\cap A_n\right)=\nu(A).\] Notably, since \(A\cap A_n\subseteq A_n\) for all \(n\), \(\mu\) and \(\nu\) agree on them. Thus, \(\mu\) is unique if it is \(\sigma\)-finite. \(\square\)

We had mentioned previously that a nice property to work with was being complete. Unfortunately, Carathéodory’s method doesn’t always produce complete measures, and in general, it’s very difficult to verify that a measure is complete to begin with.

Notably, applying Carathéodory’s theorem to \(\lambda\) doesn’t *a priori* produce a complete measure, and the tempting restriction of \(\lambda\) to the Borel \(\sigma\)-algebra is strictly not complete.

**Theorem 6. **

Let \((X,\mathcal M,\mu)\) be a measure space, and let \(\mathcal N\) be the set of subsets of \(\mu\))-null sets. Define
\[\overline{\mathcal M}=\left\lbrace M\Delta N:M\in\mathcal M, N\in\mathcal N\right\rbrace.\]
Then \(\overline{\mathcal M}\) is a \(\sigma\)-algebra, and \(\overline\mu:M\cup N\mapsto\mu(M)\) is a measure on \(\left(X,\overline{\mathcal M}\right).\) \(\left(X,\overline{\mathcal M}, \overline\mu\right)\) is a complete measure space, and it's called the *completion* of \(\left(X,\mathcal M,\mu\right)\).

Note here that \(\Delta\) represents the symmetric difference.

**Proof:** As \(\varnothing\in\mathcal N\) and \(\mathcal M\), \(\varnothing\Delta\varnothing\in\overline{\mathcal M}\). To show that \(\mathcal M\) is closed under complements, notice that if \(M\Delta N\in\overline{\mathcal M}\) for \(M\in\mathcal M\) and \(N\in\mathcal N\), we have
\[\begin{align*}
\left(M\Delta N\right)^\complement
&=\left(M\cap N\right)\cup\left(M^\complement\cap N^\complement\right) \\
&=\left(N\setminus M^\complement\right)\cup\left(M^\complement\setminus N\right) \\
&=M^\complement\Delta N.
\end{align*}\]
As \(M^\complement\in\mathcal M\), we get that \(\left(M\Delta N\right)^\complement\in\overline{\mathcal M}\). If \(\left\lbrace M_i\Delta N_i\right\rbrace _i\), then \(\bigcup_{i=0}^\infty\left(M_i\Delta N_i\right)\supseteq M\Delta N\), where \(M=\bigcup_{i=0}^\infty M_i\) and \(N=\bigcup_{i=0}^\infty N_i\).