## 1.2. Measures and Their Properties

≪ 1.1. \(\sigma\)-Algebras | Table of Contents | 1.3. Constructing Measures ≫There’s an old story I remember about weighing an elephant. This was a long time ago (at *least* a hundred years ago), when people were consistently bored enough to wonder how much an elephant weighed. Maybe they wanted to sell the elephant and charged a price per pound. Who knows. But nobody had a scale that wouldn’t get demolished by the weight of an elephant!

In comes this smart aleck kid. He loads the elephant on a boat, marks the water level, removes the elephant, and adds stones to the boat until the water level is met. He then weighed the stones individually and reasoned that the total weight of the stones was equal to the weight of the elephant.

This story highlights the important property of measures that we want to abstract away: if you can measure the weight of the stones individually, you can measure (or infer) the weight of all of the stones. Let’s formally define them:

**Definition 3. Measures**

Let \((X, \mathcal M)\) be a measurable space, and let \(\mu:\mathcal M\to [0,\infty]\) such that:

- \(\mu(\varnothing)=0.\)
- For all disjoint sequences \(\lbrace A_n\rbrace\) in \(\mathcal M\), \(\mu\left(\bigcup_{n=0}^\infty A_n\right)=\sum_{n=0}^\infty\mu\left(A_n\right).\)

It’s easy to see that the second property can be restated with finite disjoint unions: if \(A_1, A_2\in\mathcal M\) are disjoint, then \(\mu\left(A_1\cup A_2\right)=\mu\left(A_1\right)+\mu\left(A_2\right)\).

Something to note is that we’re allowing \(\mu\) to take infinite values; this ensures that the series in the second condition still has meaning even if it doesn’t converge in the traditional sense. This is a necessary evil, but it remains consistent with our intuition: the real number line \(\R\) has “infinite length”.

If \(A\in\mathcal M\) and \(\mu(A)=0\), we say \(A\) is a null set, a negligible set, or that \(A\) is \(\mu\)-negligible. If some statement is true for all \(x\in X\) except for some \(\mu\)-negligible set, we say that the statement is true \(\mu\)-almost everywhere, or just \(\mu\)-a.e. or a.e. for short. In probabilistic settings, “almost always” is used instead.

**Proposition 4. **

Let \((X, \mathcal M, \mu)\) be a measure space.

- If \(E, F\in\mathcal M\) such that \(E\subseteq F\), \(\mu(E)\leq\mu(F)\).
- If \(\lbrace A_n\rbrace\) is any sequence in \(\mathcal M\), \(\mu\left(\bigcup_{n=0}^\infty A_n\right)\leq\sum_{n=0}^\infty\mu\left(A_n\right)\).
- If \(\lbrace A_n\rbrace\) is a nondecreasing sequence in \(\mathcal M\) (i.e. \(A_n\subseteq A_{n+1}\) for all \(n\)), then \[\mu\left(\bigcup_{n=0}^\infty A_n\right)=\lim_{n\to\infty}\mu\left(A_n\right)=\sup_{n\to\infty}\mu\left(A_n\right).\]
- If \(\lbrace A_n\rbrace\) is a nonincreasing sequence in \(\mathcal M\) and \(\mu\left(A_0\right)<\infty\), then \[\mu\left(\bigcap_{n=0}^\infty A_n\right)=\lim_{n\to\infty}\mu\left(A_n\right)=\inf_{n\to\infty}\mu\left(A_n\right).\]

**Proof:** For (i), we may write \(F=E\cup(F\setminus E)\). This is a disjoint union and measures are nonnegative, so \(\mu(F)=\mu(E)+\mu(F\setminus E)\geq\mu(E)\).

For (ii), define \(B_0=A_0\) and \(B_n=A_n\setminus\left(\bigcup_{j=0}^{n-1}A_j\right)\) for \(n\geq 1\). \(\lbrace B_n\rbrace\) is a disjoint sequence in \(\mathcal M\), and \(B_n\subseteq A_n\) for all \(n\), so disjoint additivity together with (i) gives \[\mu\left(\bigcup_{n=0}^\infty A_n\right)=\mu\left(\bigcup_{n=0}^\infty B_n\right)=\sum_{n=0}^\infty\mu\left(B_n\right)\leq\sum_{n=0}^\infty\mu\left(A_n\right).\] For (iii), define \(B_n\) in a similar manner. Notice that \(\bigcup_{j=0}^nB_j=A_n\) and so \(\mu\left(\bigcup_{j=0}^nB_j\right)=\mu\left(A_n\right)\). Hence, we get that \[\mu\left(\bigcup_{j=0}^\infty A_j\right)=\mu\left(\bigcup_{j=0}^\infty B_j\right)=\sum_{j=0}^\infty\mu\left(B_j\right)=\lim_{n\to\infty}\sum_{j=0}^n\mu\left(B_j\right) =\lim_{n\to\infty}\mu\left(A_n\right).\] Since \(\mu\left(A_n\right)\) is a nondecreasing sequence, we know the limit is equal to the \(\sup\).

Finally, for (iv), define \(C_n=A_n\setminus A_{n+1}\), and let \(A=\bigcap_{n=0}^\infty A_n\) for brevity. The \(C_n\)’s are a disjoint sequence in \(\mathcal M\). Furthermore, we have \(A_0=A\cup\left(\bigcup_{n=0}^\infty C_n\right)\). Since \(A\) is disjoint from each \(C_n\), \[\mu\left(A_0\right)=\mu(A)+\sum_{n=0}^\infty\mu\left(C_n\right)\implies\mu(A)=\mu\left(A_0\right)-\sum_{n=0}^\infty\mu\left(C_n\right). \] Notably, \(\bigcup_{n=0}^\infty C_n\subseteq A_0\), which has finite measure, so this subtraction is well-defined. But notice that \(A_0=A_n\cup\bigcup_{j=0}^{n-1}C_j\), so \(\mu\left(A_n\right)=\mu\left(A_0\right)-\sum_{j=0}^{n-1}\mu\left(C_j\right)\) for all \(n\). Taking limits therefore gives us \[\mu(A)=\lim_{n\to\infty}\left(\mu\left(A_0\right)-\sum_{j=0}^{n-1}\mu\left(C_j\right)\right)=\lim_{n\to\infty}\mu\left(A_n\right).\] Since \(\mu\left(A_n\right)\) is a decreasing sequence, the limit is equal to the \(\inf\). \(\square\)

A very trivial example of a measure is given by \(\mu(A)=0\) for all \(A\in\mathcal M\); another very trivial example is \(\mu(A)=\infty\) for all nonempty \(A\in\mathcal M\). A more useful example is called the *counting measure*, where \(\mu(A)\) is the cardinality of \(A\). Note that the counting measure is defined on \(\mathcal P(X)\).

The counting measure also provides an important example that demonstrates the necessity of \(\mu\left(A_0\right)<\infty\) in part (iv) of Proposition 4. Consider the counting measure on \(\R\), and define \(A_n=\left(-\frac 1n,\frac 1n\right)\) for \(n\geq 1\). Then, the intersection of all \(A_n\)’s is just \(\lbrace 0\rbrace\), whose measure is \(1\). However, the measure of each \(A_n\) is \(\infty\) in the counting measure.

We’ll note here that many measures are not well-behaved. First, there’s the issue of infinite measures. Using disjoint additivity, it’s tempting to say that if \(A, B\) are two \(\mu\)-measurable sets for some measure \(\mu\), then \(\mu(B)=\mu(A\cup B)-\mu(A\setminus B)\). However, if both \(A\cup B\) and \(A\setminus B\) have infinite measure, we end up with \(\infty-\infty\).

Looking back at part (iv) of Proposition 4 again, we had to step around this by making assumptions about the finitude of certain measures. In general, there are many results that are easy to say about measures where \(\mu(X)<\infty\) precisely because we can avoid dealing with this indeterminate form from popping up.

However, most measures we care about aren’t like that: remember that the “length” of \(\R\) is infinite. In a way, however, lengths on \(\R\) aren’t *too* infinite, as we can break up \(\R\) into pieces of finite length. We can then work *locally* before stitching things together to make a *global* statement about lengths on \(\R\).

This pattern of the local-to-global pipeline is quite common in analysis. In complex analysis, for instance, it’s often easy to show some *local* property about holomorphic functions before concluding a *global* statement about them.

**Definition 1. **

Let \((X,\mathcal M,\mu)\) be a measure space. \(\mu\) is a finite measure if \(\mu(X)<\infty\). \(\mu\) is a \(\sigma\)-finite measure if there exists a sequence \(\left\lbrace A_n\right\rbrace\) in \(\mathcal M\) such that \(\mu\left(A_n\right)<\infty\) for all \(n\) and \(X=\bigcup_{n=0}^\infty A_n\).

An example of a measure that isn’t \(\sigma\)-finite is the counting measure on \(\R\). This illustrates that \(\sigma\)-finiteness is *not* a property of the space \(X\) or the measurable space \((X,\mathcal M)\), but rather a property of the measure itself.

The other issue with arbitrary measures can be illustrated by an example. Let \(X\) be a nonempty set with more than 1 element, let \(\mathcal M=\lbrace\varnothing, X\rbrace\), and define the measure \(\mu\) that sets \(\mu(X)=0\). Although \(\mu\) is only defined on \(\mathcal M\), one may reason by monotonicity that \(\mu\) *should* be defined on \(\mathcal P(X)\), and that the measure of *any* subset of \(X\) should be zero.

More broadly, subsets of null sets may not be measurable, even though there’s only one possible value their measure could take. Other nonmeasurable sets may only be a null set (or a subset of a null set) away from a measurable set, and one could argue that they are “close enough” to being measurable. This property turns out to be very helpful, and it has a name.

**Definition 2. **

Let \((X, \mathcal M,\mu)\) be a measure space. \(\mu\) is called a complete measure if \(A\in\mathcal M\) whenever \(A\subset N\) for some null set \(N\).

This should remind you of complete metric spaces. In a sense, the two concepts are very similar. Complete metric spaces are nice to work with because there are no “holes” in them; complete measure spaces are nice to work with because there are no “holes” in the domain.

Just like how any metric space has a completion, we can take complete measures. We’ve given an intuitive outline above, and we’ll provide a more rigorous treatment of taking completions in a future section.