What Makes a Measure Finite?
A measure can assign infinite size to some measurable sets, including the whole space. A finite measure is one for which the whole space has finite measure. This condition has immediate consequences: every measurable set has finite measure, and the measure can be rescaled to a probability measure whenever the whole space has positive measure.
As in the previous tutorial, write \((X,\mathcal{F},\mu)\) for a measure space. Here \(X\) is the underlying space, \(\mathcal{F}\) is its sigma-algebra, and \(\mu\) is the measure. The word “finite” describes the total measure, not the number of elements in \(X\). A space can contain infinitely many points and still have finite measure.
A finite measure is still a measure: in particular, it remains countably additive on pairwise disjoint measurable sets. Finiteness does not replace countable additivity, and it does not mean that \(\mathcal{F}\) contains only finitely many sets. The condition concerns just one value, \(\mu(X)\), but it constrains the values on every measurable subset of \(X\).
Proof. The measure is nonnegative, so \(\mu(A)\geq 0\). Since \(A\subseteq X\), monotonicity of measures, established earlier in the course, gives \(\mu(A)\leq\mu(X)\). The definition of a finite measure states that \(\mu(X)<\infty\). Therefore \(0\leq\mu(A)\leq\mu(X)<\infty\), as required. \(\square\)
The bound applies to every measurable set, including complements and countable unions. For example, if \(\mu(X)=M\), no measurable subset can have measure greater than \(M\). This is often the first useful check when deciding whether a proposed measure space is finite.
Finite Does Not Mean a Finite Number of Points
Counting measure gives a simple distinction. On a set \(X\), counting measure assigns to a set its number of elements when that set is finite, and assigns infinity when it is infinite. Thus, on the full power set, counting measure is finite exactly when \(X\) is finite. But many other measures assign finite total mass to spaces with infinitely many points.
Worked Example: Counting Measure on a Finite Space
Let \(X=\{a,b,c,d\}\), let \(\mathcal{F}=\mathcal{P}(X)\), and let \(\mu\) be counting measure. Then \(\mu(X)=4\), so \(\mu\) is a finite measure. For \(A=\{a,c\}\), counting measure gives \(\mu(A)=2\), which is no larger than the total mass \(4\). The whole space has finitely many points in this example, but that is a feature of the chosen space, not part of the definition of a finite measure.
Worked Example: A Finite Measure on an Infinite Space
On \((\mathbb{N},\mathcal{P}(\mathbb{N}))\), assign weight \(w_n=3\cdot 2^{-n}\) to the point \(n\), and define \(\mu(A)=\sum_{n\in A}w_n\). The weighted atomic measure theorem from “Examples of Measures” ensures that this assignment is a measure. Its total mass is $$ \mu(\mathbb{N})=\sum_{n=1}^{\infty}3\cdot 2^{-n} =3\sum_{n=1}^{\infty}2^{-n} =3. $$ Therefore this is a finite measure, even though \(\mathbb{N}\) has infinitely many elements. For instance, $$ \mu(\{4,5,6,\ldots\}) =\sum_{n=4}^{\infty}3\cdot 2^{-n} =3\frac{2^{-4}}{1-1/2} =\frac{3}{8}. $$ This value is finite and no greater than the total mass \(3\).
Finite Measures Are Sigma-Finite
A measure is called sigma-finite if the space can be covered by countably many measurable sets, each having finite measure. Finite measures meet this condition immediately: the whole space itself is already one measurable set of finite measure. This fact connects the current topic with the broader class of measures considered in the next tutorial.
Proof. For every positive integer \(n\), set \(E_n=X\). Each \(E_n\) is measurable, and \(\mu(E_n)=\mu(X)<\infty\) because \(\mu\) is finite. Also, $$ X=\bigcup_{n=1}^{\infty}E_n. $$ Thus \(X\) is a countable union of measurable sets of finite measure, which is precisely the definition of sigma-finiteness. \(\square\)
The converse is not true: a sigma-finite measure space may have infinite total measure. Sigma-finiteness only requires a countable cover by finite-measure pieces; it does not require the sum of their measures, or the measure of their union, to be finite. The distinction is useful because sigma-finiteness is a weaker condition than finiteness.
Normalizing a Finite Measure
Probability measures are finite measures with total mass one. More generally, any finite measure with positive total mass can be rescaled to have total mass one. This gives a direct way to pass from a finite measure space to a probability space without changing which measurable sets have measure zero.
Proof. Since \(\mu(A)\geq 0\) and \(\mu(X)>0\), the value \(\mathbb{P}(A)\) is nonnegative for every measurable \(A\). Let \((A_n)_{n=1}^{\infty}\) be pairwise disjoint measurable sets. Countable additivity of \(\mu\) gives $$ \mathbb{P}\left(\bigcup_{n=1}^{\infty}A_n\right) =\frac{\mu\left(\bigcup_{n=1}^{\infty}A_n\right)}{\mu(X)} =\frac{\sum_{n=1}^{\infty}\mu(A_n)}{\mu(X)} =\sum_{n=1}^{\infty}\frac{\mu(A_n)}{\mu(X)} =\sum_{n=1}^{\infty}\mathbb{P}(A_n). $$ Thus \(\mathbb{P}\) is countably additive. Finally, $$ \mathbb{P}(X)=\frac{\mu(X)}{\mu(X)}=1. $$ It is therefore a probability measure by definition. \(\square\)
The condition \(\mu(X)>0\) is essential: division by zero is not defined. If instead \(\mu(X)=0\), monotonicity shows that every measurable set \(A\) has \(0\leq\mu(A)\leq\mu(X)=0\), so \(\mu(A)=0\) for all \(A\in\mathcal{F}\). This is the zero measure, and it cannot be normalized to total mass one by dividing by its total mass.
Worked Example: Rescaling Lebesgue Measure on an Interval
Take \(X=[0,4]\), let \(\mathcal{F}\) be the Borel subsets of \(X\), and let \(\mu\) be the restriction of Lebesgue measure to \(X\). Its total mass is the length of the interval: $$ \mu(X)=4. $$ The normalization theorem defines \(\mathbb{P}(A)=\mu(A)/4\). For \(A=[1,3]\), the measure is its length, so $$ \mathbb{P}(A)=\frac{\mu([1,3])}{4} =\frac{3-1}{4} =\frac12. $$ The whole interval has probability \(4/4=1\), and the subinterval has probability one-half. The original measure records length, while its normalization records length as a fraction of the total length.
Disjoint Sets and Vanishing Tails
Finiteness also limits how much measure can be distributed among disjoint sets. If a sequence of measurable sets is pairwise disjoint, the sum of their measures cannot exceed the measure of the whole space. In particular, the measures of the remaining sets in the sequence must eventually have arbitrarily small total mass.
Proof. For each positive integer \(N\), finite additivity on the first \(N\) pairwise disjoint sets gives $$ \sum_{n=1}^{N}\mu(A_n) =\mu\left(\bigcup_{n=1}^{N}A_n\right) \leq\mu(X), $$ where the inequality follows from monotonicity. The partial sums are nondecreasing because every term is nonnegative, and they are bounded above by the finite number \(\mu(X)\). Hence the series \(\sum_{n=1}^{\infty}\mu(A_n)\) converges and is at most \(\mu(X)\). Countable additivity also gives $$ \mu(T_N)=\mu\left(\bigcup_{n>N}A_n\right) =\sum_{n>N}\mu(A_n). $$ The right side is the tail of a convergent series of nonnegative real numbers, so it tends to zero as \(N\to\infty\). Therefore \(\mu(T_N)\to 0\). \(\square\)
One consequence is that there cannot be infinitely many pairwise disjoint measurable sets each having measure at least some fixed \(\varepsilon>0\). If there were, their first \(N\) sets would have total measure at least \(N\varepsilon\), which eventually exceeds the finite bound \(\mu(X)\). Tail control is stronger: even when the individual measures become small, the total measure of all sets after the \(N\)th one must tend to zero.
Worked Example: Tail Control for Weighted Points
Return to the measure on \(\mathbb{N}\) with \(\mu(\{n\})=3\cdot 2^{-n}\). The singleton sets are pairwise disjoint, and their union is \(\mathbb{N}\). The tail after \(N\) has measure $$ \mu(\{N+1,N+2,\ldots\}) =\sum_{n=N+1}^{\infty}3\cdot 2^{-n} =3\frac{2^{-(N+1)}}{1-1/2} =3\cdot 2^{-N}. $$ This tends to zero as \(N\to\infty\). For example, after the first four points the remaining mass is \(3\cdot 2^{-4}=3/16\). The underlying space is infinite, but the mass left in the tail can be made as small as desired by taking enough initial points.
Why Finiteness Matters
Finiteness provides a uniform bound for all measurable sets and makes normalization possible when the total mass is positive. It also prevents disjoint pieces from carrying an unbounded amount of mass and ensures that the tails of any disjoint measurable decomposition have measure tending to zero. These features make finite measures a particularly manageable setting for many arguments.
A common pitfall is to confuse “finite measure” with “finite sample space,” or to assume that every finite measure is already a probability measure. The first is false because infinitely many points can have summable weights. The second is false because a finite measure may have total mass other than one. A measure becomes a probability measure only after its total mass is one; positive finite total mass allows this by normalization, while zero total mass does not.
Check Your Understanding
Use the definitions and results in this tutorial to answer the following questions.
- What condition on \(\mu(X)\) makes a measure finite?
- Why does a finite measure on an infinite set of points not contradict the definition of a finite measure?
- How can a finite measure with positive total mass be converted into a probability measure?
- Why is a finite measure automatically sigma-finite?
- For pairwise disjoint measurable sets in a finite measure space, what happens to the measure of the union of the sets after the \(N\)th one as \(N\) tends to infinity?