From a Partition Supremum to a Measure
For a signed measure, the value \(|\nu(E)|\) records the net mass of a measurable set \(E\): positive and negative contributions can cancel. Total variation measures something different. It records the largest total absolute mass that can be exposed by splitting \(E\) into measurable pieces. The definition takes a supremum over finite partitions, so countable additivity is not automatic. Its proof must show both that splitting a set cannot create too much variation and that variations on disjoint pieces can be combined.
Throughout, \((X,\mathcal M)\) is a measurable space and \(\nu\) is a finite, real-valued signed measure on it. Here “finite” means that its positive and negative Jordan parts \(\nu^+\) and \(\nu^-\), established in the earlier material on the Jordan decomposition, satisfy \(\nu^+(X)<\infty\) and \(\nu^-(X)<\infty\). In particular, \(\nu=\nu^+-\nu^-\), and \(M=\nu^+(X)+\nu^-(X)<\infty\). We use countable additivity of these positive measures. All sets in partitions below belong to \(\mathcal M\).
The quantity is finite: for any finite partition \(E_1,\ldots,E_k\) of \(E\), the Jordan decomposition and the triangle inequality give $$ \sum_{j=1}^{k}|\nu(E_j)| \le \sum_{j=1}^{k}\bigl(\nu^+(E_j)+\nu^-(E_j)\bigr) =\nu^+(E)+\nu^-(E) \le M. $$ Thus \(0\le|\nu|(E)\le M\). This bound also explains why the finite signed-measure assumption is useful: it controls sums of absolute values over countably many disjoint sets.
Two Bounds That Give Countable Additivity
Proof. Nonnegativity follows from the definition, and the empty-set value is zero. It remains to prove countable additivity. Let \(E_1,E_2,\ldots\in\mathcal M\) be pairwise disjoint and let \(E=\bigcup_{n=1}^{\infty}E_n\). We prove the two inequalities separately.
First fix a finite measurable partition \(A_1,\ldots,A_k\) of \(E\). For each \(i\), the sets \(A_i\cap E_n\), \(n\ge1\), are disjoint and have union \(A_i\). Countable additivity of \(\nu\) gives $$ \nu(A_i)=\sum_{n=1}^{\infty}\nu(A_i\cap E_n). $$ This series is absolutely convergent: since \(\nu=\nu^+-\nu^-\), $$ \sum_{n=1}^{\infty}|\nu(A_i\cap E_n)| \le \sum_{n=1}^{\infty}\bigl(\nu^+(A_i\cap E_n)+\nu^-(A_i\cap E_n)\bigr) \le M. $$ The triangle inequality for an absolutely convergent series therefore yields $$ |\nu(A_i)|\le\sum_{n=1}^{\infty}|\nu(A_i\cap E_n)|. $$ For each fixed \(n\), the nonempty sets among \(A_1\cap E_n,\ldots,A_k\cap E_n\) form a finite partition of \(E_n\). Hence $$ \sum_{i=1}^{k}|\nu(A_i\cap E_n)|\le|\nu|(E_n). $$ Summing the preceding inequalities over the finite index \(i\), and then summing over \(n\), gives $$ \sum_{i=1}^{k}|\nu(A_i)| \le\sum_{n=1}^{\infty}\sum_{i=1}^{k}|\nu(A_i\cap E_n)| \le\sum_{n=1}^{\infty}|\nu|(E_n). $$ Taking the supremum over all finite partitions \(A_1,\ldots,A_k\) of \(E\) proves $$ |\nu|(E)\le\sum_{n=1}^{\infty}|\nu|(E_n). $$
For the reverse inequality, fix \(N\ge1\) and \(\varepsilon>0\). For each \(1\le n\le N\), the definition of supremum gives a finite measurable partition \(B_{n,1},\ldots,B_{n,k_n}\) of \(E_n\) such that $$ \sum_{j=1}^{k_n}|\nu(B_{n,j})|>|\nu|(E_n)-\frac{\varepsilon}{N}. $$ If \(|\nu|(E_n)=0\), any finite partition suffices for this strict inequality. Combine all these partitions and add the remainder \(R=E\setminus\bigcup_{n=1}^{N}E_n\), omitting \(R\) if it is empty. The resulting finite family is a partition of \(E\), since the \(E_n\) are pairwise disjoint. Its sum of absolute values is at least the sum over the selected \(B_{n,j}\); the additional term \(|\nu(R)|\), if present, is nonnegative. Consequently, $$ |\nu|(E)\ge\sum_{n=1}^{N}\sum_{j=1}^{k_n}|\nu(B_{n,j})| >\sum_{n=1}^{N}|\nu|(E_n)-\varepsilon. $$ Because this holds for every \(\varepsilon>0\), it implies \(|\nu|(E)\ge\sum_{n=1}^{N}|\nu|(E_n)\). This is true for every \(N\), so taking the increasing limit of the nonnegative partial sums gives $$ |\nu|(E)\ge\sum_{n=1}^{\infty}|\nu|(E_n). $$ The two inequalities prove countable additivity. Together with nonnegativity and \(|\nu|(\varnothing)=0\), this shows that \(|\nu|\) is a positive measure. Its finiteness follows from \(|\nu|(X)\le M<\infty\). \(\square\)
Worked Calculations
Worked Example: A Four-Point Space
Let \(X=\{1,2,3,4\}\), let \(\mathcal M\) be its power set, and define \(\nu(\{1\})=4\), \(\nu(\{2\})=-3\), \(\nu(\{3\})=1\), and \(\nu(\{4\})=-2\). For any \(A\subseteq X\), the triangle inequality shows that every finite partition \(A_1,\ldots,A_k\) of \(A\) satisfies $$ \sum_{j=1}^{k}|\nu(A_j)| \le\sum_{j=1}^{k}\sum_{i\in A_j}|\nu(\{i\})| =\sum_{i\in A}|\nu(\{i\})|. $$ The singleton partition of \(A\) attains this upper bound. Thus \(|\nu|(A)=\sum_{i\in A}|\nu(\{i\})|\) in this example.
Take \(A=\{1,2\}\) and \(C=\{3,4\}\). The singleton partitions give $$ |\nu|(A)=|4|+|-3|=7,\qquad |\nu|(C)=|1|+|-2|=3. $$ The whole-space singleton partition gives \(|\nu|(X)=4+3+1+2=10\), so \(|\nu|(A\cup C)=10=7+3\). In contrast, \(\nu(A)=4-3=1\), \(\nu(C)=1-2=-1\), and \(\nu(X)=0\). The check confirms both additivity of variation on these disjoint sets and the fact that variation is not simply the absolute value of the signed mass.
Worked Example: A Countable Partition of the Integers
Let \(X=\mathbb N=\{1,2,\ldots\}\), \(\mathcal M\) its power set, and define $$ \nu(A)=\sum_{n\in A}(-1)^{n+1}2^{-n}. $$ The series is absolutely convergent for every \(A\), since \(\sum_{n=1}^{\infty}2^{-n}=1\). For disjoint \(A_j\), the series over \(\bigcup_j A_j\) can be grouped by the sets \(A_j\): absolute convergence ensures that regrouping does not change the sum. Thus \(\nu\) is countably additive. Its positive and negative masses are finite, so the theorem applies.
Let \(O\) be the odd positive integers and \(Q\) the even positive integers. The geometric-series calculations are $$ \nu(O)=\sum_{k=1}^{\infty}2^{-(2k-1)} =\frac{1/2}{1-1/4}=\frac23, \qquad \nu(Q)=-\sum_{k=1}^{\infty}2^{-2k} =-\frac{1/4}{1-1/4}=-\frac13. $$ For any finite partition of \(X\), the sum of the absolute signed masses is at most \(\sum_{n=1}^{\infty}2^{-n}=1\), by the triangle inequality and absolute convergence. The two-set partition \(\{O,Q\}\) attains \(2/3+1/3=1\), so \(|\nu|(X)=1\). The same argument on \(O\) and \(Q\), using each set alone as a partition, gives \(|\nu|(O)=2/3\) and \(|\nu|(Q)=1/3\). Therefore \(|\nu|(O\cup Q)=1=|\nu|(O)+|\nu|(Q)\). Also, the singleton partition gives \(\sum_{n\ge1}|\nu(\{n\})|=\sum_{n\ge1}2^{-n}=1\), an independent check of the value.
Worked Example: Two Opposite Dirac Masses
On \(\mathbb R\) with its Borel sigma-algebra, let \(\nu= \delta_0-\delta_1\), where \(\delta_x(E)=1\) if \(x\in E\) and \(0\) otherwise. Consider \(E=\{0,1\}\). The one-piece partition gives \(|\nu(E)|=|1-1|=0\), but the partition into \(\{0\}\) and \(\{1\}\) gives \(|1|+|-1|=2\). No partition can give more: pieces containing neither point have mass zero, and the two nonzero contributions have absolute values summing to at most \(2\). Hence \(|\nu|(E)=2\).
For the disjoint sets \(E_0=\{0\}\) and \(E_1=\{1\}\), the only nonzero contribution on each is respectively \(1\) and \(-1\), so \(|\nu|(E_0)=1\) and \(|\nu|(E_1)=1\). The result \(|\nu|(E_0\cup E_1)=2\) agrees with their sum. The calculation checks concretely that the partition supremum detects cancellation that the one-piece partition misses.
A Common Mistake and Why the Measure Property Matters
A tempting but incorrect calculation is to take the one-piece partition of \(E\) and conclude that \(|\nu|(E)=|\nu(E)|\). In the Dirac example, this gives the wrong value \(0\) for \(E=\{0,1\}\); the two-piece partition gives \(2\). The correction is to take the supremum over all finite measurable partitions, not to evaluate only the unsplit set. Likewise, countable additivity does not follow just by moving a supremum through an infinite sum. The proof above instead uses two separate arguments: every partition of the union is bounded by the variations of its intersections with the disjoint pieces, while near-maximizing partitions of finitely many pieces can be joined into a partition of the union.
The result makes \(|\nu|\) available as an ordinary finite positive measure. In particular, it permits measure-theoretic arguments to control the size of signed contributions without relying on cancellation. The finiteness assumption enters explicitly when the proof represents \(\nu(A_i)\) as an absolutely convergent series over the disjoint pieces. For an arbitrary signed set function without finite positive and negative masses, that absolute-convergence justification cannot simply be omitted.
Check Your Understanding
For each question, use the finite-partition definition and state where countable additivity or finiteness is needed.
- For a finite partition \(E_1,\ldots,E_k\) of \(E\), explain why \(\sum_j|\nu(E_j)|\le\nu^+(E)+\nu^-(E)\).
- In the four-point example, compute \(|\nu|(\{1,3,4\})\) and compare it with \(|\nu(\{1,3,4\})|\).
- Why does the proof of \(|\nu|(\bigcup_n E_n)\le\sum_n|\nu|(E_n)\) need absolute convergence of \(\sum_n\nu(A_i\cap E_n)\)?
- For the countable example, compute \(\nu(O)\) and \(\nu(Q)\) using geometric series and verify that the partition \(\{O,Q\}\) attains \(|\nu|(X)\).
- For \(\nu=\delta_0-\delta_1\), why does the one-piece partition of \(\{0,1\}\) fail to attain the supremum?