Signed and complex measures allow cancellation, but their structure is still controlled by positivity through Hahn, Jordan, and total variation decompositions.
Signed measures
A signed measure is countably additive but may take negative values.
Definition. A finite signed measure \(\nu\) on \((X,\mathcal A)\) is a countably additive map \(\nu:\mathcal A\to\mathbb R\) with \(\nu(\varnothing)=0\). More generally, an extended signed measure may take values in \([-\infty,\infty]\), but it is not allowed to take both \(+\infty\) and \(-\infty\); this prevents undefined expressions under additivity.
Definition. A set \(P\) is positive for \(\nu\) if every measurable subset of \(P\) has nonnegative \(\nu\)-measure; negative sets are defined analogously.
Hahn decomposition
Every signed measure splits the space into positive and negative regions.
Theorem. There exist disjoint measurable sets \(P,N\) with \(P\cup N=X\), where \(P\) is positive and \(N\) is negative for \(\nu\). The decomposition is unique modulo \(\nu\)-null sets.
Proof. We prove the finite case. Let \[ \alpha=\sup\{\nu(A):A\in\mathcal A\}<\infty. \] Choose \(A_n\) with \(\nu(A_n)>\alpha-2^{-n}\). If \(B\subset A_n\), then \[ \nu(A_n\setminus B)=\nu(A_n)-\nu(B)\le\alpha, \] so \(\nu(B)\ge\nu(A_n)-\alpha>-2^{-n}\). Put \[ P_k=\bigcup_{n\ge k}A_n,\qquad P=\bigcap_{k\ge1}P_k. \] By adding the \(A_n\)'s one at a time and using the lower bound on every subset of \(A_n\), \[ \nu(P_k)\ge\alpha-\sum_{n\ge k}2^{-n}. \] Since \(P_k\downarrow P\), continuity of a finite signed measure gives \(\nu(P)=\alpha\). If \(B\subset P\), then \(B\subset P_k\) for every \(k\); the same successive-union bound gives \(\nu(B)\ge-\sum_{n\ge k}2^{-n}\), hence \(\nu(B)\ge0\). Thus \(P\) is positive.
Let \(N=X\setminus P\). If some \(B\subset N\) had \(\nu(B)>0\), then \(P\cap B=\varnothing\) and \[ \nu(P\cup B)=\alpha+\nu(B)>\alpha, \] contradicting the definition of \(\alpha\). Hence \(N\) is negative. For uniqueness, if \(P,N\) and \(P',N'\) are two decompositions, then \(P\cap N'\) is both positive and negative, so every subset has measure zero; similarly for \(P'\cap N\). The two decompositions differ only by such null sets.
Jordan decomposition
A signed measure is the difference of two mutually singular positive measures.
Theorem. There are unique mutually singular positive measures \(\nu^+\) and \(\nu^-\) such that \(\nu=\nu^+-\nu^-\).
Proof. Given a Hahn decomposition \(X=P\sqcup N\), define \[ \nu^+(A)=\nu(A\cap P),\qquad \nu^-(A)=-\nu(A\cap N). \] These are positive measures, are concentrated on disjoint sets, and satisfy \(\nu=\nu^+-\nu^-\). The null-set uniqueness of Hahn decompositions makes them independent of the chosen \(P,N\). Alternatively, if \(\nu=\alpha-\beta\) with \(\alpha\perp\beta\), choose disjoint supports \(P,N\). On \(P\), positivity of \(\nu\) forces \(\beta(P)=0\) and \(\alpha(A)=\nu(A)\) for \(A\subset P\); similarly \(\beta(A)=-\nu(A)\) on \(N\). Thus \(\alpha=\nu^+\) and \(\beta=\nu^-\).
Total variation
Total variation measures the size of a signed measure without cancellation.
Definition. The total variation measure is \(|\nu|=\nu^++\nu^-\). Equivalently, \[ |\nu|(A)=\sup\sum_i|\nu(A_i)|, \] where the supremum is over finite measurable partitions of \(A\).
Proof. The Jordan formula gives \(\sum_i|\nu(A_i)|\le\sum_i(\nu^+(A_i)+\nu^-(A_i))=|\nu|(A)\). Taking the partition by \(A\cap P\) and \(A\cap N\) attains equality.
Proposition. For every bounded measurable \(f\), \[ \left|\int f\,\mathrm{d}\nu\right| \le\int|f|\,\mathrm{d}|\nu|. \] Moreover, \(|\nu|\) is the smallest positive measure \(\lambda\) satisfying \(|\nu(A)|\le\lambda(A)\) for every \(A\).
Proof. From \(\nu=\nu^+-\nu^-\), \[ \left|\int f\,\mathrm{d}\nu\right| \le\int|f|\,\mathrm{d}\nu^++\int|f|\,\mathrm{d}\nu^- =\int|f|\,\mathrm{d}|\nu|. \] If \(|\nu(A)|\le\lambda(A)\), apply this inequality to every cell in a finite partition of \(A\): \[ \sum_i|\nu(A_i)|\le\sum_i\lambda(A_i)=\lambda(A). \] Taking the supremum over partitions gives \(|\nu|(A)\le\lambda(A)\).
Complex measures
Complex measures reduce to signed measures through real and imaginary parts.
Remark. A complex measure is a countably additive \(\nu:\mathcal A\to\mathbb C\). Countable additivity forces its total variation to be finite on \(X\). Unlike the signed case, two pieces do not suffice to attain the variation; nevertheless \[ |\nu|(A)=\sup_{\{A_i\}}\sum_i|\nu(A_i)| \] defines the minimal dominating positive measure and yields the same integral inequality.