The Hahn Decomposition Theorem

From an Infimum to a Decomposition

A signed measure can assign positive values to some measurable sets and negative values to others. The challenge in a Hahn decomposition is to divide the whole space into two measurable pieces so that every measurable subset of one piece has nonnegative measure, while every measurable subset of the other has nonpositive measure. The conclusion is stronger than merely finding one set with positive measure and another with negative measure: it controls all measurable subsets of each piece.

Let \((X,\mathcal{M},\nu)\) be a measurable space with a finite signed measure. Here “finite” means that \(\nu=\lambda-\rho\) for finite positive measures \(\lambda\) and \(\rho\); in particular, \(\nu\) takes finite real values on measurable sets. No assumption that \(\nu\) is a probability measure, or that the underlying space is complete or sigma-finite, is needed. For disjoint measurable sets, countable additivity holds, and the series of their signed values is absolutely convergent: if \(E_j\) are disjoint, then \(\sum_j|\nu(E_j)|\le\sum_j(\lambda(E_j)+\rho(E_j))\le\lambda(X)+\rho(X)<\infty\).

The value of a set alone does not determine whether it is positive or negative: the defining condition tests every measurable subset. The proof will start instead with the infimum of all set values. Define \(\alpha=\inf\{\nu(A):A\in\mathcal{M}\}\). This is a finite real number: \(\nu(\varnothing)=0\) gives \(\alpha\le0\), while \(\nu(A)=\lambda(A)-\rho(A)\ge-\rho(X)\) gives \(\alpha\ge-\rho(X)>-\infty\).

Continuity for a Finite Signed Measure

We will use continuity from below and above for \(\nu\). If \(E_n\) increases to \(E=\bigcup_nE_n\), continuity from below for the finite positive measures \(\lambda,\rho\) gives \(\nu(E_n)=\lambda(E_n)-\rho(E_n)\to\lambda(E)-\rho(E)=\nu(E)\). If \(E_n\) decreases to \(E=\bigcap_nE_n\), the same conclusion follows from continuity from above for the finite positive measures \(\lambda,\rho\). Thus these continuity properties apply here even though \(\nu\) need not be nonnegative.

Constructing a Negative Set at the Infimum

The next proposition is the key construction. Its point is that a sequence of nearly minimizing sets need not itself consist of negative sets. Taking unions of the tails controls their values, and intersecting those tails produces a set that is negative.

Proof. For each integer \(n\ge1\), the definition of the infimum gives a measurable set \(A_n\) with \(\nu(A_n)<\alpha+2^{-n}\). Define the measurable tail unions \(B_n=\bigcup_{k=n}^{\infty}A_k\). For \(m\ge n\), let \(U_m=\bigcup_{k=n}^{m}A_k\). We claim \(\nu(U_m)<\alpha+\sum_{k=n}^{m}2^{-k}\). For \(m=n\) this is the choice of \(A_n\). If the bound holds at \(m-1\), then \(\nu(U_m)=\nu(U_{m-1})+\nu(A_m)-\nu(U_{m-1}\cap A_m)\). The intersection is measurable, so its value is at least \(\alpha\). Using the inductive bound and \(\nu(A_m)<\alpha+2^{-m}\) gives \(\nu(U_m)<\alpha+\sum_{k=n}^{m-1}2^{-k}+2^{-m}\), proving the claim. As \(m\to\infty\), \(U_m\) increases to \(B_n\); continuity from below therefore yields \(\nu(B_n)\le\alpha+\sum_{k=n}^{\infty}2^{-k}=\alpha+2^{1-n}\).

Now put \(N=\bigcap_{n=1}^{\infty}B_n\). The sets \(B_n\) decrease, so continuity from above gives \(\nu(N)=\lim_n\nu(B_n)\). Since \(\alpha\le\nu(N)\) by the definition of \(\alpha\), and \(\nu(B_n)\le\alpha+2^{1-n}\) for every \(n\), it follows that \(\nu(N)=\alpha\).

It remains to prove that \(N\) is negative. Suppose, to the contrary, that some measurable \(E\subseteq N\) satisfies \(\nu(E)>0\). Fix \(n\). Because \(E\subseteq N\subseteq B_n\), the sets \(D_k=E\cap\bigl(A_k\setminus\bigcup_{j=n}^{k-1}A_j\bigr)\), for \(k\ge n\), are disjoint and have union \(E\); for \(k=n\), the union over \(j=n,\ldots,k-1\) is empty. Countable additivity gives \(\nu(E)=\sum_{k=n}^{\infty}\nu(D_k)\). This series is absolutely convergent, since the \(D_k\) are disjoint and \(\nu=\lambda-\rho\) with \(\lambda,\rho\) finite. If \(\nu(D_k)\le2^{-k}\) for every \(k\ge n\), then \(\nu(E)\le\sum_{k=n}^{\infty}2^{-k}=2^{1-n}\). Choose \(n\) so large that \(2^{1-n}<\nu(E)\); this contradicts that inequality. Thus for some \(k\ge n\), \(\nu(D_k)>2^{-k}\). Since \(D_k\subseteq A_k\), the set \(A_k\setminus D_k\) is measurable, and \(\nu(A_k\setminus D_k)=\nu(A_k)-\nu(D_k)<(\alpha+2^{-k})-2^{-k}=\alpha\). This contradicts the definition of \(\alpha\) as a lower bound for all measurable-set values. No such \(E\) exists, so every measurable subset of \(N\) has value at most zero. Therefore \(N\) is negative and \(\nu(N)=\alpha\). \(\square\)

The Hahn Decomposition Theorem

Proof. Apply the proposition to obtain a negative set \(N\) with \(\nu(N)=\alpha\), where \(\alpha=\inf_{A\in\mathcal{M}}\nu(A)\). Set \(P=X\setminus N\). The sets are measurable, disjoint, and have union \(X\). To check positivity, take any measurable \(F\subseteq P\). If \(\nu(F)<0\), then \(F\cap N=\varnothing\), so additivity gives \(\nu(N\cup F)=\nu(N)+\nu(F)=\alpha+\nu(F)<\alpha\). But \(N\cup F\in\mathcal{M}\), contradicting the definition of \(\alpha\). Hence \(\nu(F)\ge0\) for every measurable \(F\subseteq P\), so \(P\) is positive. The set \(N\) is negative by the proposition. This proves the required decomposition. \(\square\)

Common Pitfall and Significance

Common mistake: “Since \(\nu(X)=0\), the whole space is a negative set.” The two-point example shows the error: a negative set must satisfy \(\nu(E)\le0\) for every measurable subset \(E\), not merely have a nonpositive value itself. The correction is to test the sign on all subsets, or in the theorem’s proof to use the infimum construction and then verify the complement by contradiction.

The minimization step is essential. A set whose value is close to the infimum may still contain a subset of positive value; deleting that subset would lower the set’s value. The tail-union construction controls this possibility across a sequence of approximations, and the resulting minimizing set is negative precisely because any positive-valued subset could be removed to contradict minimality. Once the negative set is obtained, the same minimality argument makes its complement positive. The theorem is thus an existence result grounded in the lower bound and additivity of a finite signed measure; it does not require the original measure to have one sign.