Reading a Signed Measure from Its Density
A signed measure defined by a density carries its sign locally in the density: regions where the density is nonnegative give nonnegative values on all measurable subsets, and regions where it is negative give nonpositive values. The zero set needs a deliberate assignment. It may be placed on either side, or divided measurably between them, without affecting the sign conditions.
Let \((X,\mathcal{M},\mu)\) be any measure space; no finiteness or sigma-finiteness assumption on \(\mu\) is needed. Let \(f:X\to\mathbb{R}\) be measurable and integrable, meaning \(\int_X |f|\,d\mu<\infty\). Define the finite signed measure \(\nu\) on \(\mathcal{M}\) by \(\nu(E)=\int_E f\,d\mu\). The integral is finite for every \(E\in\mathcal{M}\), since \(\int_E|f|\,d\mu\le\int_X|f|\,d\mu\). We use the established fact that this formula defines a signed measure.
The conditions concern every measurable subset, not just the value of the signed measure on the whole region. A region can have total signed measure zero while containing both positive and negative contributions, so checking only \(\nu(P)\) or \(\nu(N)\) is not enough.
The Sign-Region Construction
Proof. The sets \(P\) and \(N\) are measurable because \(f\) is measurable and \([0,\infty)\) and \((-\infty,0)\) are Borel sets. They are disjoint and their union is \(X\). If \(A\in\mathcal{M}\) and \(A\subseteq P\), then \(f\chi_A\ge0\) everywhere, so the nonnegative Lebesgue integral gives \(\nu(A)=\int_A f\,d\mu\ge0\). Thus \(P\) is positive. If \(A\in\mathcal{M}\) and \(A\subseteq N\), then \(f\chi_A\le0\), so \(\nu(A)=\int_A f\,d\mu\le0\). Thus \(N\) is negative. For the general assertion, write \(Z=Z_P\cup Z_N\) as a disjoint measurable partition and take \(P'=\{f>0\}\cup Z_P\), \(N'=\{f<0\}\cup Z_N\). These sets partition \(X\); on \(P'\), \(f\ge0\), and on \(N'\), \(f\le0\). The same integral inequalities show that \(P'\) is positive and \(N'\) is negative. \(\square\)
This result uses the actual measurable representative \(f\) in the displayed definitions. If \(f\) is changed on a \(\mu\)-null set, the resulting density defines the same signed measure, because its integral over every measurable set is unchanged. The sign regions themselves may change on that null set, but the sign conclusions remain valid.
Worked Example: The Density f(x) = x on a Symmetric Interval
Take \(X=[-2,2]\), \(\mathcal{M}\) to be the Lebesgue-measurable sets in \(X\), \(\mu=m\), and \(f(x)=x\). The signed measure is \(\nu(E)=\int_E x\,dm\). The sign regions are \(P=[0,2]\) and \(N=[-2,0)\); the zero \(x=0\) is assigned to \(P\).
For every measurable \(A\subseteq P\), \(x\ge0\) on \(A\), hence \(\nu(A)=\int_A x\,dm\ge0\). For every measurable \(B\subseteq N\), \(x<0\) on \(B\), hence \(\nu(B)=\int_B x\,dm\le0\). The values on the full regions are
Check: adding the two region values gives \(2+(-2)=0\), and direct integration gives \(\nu(X)=\int_{-2}^2x\,dx=[x^2/2]_{-2}^2=2-2=0\). Assigning \(0\) to \(N\) instead would not change any integral, since a singleton has Lebesgue measure zero; the sign conditions also hold at \(0\).
Worked Example: A Density with a Whole Interval of Zeros
Let \(X=[0,3]\) with Lebesgue measure and define \(f=2\) on \([0,1)\), \(f=0\) on \([1,2)\), and \(f=-3\) on \([2,3]\). This measurable function is integrable, with \(\int_X|f|\,dm=2\cdot1+3\cdot1=5\). Using \(P=\{f\ge0\}\) and \(N=\{f<0\}\) gives \(P=[0,2)\) and \(N=[2,3]\).
If \(A\subseteq P\) is measurable, \(f\) equals either \(2\) or \(0\) on \(A\), so \(\nu(A)=2m(A\cap[0,1))\ge0\). If \(B\subseteq N\) is measurable, \(f=-3\) on \(B\), so \(\nu(B)=-3m(B)\le0\). In particular, \(\nu(P)=2m([0,1))=2\), and \(\nu(N)=-3m([2,3])=-3\).
For an explicit alternative assignment, put \(Z_P=[1,3/2)\) and \(Z_N=[3/2,2)\). Then \(P'=[0,1)\cup[1,3/2)=[0,3/2)\) and \(N'=[2,3]\cup[3/2,2)=[3/2,3]\). On \(P'\), \(f\) is nonnegative, and on \(N'\), \(f\) is nonpositive, including where it is zero. Thus the subset sign conditions still hold. Check: \(\nu(P')=2m([0,1))=2\) and \(\nu(N')=-3m([2,3])=-3\), so \(\nu(P')+\nu(N')=-1\). Directly, \(\nu(X)=2\cdot1+0\cdot1+(-3)\cdot1=-1\), as required.
Worked Example: A Density That Changes Sign at Two Points
Let \(X=[-2,2]\) with Lebesgue measure and \(f(x)=x^2-1\). This polynomial is measurable and integrable on the finite interval. Since \(x^2-1\ge0\) when \(|x|\ge1\) and \(x^2-1<0\) when \(|x|<1\), choose \(P=[-2,-1]\cup[1,2]\) and \(N=(-1,1)\). The zeros \(-1\) and \(1\) are assigned to \(P\).
Every measurable subset of \(P\) has nonnegative density throughout, and every measurable subset of \(N\) has negative density throughout. To calculate the full-region values, an antiderivative is \(F(x)=x^3/3-x\). On the right component of \(P\),
The left component has the same integral because the density is even: substituting \(u=-x\) converts \(\int_{-2}^{-1}(x^2-1)\,dx\) to \(\int_1^2(u^2-1)\,du=4/3\). Therefore \(\nu(P)=8/3\). On the negative region,
Check: the region values sum to \(8/3-4/3=4/3\). Direct integration over all of \(X\) gives \(\int_{-2}^2(x^2-1)\,dx=2\int_0^2(x^2-1)\,dx=2(8/3-2)=4/3\), the same value.
What Other Hahn Decompositions Can Look Like
The sign regions give a direct construction, and they also constrain any other Hahn decomposition for this density. The only freedom, apart from \(\mu\)-null sets, lies where the density is zero.
Proof. Set \(A=Q\cap\{f<0\}\). If \(\mu(A)>0\), then \(A=\bigcup_{n=1}^{\infty}A_n\), where \(A_n=Q\cap\{f\le-1/n\}\). At least one \(A_n\) has positive measure, since a countable union of null sets is null. For that \(n\), the negative-set integral bound gives \(\nu(A_n)=\int_{A_n}f\,d\mu\le-\mu(A_n)/n<0\). This contradicts that \(Q\) is positive, because \(A_n\subseteq Q\). Therefore \(\mu(Q\cap\{f<0\})=0\). For \(B=R\cap\{f>0\}\), if \(\mu(B)>0\), then \(B=\bigcup_{n=1}^{\infty}(R\cap\{f\ge1/n\})\), so some \(B_n=R\cap\{f\ge1/n\}\) has positive measure. Then \(\nu(B_n)\ge\mu(B_n)/n>0\), contradicting that \(R\) is negative. Hence \(\mu(R\cap\{f>0\})=0\). \(\square\)
The strict sign in the level sets matters: on \(\{f=0\}\), neither sign is forced. The proposition says that a positive side cannot contain a set of positive \(\mu\)-measure where the density is strictly negative, and the negative side cannot contain a set of positive measure where it is strictly positive. It does not say that the zero set is \(\mu\)-null.
A Common Pitfall: Forgetting the Zero Set
For \(f(x)=x\) on \([-1,1]\), the sets \(\{f>0\}=(0,1]\) and \(\{f<0\}=[-1,0)\) are positive and negative, respectively, but they do not partition \(X\): they omit \(0\). Calling them a Hahn decomposition is therefore incorrect, even though the omitted point has measure zero. The correction is to assign that point, for example \(P=\{f\ge0\}=[0,1]\) and \(N=\{f<0\}=[-1,0)\). More generally, a Hahn decomposition must cover the entire space; nullity of an omitted set does not turn two sets into a partition.
A related mistake is to use both \(\{f\ge0\}\) and \(\{f\le0\}\) as the two sides: they overlap on \(\{f=0\}\), so they are not disjoint unless the zero set is explicitly assigned to just one side or split into disjoint pieces. This construction concerns Hahn decompositions only; the positive and negative measures built from such a decomposition are a separate step.
Check Your Understanding
In each question, take the signed measure to be defined by integrating the stated real-valued density against the specified measure.
- On \([0,1]\) with Lebesgue measure, find a Hahn decomposition for \(f(x)=x-\tfrac12\), and compute the signed measure of each side.
- If \(f=0\) on a measurable set \(Z\), explain why any measurable partition of \(Z\) can be assigned to the two sides of the sign-region decomposition.
- For an arbitrary measurable \(A\subseteq\{f<0\}\), which inequality does \(\int_A f\,d\mu\) satisfy, and why does that verify the negative-set condition?
- Explain why the sets \(\{f>0\}\) and \(\{f<0\}\) alone need not be a Hahn decomposition, even when \(\{f=0\}\) is null.
- In the proposition, why must some set \(Q\cap\{f\le-1/n\}\) have positive measure if \(\mu(Q\cap\{f<0\})>0\)?