Signed Measures Defined by Integrable Densities

From an Integrable Density to a Signed Measure

A real-valued integrable function can assign positive weight on some parts of a space and negative weight on others. Integrating that function over measurable sets turns these local signs into a set function with the defining countable-additivity property of a signed measure. The construction also gives a direct way to locate sets on which the signed measure is nonnegative or nonpositive.

Let \((X,\mathcal{M},\mu)\) be any measure space; no assumption of finiteness, sigma-finiteness, or completeness is needed. Let \(f:X\to\mathbb{R}\) be measurable and integrable, meaning \(\int_X |f|\,d\mu<\infty\). All statements about \(f\) below refer to this chosen measurable representative and hold pointwise unless “almost everywhere” is stated. Define a set function \(\nu:\mathcal{M}\to\mathbb{R}\) by

The integral is finite for every \(A\), since \(|f\chi_A|\le |f|\) and \(\int_X|f|\,d\mu<\infty\). We use the standard definition: a signed measure is a countably additive set function whose values may in general be extended real, but it cannot take both \(+\infty\) and \(-\infty\). Here the values are all real and bounded in absolute value by \(\int_X|f|\,d\mu\), so \(\nu\) is a finite signed measure.

Here a measurable set \(P\) is called a positive set for \(\nu\) if \(\nu(E)\ge0\) for every measurable \(E\subseteq P\). A negative set \(N\) is one for which \(\nu(E)\le0\) for every measurable \(E\subseteq N\). These definitions concern all measurable subsets, not merely the value of the measure on the whole set.

Why the Construction Gives a Finite Signed Measure

For an integrable real-valued function \(h\), its signed integral is defined by \(\int h\,d\mu=\int h^+\,d\mu-\int h^-\,d\mu\), where \(h^+=\max(h,0)\) and \(h^-=\max(-h,0)\). Both integrals on the right are finite when \(\int|h|\,d\mu<\infty\). The usual integral inequality \(\left|\int h\,d\mu\right|\le\int|h|\,d\mu\) follows because \(-|h|\le h\le |h|\) and the integral is monotone. Linearity of the integral for integrable functions and the Dominated Convergence Theorem will be used below; the theorem applies to a sequence converging pointwise and bounded in absolute value by one integrable function.

Proof. First, \(\nu(\varnothing)=\int f\chi_{\varnothing}\,d\mu=0\). Let \((A_k)_{k\ge1}\) be pairwise disjoint measurable sets, and put \(A=\bigcup_{k=1}^{\infty}A_k\). Define \(S_n=\sum_{k=1}^n f\chi_{A_k}\). At each \(x\), at most one of the indicators \(\chi_{A_k}(x)\) is nonzero. Thus \(S_n(x)\to f(x)\chi_A(x)\), and \(|S_n(x)|\le |f(x)|\) for every \(n\) and \(x\). The function \(|f|\) is integrable, so the Dominated Convergence Theorem gives \(\int S_n\,d\mu\to\int f\chi_A\,d\mu=\nu(A)\). By finite linearity, \(\int S_n\,d\mu=\sum_{k=1}^n\nu(A_k)\).

To verify that this series is absolutely convergent, apply the integral inequality to each \(f\chi_{A_k}\). For every \(n\), disjointness gives

The partial sums on the left are increasing and bounded, so their limit is finite. Therefore the series \(\sum_k\nu(A_k)\) converges absolutely, and the limit of its partial sums, already shown to equal \(\nu(A)\), proves countable additivity. The same integral inequality gives \(|\nu(A)|\le\int_A|f|\,d\mu\). Finally, \(P\) and \(N\) are measurable because \(f\) is measurable. For every measurable \(E\subseteq P\), \(f\chi_E\ge0\), so \(\nu(E)=\int_E f\,d\mu\ge0\). For every measurable \(E\subseteq N\), \(f\chi_E\le0\), so \(\nu(E)\le0\). Thus \(P\) is positive and \(N\) is negative. \(\square\)

The sets \(P\) and \(N\) partition \(X\); the zero set \(\{f=0\}\) is included in \(P\). One could instead include that zero set in \(N\), since its contribution to the integral on any subset is zero. If a density is changed on a \(\mu\)-null set, the resulting signed measure does not change, though the pointwise sign sets may change there.

Signed Values Are Not Absolute Masses

A signed measure records cancellation. In the linear-density example, the integral over \([0,2]\) was zero because the negative and positive contributions each had magnitude \(\tfrac12\). It would be incorrect to conclude that the density vanishes or that its absolute integral is zero: the absolute integral is \(\int_0^2|x-1|\,dx=\tfrac12+\tfrac12=1\).

A common wrong calculation is to replace \(f\) by \(|f|\) when computing \(\nu(A)\). For \(f(x)=x-1\) on \([0,2]\), this gives the incorrect value \(\int_0^2|x-1|\,dx=1\) for \(\nu([0,2])\). The correction is to retain the sign: \(\nu([0,2])=\int_0^1(x-1)\,dx+\int_1^2(x-1)\,dx=-\tfrac12+\tfrac12=0\). The absolute value instead provides a bound, \(|\nu(A)|\le\int_A|f|\,d\mu\).

Integrability is essential to this finite-measure construction. For example, on \((\mathbb{R},\mathcal{L},m)\), the nonnegative measurable function \(f(x)=1/x\) on \((0,1)\), extended by zero elsewhere, is not integrable: \(\int_0^1(1/x)\,dx=\infty\). Assigning \(\nu((0,1))=\infty\) does not produce a finite signed measure from an integrable density. In contrast, when \(f\) is integrable, the bound \(|\nu(A)|\le\int_A|f|\,d\mu\le\int_X|f|\,d\mu\) verifies finiteness on every measurable set, including \(X\).