From a Density to an Integral
A Radon–Nikodym density does more than describe the values of a measure on sets: it also converts integration with respect to that measure into integration with respect to the reference measure. The key identity is simple to write, but its domain matters. For a positive measure it applies to nonnegative measurable functions, including ones whose integrals are infinite. For a signed measure, the integral must be defined by absolute integrability with respect to its total variation.
Let \((X,\mathcal{M},\mu)\) be a sigma-finite measure space. Let \(\nu\) be a finite signed measure on \((X,\mathcal{M})\) that is absolutely continuous with respect to \(\mu\), meaning \(\mu(E)=0\) implies \(\nu(E)=0\) for every \(E\in\mathcal{M}\). By the Radon–Nikodym theorem for finite signed measures, there is a real-valued measurable function \(h\in L^1(\mu)\), unique \(\mu\)-a.e., such that \(\nu(E)=\int_E h\,d\mu\) for every \(E\in\mathcal{M}\). We write \(h=d\nu/d\mu\). For a positive \(\nu\), \(h\ge0\) \(\mu\)-a.e.
Here \(|\nu|\) is the total variation measure of \(\nu\), which records the size of a signed measure without cancellation. In particular, for a real-valued \(g\), the second identity gives an exact test: \(g\) is integrable with respect to the signed measure \(\nu\) if and only if \(|g||h|\) is integrable with respect to \(\mu\).
Why the Transfer Identity Holds
Proof. First consider the positive case. For every measurable set \(E\), the definition of \(h\) gives \(\nu(E)=\int_E h\,d\mu\). Thus the identity holds for indicators: \(\int\chi_E\,d\nu=\int\chi_Eh\,d\mu\). By finite additivity of the integral, it holds for every nonnegative simple function \(\varphi=\sum_{j=1}^k a_j\chi_{E_j}\), where the \(E_j\) are disjoint and \(a_j\ge0\). For general measurable \(g:X\to[0,\infty]\), choose nonnegative simple functions \(\varphi_n\uparrow g\). The Monotone Convergence Theorem applies on both sides: \(\nu\) and \(\mu\) are measures, and both sequences \(\varphi_n\) and \(\varphi_nh\) increase pointwise because \(h\ge0\) \(\mu\)-a.e. Therefore $$ \int g\,d\nu=\lim_n\int\varphi_n\,d\nu =\lim_n\int\varphi_nh\,d\mu=\int gh\,d\mu. $$ If \(h\ge0\) only almost everywhere, choose a nonnegative version of \(h\); this does not change any of the integrals.
For the signed case, first note that for every \(E\in\mathcal{M}\), $$ |\nu|(E)=\int_E|h|\,d\mu. $$ Indeed, for any finite measurable partition \(E=\bigcup_{j=1}^k E_j\), the triangle inequality gives $$ \sum_{j=1}^k|\nu(E_j)| =\sum_{j=1}^k\left|\int_{E_j}h\,d\mu\right| \le\int_E|h|\,d\mu. $$ Taking the supremum over such partitions proves \(|\nu|(E)\le\int_E|h|\,d\mu\). For the reverse inequality, partition \(E\) into \(E\cap\{h\ge0\}\) and \(E\cap\{h<0\}\). On each piece \(h\) has a fixed sign, so the sum of the absolute values of the two integrals is \(\int_E|h|\,d\mu\). This partition is allowed in the supremum defining \(|\nu|(E)\), proving the reverse inequality.
For a measurable simple function \(s\), linearity and the identity on sets give \(\int s\,d\nu=\int sh\,d\mu\). Now let \(g\) be real-valued and satisfy \(\int|g|\,d|\nu|<\infty\). Choose simple functions \(s_n\) with \(s_n\to g\) \(|\nu|\)-a.e. and \(|s_n|\le|g|\); such approximations are obtained by truncating \(g\) and rounding its magnitude down to increasingly fine dyadic levels. The equality just proved for total variation gives \(\int |g||h|\,d\mu=\int|g|\,d|\nu|<\infty\). Thus \(|s_n|\le|g|\) provides an integrable dominating function both for integration with respect to \(|\nu|\) and for integration with respect to \(|h|\,d\mu\). The Dominated Convergence Theorem applies on these two measure spaces. Its hypotheses hold because \(s_n\to g\) almost everywhere for each measure, and the stated dominating function is integrable. Passing to the limit in \(\int s_n\,d\nu=\int s_nh\,d\mu\) yields \(\int g\,d\nu=\int gh\,d\mu\). Applying the same argument to \(|g|\), or using the total variation identity, gives the claimed integrability equality. \(\square\)
Computing with Positive Densities
Worked Example: A Probability Density and a Moment
On \(X=[0,1]\) with Lebesgue measure \(m\), define the positive measure \(\nu\) by \(d\nu/dm=2x\). Compute \(\int_X x^2\,d\nu\).
The density is nonnegative, and its total mass is $$ \nu(X)=\int_0^1 2x\,dx=[x^2]_0^1=1. $$ Thus \(\nu\) is a probability measure. The transfer identity gives $$ \int_X x^2\,d\nu=\int_0^1 x^2(2x)\,dx =2\int_0^1x^3\,dx =2\left[\frac{x^4}{4}\right]_0^1 =\frac12. $$ The integrand is nonnegative, so the positive-measure version applies without a separate integrability assumption. As a check, \(x^2\) lies between \(0\) and \(1\), so its integral under a probability measure must lie between \(0\) and \(1\); the computed value \(1/2\) satisfies this bound.
Worked Example: Integrable Under One Measure but Not Another
Take \(X=(0,1]\), \(\mu=m\), and define the positive measure \(\nu\) by \(d\nu/dm=x\). For \(g(x)=1/x\), determine whether \(g\) is integrable under \(\mu\) and under \(\nu\), and compute its \(\nu\)-integral.
First, \(\nu(X)=\int_0^1x\,dx=1/2\), so \(\nu\) is finite. Under \(\mu\), $$ \int_X |g|\,d\mu=\int_0^1\frac{1}{x}\,dx=\infty. $$ For example, the integrals over \([1/n,1]\) equal \(\int_{1/n}^1 dx/x=\ln n\), which grow without bound as \(n\to\infty\). Thus \(g\notin L^1(\mu)\). Under \(\nu\), the density test gives $$ \int_X |g|\,d\nu=\int_0^1\frac{1}{x}\,x\,dx =\int_0^1 1\,dx=1. $$ Hence \(g\in L^1(\nu)\), and \(\int_Xg\,d\nu=1\). The result is consistent with truncation: for \(g_N(x)=\min(1/x,N)\), the transferred integrand is \(g_N(x)x=\min(1,Nx)\), which increases to \(1\) on \((0,1]\); Monotone Convergence applies and gives the same limit \(1\).
Signed Densities and Cancellation
For a signed measure, positive and negative parts of the density can cancel in \(\int g\,d\nu\). That cancellation does not remove the absolute-integrability requirement: the integral against a signed measure is defined through its positive and negative parts only when \(\int|g|\,d|\nu|<\infty\). The density test is therefore made with \(|h|\), not with \(h\).
Worked Example: Integrating Against a Signed Measure
On \(X=[0,1]\) with Lebesgue measure \(m\), let \(h(x)=1\) on \([0,1/2]\) and \(h(x)=-1\) on \((1/2,1]\), and define \(\nu(E)=\int_Eh\,dm\). Compute \(\int_X x\,d\nu\) and verify integrability.
The total variation density is \(|h|=1\) on all of \(X\). Consequently, $$ \int_X|x|\,d|\nu|=\int_0^1x|h(x)|\,dx =\int_0^1x\,dx=\frac12<\infty, $$ so \(x\) is integrable with respect to \(\nu\). By transfer, $$ \int_Xx\,d\nu =\int_0^{1/2}x\,dx-\int_{1/2}^1x\,dx =\left[\frac{x^2}{2}\right]_0^{1/2} -\left[\frac{x^2}{2}\right]_{1/2}^1 =\frac18-\frac38=-\frac14. $$ As an independent check, \(\nu([0,1])=\int_0^{1/2}1\,dx-\int_{1/2}^1 1\,dx=1/2-1/2=0\), but the integral of \(x\) need not vanish: the two halves have different \(x\)-values on average. The direct calculation above gives that difference as \(-1/4\).
A Common Mistake: Forgetting the Density
One cannot generally compute \(\int g\,d\nu\) by integrating \(g\) against the reference measure \(\mu\) alone. For example, on \([0,1]\) let \(d\nu/dm=4x\) and take \(g(x)=x\). The incorrect calculation omits the density and gives \(\int_0^1x\,dx=1/2\). The correct calculation is $$ \int_0^1x\,d\nu=\int_0^1x(4x)\,dx =4\left[\frac{x^3}{3}\right]_0^1=\frac43. $$ The check \(\nu([0,1])=\int_0^1 4x\,dx=2\) confirms that \(\nu\) is not the reference Lebesgue measure; in particular, it is not a probability measure that would force the integral of \(x\) to be at most \(1\).
A related error is to use \(h\), rather than \(|h|\), to test integrability for a signed measure. A signed integral can be small because positive and negative contributions cancel, even when the absolute integral is infinite. The criterion is \(\int |g||h|\,d\mu<\infty\); cancellation is considered only after that condition is met.
The transfer identity is useful whenever a measure is given through a density: it turns a question about integration under \(\nu\) into a weighted integral under \(\mu\). For positive measures it also permits nonnegative integrals with value \(+\infty\), while for finite signed measures it precisely identifies the functions for which the signed integral exists. The density is meaningful only \(\mu\)-a.e., and all these integral identities respect that convention.
Check Your Understanding
Use the transfer identity, and state the integrability condition whenever the measure is signed.
- If \(d\nu/dm=3x^2\) on \([0,1]\), compute \(\nu([0,1])\) and \(\int_0^1x\,d\nu\).
- Suppose \(d\nu/d\mu=h\) for a positive measure \(\nu\). State the condition on \(h\) and \(g\) that ensures \(\int|g|\,d\nu<\infty\).
- For the signed density \(h=1\) on \([0,1/3]\) and \(h=-1\) on \((1/3,1]\), compute \(\int_0^1 1\,d\nu\).
- Explain why \(\int g\,d\nu=\int gh\,d\mu\) for a signed measure does not by itself imply that \(g\) is integrable.
- On \((0,1]\), take \(d\nu/dm=x^2\) and \(g(x)=1/x\). Determine whether \(g\) is integrable under \(\nu\), using the density criterion.