A Measure Density Acts on Every Set
A Radon–Nikodym derivative is useful because it replaces a measure by a measurable function inside an integral. The defining identity concerns every measurable set, not just the total mass of the space. Once that identity is established, it also determines integrals of nonnegative measurable functions, including functions that are not simple.
Throughout, \((X,\mathcal{M},\mu)\) is a measure space with \(\mu\) sigma-finite; \(\mu\) need not be finite or complete. A positive measure \(\nu\) will be assumed finite when we invoke the finite-measure form of the Radon–Nikodym theorem. For signed measures, we assume \(\nu\) is finite, meaning its total variation satisfies \(|\nu|(X)<\infty\). A measure \(\nu\) is absolutely continuous with respect to \(\mu\), written \(\nu\ll\mu\), if \(\mu(A)=0\) implies \(\nu(A)=0\) for every \(A\in\mathcal{M}\).
The standard Radon–Nikodym theorem says that if \(\mu\) is sigma-finite, \(\nu\) is a finite positive measure, and \(\nu\ll\mu\), then such a nonnegative measurable \(f\) exists and \(\int_X f\,d\mu=\nu(X)<\infty\). The corresponding theorem for finite signed measures gives an integrable real-valued density when \(\nu\ll\mu\). These are the existence results from the preceding material; here the emphasis is on what the defining identity implies for integrals.
From Sets to Nonnegative Functions
Proof. First suppose \(g=\sum_{j=1}^{k}a_j\chi_{E_j}\) is a nonnegative simple function, with disjoint \(E_j\in\mathcal{M}\) and finite coefficients \(a_j\ge0\). By the definition of the integral of a simple function and the assumed identity on measurable sets,
Now let \(g:X\to[0,\infty]\) be measurable. There is an increasing sequence of nonnegative measurable simple functions \(g_n\) with \(g_n\uparrow g\) pointwise. The Monotone Convergence Theorem applies to \(g_n\) on \((X,\mathcal{M},\nu)\), since \(\nu\) is a measure and each \(g_n\) is nonnegative and measurable. It also applies to \(g_nf\) on \((X,\mathcal{M},\mu)\): these functions are measurable, nonnegative, and increase pointwise to \(gf\). Using the simple-function identity for every \(n\) gives
The equalities hold in \([0,\infty]\), so no finiteness of these integrals is required. \(\square\)
This argument explains why the set identity is the central one: indicators give the defining identity, finite nonnegative combinations give the formula for simple functions, and Monotone Convergence extends it to every nonnegative measurable function. No interchange of an unrestricted limit and an integral is being assumed; the named convergence theorem applies to each increasing sequence.
Worked Example: A Density on a Finite Counting Space
Let \(X=\{1,2,3\}\), let \(\mathcal{M}\) be all subsets of \(X\), and let \(\mu\) be counting measure, so \(\mu(A)\) is the number of points in \(A\). Define \(f(1)=\tfrac12\), \(f(2)=\tfrac34\), \(f(3)=\tfrac14\), and define \(\nu(A)=\int_A f\,d\mu\). This is a finite positive measure, and by definition \(d\nu/d\mu=f\).
Take \(g(1)=1\), \(g(2)=3\), and \(g(3)=0\). Since each point has counting measure one, the density formula gives
Directly from the masses, \(\nu(\{1\})=\tfrac12\), \(\nu(\{2\})=\tfrac34\), and \(\nu(\{3\})=\tfrac14\). Thus \(\int_Xg\,d\nu=1\cdot\tfrac12+3\cdot\tfrac34+0\cdot\tfrac14=\tfrac{11}{4}\) again. As a separate check, the total mass is \(\nu(X)=\tfrac12+\tfrac34+\tfrac14=\tfrac32=\int_Xf\,d\mu\), as the set identity requires for \(A=X\).
Worked Example: A Lebesgue Density and a Step Function
On \(X=[0,1]\) with its Borel sigma-algebra and Lebesgue measure \(m\), set \(f(x)=2x\) and \(\nu(A)=\int_A2x\,dm\). This defines a finite positive measure: its total mass is \(\int_0^1 2x\,dx=1\). Let \(g=\chi_{[0,1/2]}+2\chi_{(1/2,1]}\). The endpoint \(1/2\) belongs to the first set, not the second; it has Lebesgue measure zero and \(\nu\)-measure zero.
The density formula gives
To check directly, the measure of the first piece is \(\nu([0,1/2])=\int_0^{1/2}2x\,dx=\tfrac14\), while that of the second is \(\nu((1/2,1])=\int_{1/2}^1 2x\,dx=\tfrac34\). Integrating the step function against these two masses gives \(1\cdot\tfrac14+2\cdot\tfrac34=\tfrac74\). The two computations agree, and \(\nu(X)=\tfrac14+\tfrac34=1\) checks the normalization.
Worked Example: Integrating a Nonnegative Function Against a Density
On \([0,1]\) with Borel sigma-algebra and Lebesgue measure \(m\), let \(f(x)=1+x\) and define \(\nu(A)=\int_A(1+x)\,dm\). This is a finite positive measure because \(\nu([0,1])=\int_0^1(1+x)\,dx=\tfrac32\). For \(g(x)=1+x\), both \(f\) and \(g\) are nonnegative and measurable, so the theorem applies even though \(g\) is not a simple function.
Using the density formula and an antiderivative,
As a direct check of the arithmetic, \(\int_0^1(1+x)^2\,dx=1+1+\tfrac13=\tfrac73\). Also, setting \(g=1\) in the theorem recovers \(\int_X1\,d\nu=\int_Xf\,dm=\tfrac32=\nu(X)\). This separate total-mass check confirms that the density is being used once to specify \(\nu\), not inserted again after changing the measure.
Signed Densities and Their Integral Formula
For a signed measure, the density may take both positive and negative values. Write \(f^+=\max(f,0)\) and \(f^-=\max(-f,0)\), so \(f=f^+-f^-\) and \(|f|=f^++f^-\). If \(\nu\) is a finite signed measure on \((X,\mathcal{M})\), \(\mu\) is sigma-finite, and \(\nu\ll\mu\), the signed Radon–Nikodym theorem gives an integrable real-valued density \(f\) satisfying \(\nu(A)=\int_Af\,d\mu\) for every \(A\in\mathcal{M}\).
Proof. Define positive measures \(\nu_+(A)=\int_Af^+\,d\mu\) and \(\nu_-(A)=\int_Af^-\,d\mu\). They are finite because \(f\) is integrable. For every \(A\in\mathcal{M}\), the density identity gives \(\nu(A)=\nu_+(A)-\nu_-(A)\). The measures \(\nu_+\) and \(\nu_-\) are concentrated on the disjoint measurable sets \(\{f\ge0\}\) and \(\{f<0\}\), respectively. Thus they are the positive and negative parts in the Jordan decomposition of \(\nu\), by uniqueness of that decomposition. Apply the positive integral formula already proved to \(\nu_+\) with density \(f^+\), and to \(\nu_-\) with density \(f^-\). This yields \(\int g\,d\nu_+=\int gf^+\,d\mu\) and \(\int g\,d\nu_-=\int gf^-\,d\mu\). Subtracting gives the asserted signed formula whenever the two positive integrals are not both infinite, exactly the condition that makes the signed integral defined. \(\square\)
The sigma-finiteness assumption on \(\mu\) is retained here. In particular, the positive integral formula is applied only within its stated hypotheses; this proof does not silently extend that theorem to an arbitrary measure space.
A Common Pitfall: Counting the Density Twice
The notation \(d\nu=f\,d\mu\) is shorthand for \(\nu(A)=\int_Af\,d\mu\). Once the integration measure is \(\nu\), the density has already been included. The formula is \(\int g\,d\nu=\int gf\,d\mu\), not \(\int gf\,d\nu\).
For a concrete check, use the preceding example \(f(x)=1+x\) and \(g(x)=1+x\) on \([0,1]\). The correct value is \(\int gf\,dm=\int_0^1(1+x)^2\,dx=\tfrac73\). A mistaken second insertion would give \(\int gf\,d\nu\); applying the density formula to that expression gives \(\int_0^1gf^2\,dm=\int_0^1(1+x)^3\,dx\). Expanding and integrating, this is \(1+\tfrac32+1+\tfrac14=\tfrac{15}{4}\), not \(\tfrac73\). The error is not in either integral calculation: it is the extra factor of \(f\). When checking any proposed density formula, first identify the measure after the differential \(d\), then insert the density exactly once if converting that measure to \(d\mu\).
Check Your Understanding
For each question, state the measure space and verify that the relevant integral is defined before calculating.
- If \(\nu(A)=\int_A f\,d\mu\) for every \(A\in\mathcal{M}\), how does the density identity prove the integral formula first for a nonnegative simple function?
- For counting measure on \(X=\{1,2\}\), let \(f(1)=2\), \(f(2)=1\), and \(\nu(A)=\int_Af\,d\mu\). Compute \(\int g\,d\nu\) for \(g(1)=3\), \(g(2)=4\), using both the point masses and the density formula.
- Explain why Monotone Convergence can be applied to both sides of the simple-function identity when \(g_n\uparrow g\) and \(f\ge0\).
- For a signed density \(f\), what condition prevents the expression \(\int gf^+\,d\mu-\int gf^-\,d\mu\) from being undefined as \(+\infty-\infty\)?
- On \([0,1]\), take \(f(x)=1+x\) and \(g(x)=1\). Compute \(\int g\,d\nu\) and explain why converting to \(d\mu\) introduces one factor of \(f\), not two.