Representing Measures Given by Densities

From a Density to a Representing Measure

A positive linear functional built by integrating against a nonnegative density has a natural candidate representing measure: assign to each set the integral of the density over that set. On a compact interval, an integrable density produces a finite measure and a bounded functional on continuous functions. On \(\mathbb{R}\), local integrability is enough to define a measure finite on compact sets and a functional on compactly supported continuous functions, even when the measure of all of \(\mathbb{R}\) is infinite.

Throughout, \(m\) is Lebesgue measure on \(\mathbb{R}\), and \(C([a,b])\) denotes the real-valued continuous functions on a compact interval. The space \(C_c(\mathbb{R})\) consists of real-valued continuous functions with compact support. Densities below are nonnegative, Lebesgue-measurable functions; their integrals are Lebesgue integrals. Measures we define are on the Borel sigma-algebra, not on the completed Lebesgue sigma-algebra. Changing a density on a Lebesgue-null set does not change its integrals or the resulting Borel measure.

Definition: If \(g\ge 0\) is Lebesgue measurable and \(E\) is Borel, define \(\nu_g(E)=\int_E g\,dm\). On a compact interval \([a,b]\), this gives a finite Borel measure if \(\int_a^b g\,dm<\infty\). On \(\mathbb{R}\), it gives a measure finite on every compact set if \(g\) is locally integrable, meaning \(\int_K g\,dm<\infty\) for every compact \(K\subset\mathbb{R}\).

The integral is countably additive over disjoint sets: if \((E_j)\) are pairwise disjoint Borel sets, then \(\chi_{\cup_j E_j}g=\sum_j \chi_{E_j}g\). The partial sums increase to the full sum, so the Monotone Convergence Theorem gives \(\nu_g(\cup_j E_j)=\sum_j\nu_g(E_j)\). Thus \(\nu_g\) is a measure. It is finite on \([a,b]\) when \(g\) is integrable there, and locally finite on \(\mathbb{R}\) when \(g\) is locally integrable; in the latter case it is also sigma-finite, since \(\mathbb{R}=\cup_{n\ge1}[-n,n]\).

Why These Measures Represent the Functionals

Theorem: Let \(g\ge0\) be Lebesgue measurable. On a compact interval \([a,b]\), suppose \(\int_a^b g\,dm<\infty\), and define \(L(f)=\int_{[a,b]} f g\,dm\) for \(f\in C([a,b])\). Then \(L\) is a bounded positive linear functional, and its representing finite regular Borel measure is \(\nu_g\). On \(\mathbb{R}\), suppose \(g\) is locally integrable and define \(L(f)=\int_{\mathbb{R}} f g\,dm\) for \(f\in C_c(\mathbb{R})\). This integral is finite for every \(f\in C_c(\mathbb{R})\), and its representing locally finite regular Borel measure is \(\nu_g\). The measure on \(\mathbb{R}\) need not have finite total mass.

Proof. On \([a,b]\), a continuous \(f\) is bounded, so \(|L(f)|\le \|f\|_\infty\int_{[a,b]}g\,dm<\infty\). Linearity follows from linearity of the integral, and if \(f\ge0\), then \(L(f)\ge0\). By the definition of \(\nu_g\), \(L(f)=\int f\,d\nu_g\). On \(\mathbb{R}\), the support of \(f\in C_c(\mathbb{R})\) is contained in some compact set \(K\); hence \(|L(f)|\le\|f\|_\infty\int_Kg\,dm<\infty\), and the same integral properties give linearity, positivity, and \(L(f)=\int f\,d\nu_g\).

For regularity, use the regularity of Lebesgue measure and the absolute continuity of the integral of an integrable function: on any bounded region where \(g\) is integrable, for every \(\varepsilon>0\) there is a \(\delta>0\) such that \(m(A)<\delta\) implies \(\int_A g\,dm<\varepsilon\). This implication follows by choosing \(M>0\) with \(\int_{\{g>M\}}g\,dm<\varepsilon/2\), and then taking \(\delta=\varepsilon/(2M)\): the integral over \(A\cap\{g\le M\}\) is at most \(M m(A)<\varepsilon/2\). Lebesgue regularity therefore transfers to \(\nu_g\) on bounded regions: approximate a Borel set from outside by open sets, or from inside by compact sets, with the error in \(\nu_g\) as small as desired. For \(\mathbb{R}\), apply this argument on bounded intervals; countable exhaustion by \([-n,n]\) gives outer regularity for Borel sets and inner regularity on open sets. Thus \(\nu_g\) is a regular Borel measure, locally finite in the real-line case. The representation is exactly the displayed integral identity. Uniqueness is the standard uniqueness theorem for representing regular Borel measures by their continuous test-function integrals, established in the earlier material on uniqueness from continuous test functions. \(\square\)

Proposition: In the compact-interval setting of the theorem, the operator norm of \(L\) is \(\nu_g([a,b])=\int_{[a,b]}g\,dm\).

Proof. For every \(f\in C([a,b])\) with \(\|f\|_\infty\le1\), the estimate in the theorem gives \(|L(f)|\le\int_{[a,b]}g\,dm\). Therefore \(\|L\|\le\int_{[a,b]}g\,dm\). The constant function \(f=1\) has \(\|f\|_\infty=1\) and \(L(1)=\int_{[a,b]}g\,dm\), so \(\|L\|\ge L(1)\). The two inequalities prove the equality, including the case \(g=0\) almost everywhere. \(\square\)

Worked Densities on Compact Intervals

Worked Example: The Density \(g(x)=2x\) on \([0,1]\)

Let \(L(f)=\int_0^1 2x f(x)\,dx\) for \(f\in C([0,1])\). Since \(2x\ge0\) on this interval and \(\int_0^1 2x\,dx=1\), the representing measure \(\nu_g(E)=\int_E2x\,dx\) is finite, with total mass \(1\).

For \(0\le u\le v\le1\), endpoints have Lebesgue measure zero, so the choice of open or closed endpoints does not affect the answer. Direct integration gives

\nu_g([u,v])=\int_u^v2x\,dx=v^2-u^2.

For example, \(\nu_g([1/4,3/4])=(3/4)^2-(1/4)^2=9/16-1/16=1/2\). To check the representation on a nonconstant test function, take \(f(x)=x\). Then

L(f)=\int_0^1 2x^2\,dx=\left[\frac{2x^3}{3}\right]_0^1=\frac23.

As an independent check, \(\int f\,d\nu_g\) is the same density integral, \(\int_0^1x(2x)\,dx=2/3\). Also \(L(1)=1=\nu_g([0,1])\), consistent with the total mass and the norm formula.

Worked Example: The Density \(g(x)=1+x\) on \([0,2]\)

Set \(L(f)=\int_0^2(1+x)f(x)\,dx\). The density is nonnegative throughout \([0,2]\), and its total integral is \(2+2=4\). Thus the representing measure has total mass \(4\) and \(L\) is bounded with norm \(4\).

For \(0\le u\le v\le2\),

\nu_g([u,v])=\int_u^v(1+x)\,dx=(v-u)+\frac{v^2-u^2}{2}.

On \([1/2,3/2]\), this is \((3/2-1/2)+((3/2)^2-(1/2)^2)/2=1+(9/4-1/4)/2=2\). The interval has length \(1\), and the average value of \(1+x\) there is \(2\), which independently gives mass \(2\).

For the continuous test function \(f(x)=1+x\), the functional value is

L(f)=\int_0^2(1+x)^2\,dx =\int_0^2(1+2x+x^2)\,dx =2+4+\frac83=\frac{26}{3}.

Checking by the representing measure gives \(\int f\,d\nu_g=\int_0^2(1+x)(1+x)\,dx=26/3\), as required.

Locally Finite Measures on the Real Line

Worked Example: Constant Density and Infinite Total Mass

Take \(g(x)=1\) on \(\mathbb{R}\), and define \(L(f)=\int_{\mathbb{R}} f(x)\,dx\) for \(f\in C_c(\mathbb{R})\). Every such integral is finite because \(f\) has compact support. The representing measure is Lebesgue measure on the Borel sets: \(\nu_g(E)=m(E)\). It is locally finite, but not finite, since \(\nu_g([-n,n])=2n\) for each positive integer \(n\), and \(\nu_g(\mathbb{R})=\infty\).

For a test function, let \(f(x)=\max(1-|x|,0)\). It is continuous, supported on \([-1,1]\), and nonnegative. Its functional value is the sum of the two triangular areas:

L(f)=\int_{-1}^{0}(1+x)\,dx+\int_0^1(1-x)\,dx =\frac12+\frac12=1.

The same computation is \(\int f\,d\nu_g=1\), verifying the representation directly. This functional cannot be bounded in the uniform norm on \(C_c(\mathbb{R})\): for each positive integer \(n\), choose a continuous compactly supported \(f_n\) with \(0\le f_n\le1\), equal to \(1\) on \([-n,n]\), and decreasing linearly to \(0\) on each of \([-n-1,-n]\) and \([n,n+1]\). Then \(\|f_n\|_\infty=1\) but \(L(f_n)\ge2n\). Thus local finiteness and a well-defined functional on each compactly supported test function do not imply a global uniform-norm bound.

Worked Example: An Unbounded Density with Infinite Mass

Let \(g(x)=e^{-x}\) on \(\mathbb{R}\). This density is nonnegative and integrable on every compact interval, so it defines a locally finite regular Borel measure and a functional \(L(f)=\int f(x)e^{-x}\,dx\) on \(C_c(\mathbb{R})\). For finite \(a<b\),

\nu_g([a,b])=\int_a^b e^{-x}\,dx=e^{-a}-e^{-b}.

In particular, \(\nu_g([0,\ln 2])=1-1/2=1/2\). But for each positive integer \(n\), \(\nu_g([-n,0])=e^n-1\), which grows without bound; hence \(\nu_g(\mathbb{R})=\infty\). This is a second example where the representing measure is valid on \(\mathbb{R}\) despite having infinite total mass. The density is unbounded toward the left, but its integral on every compact set remains finite, the condition needed for the \(C_c(\mathbb{R})\) functional.

A Common Mistake

A common error is to treat every measure representing an integral functional as necessarily finite, or to claim that \(f\mapsto\int_{\mathbb{R}}f\,dm\) is bounded by \(\|f\|_\infty\) on \(C_c(\mathbb{R})\). The incorrect estimate is \(\left|\int f\,dm\right|\le\|f\|_\infty\), as if the whole line had finite measure. For the functions \(f_n\) above, the left side is at least \(2n\), while the right side is \(1\); the estimate fails. The correction is to use the support: \(\left|\int f\,dm\right|\le\|f\|_\infty m(\operatorname{supp}f)\), which is finite for each compact support but has no uniform bound as the support varies. The representing measure for a functional on \(C_c(\mathbb{R})\) may therefore be locally finite with infinite total mass. This is different from the compact-interval case, where the constant test function \(1\) belongs to \(C([a,b])\) and detects the finite total mass.

Check Your Understanding

In each question, distinguish the measure on Borel sets from the Lebesgue integral used to define it.

For \(g(x)=3x^2\) on \([0,1]\), compute the total mass and the measure of \([1/3,2/3]\).
For \(g(x)=2\) on \(\mathbb{R}\), define the representing measure and determine whether it is finite, locally finite, and sigma-finite.
Explain why \(g(x)=e^{-x}\) defines a functional on \(C_c(\mathbb{R})\) even though its total mass is infinite.
For \(g(x)=1+x\) on \([0,2]\), use the constant test function to recover the operator norm of the functional.
Give the support-dependent estimate that ensures \(\int f g\,dm\) is finite for \(f\in C_c(\mathbb{R})\) when \(g\) is locally integrable.