Integrating Continuous Functions Against Radon Measures

From a Measure to a Functional

Continuous functions with compact support are natural test functions for a measure on a locally compact space. Their supports lie in compact sets, where a Radon measure has finite mass; this makes their integrals finite even when the whole space has infinite measure. The resulting map from functions to numbers is a positive linear functional, and its concrete form can be read directly from the measure.

Throughout, \(X\) is a locally compact Hausdorff space, and \(\mathcal{B}_X\) is its Borel sigma-algebra. The measure \(\mu\) is a positive Radon measure on \((X,\mathcal{B}_X,\mu)\): it is finite on compact sets, outer regular on Borel sets, and inner regular on open sets. We do not assume that \(\mu\) is finite on \(X\), complete, or sigma-finite. Functions in \(C_c(X)\) are real-valued continuous functions with compact support, where \(\operatorname{supp} f\) is the closure of \(\{x:f(x)\ne0\}\).

Definition: A linear functional \(T:C_c(X)\to\mathbb{R}\) is positive if \(T(f)\ge0\) whenever \(f(x)\ge0\) for every \(x\in X\). The functional induced by \(\mu\) is \(T_\mu(f)=\int_X f\,d\mu\), for \(f\in C_c(X)\).

Continuity makes \(f\) Borel measurable. If \(K=\operatorname{supp}f\), then \(f=0\) on \(X\setminus K\), \(f\) is bounded on the compact set \(K\), and \(\mu(K)<\infty\). Thus \(f\) is integrable: its absolute integral is finite. This conclusion uses only finiteness on compact sets, not finiteness of the total measure.

The Induced Functional

Proposition: For a positive Radon measure \(\mu\) on \((X,\mathcal{B}_X,\mu)\), the map \(T_\mu:C_c(X)\to\mathbb{R}\) given by \(T_\mu(f)=\int_X f\,d\mu\) is well-defined, linear, and positive.

Proof. Let \(f\in C_c(X)\) and set \(K=\operatorname{supp}f\). Continuity on the compact set \(K\) gives \(M=\sup_{x\in K}|f(x)|<\infty\) (if \(K\) is empty, \(f=0\)). Since \(f=0\) outside \(K\), \(\int_X|f|\,d\mu\le M\mu(K)<\infty\). Therefore \(\int_X f\,d\mu\) exists as a finite real number, so \(T_\mu\) is well-defined. For \(f,g\in C_c(X)\) and \(a,b\in\mathbb{R}\), the function \(af+bg\) is in \(C_c(X)\), and linearity of the Lebesgue integral for integrable real-valued functions gives \(T_\mu(af+bg)=aT_\mu(f)+bT_\mu(g)\). If \(f\ge0\) everywhere, the integral of the nonnegative measurable function \(f\) is nonnegative, so \(T_\mu(f)\ge0\). These facts prove the claim. \(\square\)

The integral also gives a useful estimate on any fixed compact support. If \(f\in C_c(X)\) and \(K\) is a compact set containing \(\operatorname{supp}f\), then

|T_\mu(f)|\le \int_X|f|\,d\mu \le \left(\sup_{x\in K}|f(x)|\right)\mu(K).

The first inequality follows from the basic inequality \(\left|\int f\,d\mu\right|\le\int|f|\,d\mu\) for integrable real-valued functions. The second follows pointwise from \(|f|\le(\sup_K|f|)\chi_K\), followed by monotonicity of the integral. In particular, for functions supported in the same compact \(K\), the functional is controlled by the uniform size of the function and the finite number \(\mu(K)\). This is a local estimate; it does not assert that \(T_\mu\) is bounded using a single finite total mass on all of \(X\).

Worked Example: A Dirac Measure Evaluates at One Point

Take \(X=\mathbb{R}\) with its usual topology and let \(\mu=\delta_2\), the Dirac measure at \(2\), defined on Borel sets by \(\delta_2(E)=1\) if \(2\in E\) and \(0\) otherwise. This is a finite Radon measure. For \(f(x)=\max(1-|x-2|,0)\), compute \(T_{\delta_2}(f)\).

The function is continuous and vanishes for \(|x-2|\ge1\), so its support is \([1,3]\), a compact set. At the atom, \(f(2)=\max(1-0,0)=1\). The defining integral against a Dirac measure is \(\int f\,d\delta_2=f(2)\), hence \(T_{\delta_2}(f)=1\). For a direct check, \(\delta_2\) assigns mass \(1\) to \(\{2\}\) and mass \(0\) to its complement; the function has value \(1\) on that atom, so its integral is \(1\cdot1+0=1\).

Worked Example: A Weighted Atomic Measure

On \(\mathbb{R}\), let \(\mu=2\delta_{-1}+3\delta_0+\delta_2\). This is a finite Radon measure, with total mass \(2+3+1=6\). Define \(f\) by linear interpolation through the points \((-2,0),(-1,2),(0,1),(2,3),(3,0)\), and set \(f=0\) outside \([-2,3]\). Compute its integral against \(\mu\).

The interpolation gives a continuous, compactly supported function. In particular, its values at the three atoms are \(f(-1)=2\), \(f(0)=1\), and \(f(2)=3\). Integration against each Dirac measure evaluates the function at that atom, so

T_\mu(f)=2f(-1)+3f(0)+f(2) =2\cdot2+3\cdot1+1\cdot3 =4+3+3=10.

Check: the measure gives masses \(2,3,1\) to the distinct singleton sets \(\{-1\},\{0\},\{2\}\), respectively, and mass zero to their complement. The contributions are therefore \(4,3,3\), whose sum is \(10\), as above.

Worked Example: A Measure Given by a Density

Let \(X=\mathbb{R}\), and on \((\mathbb{R},\mathcal{B}_{\mathbb{R}})\) define \(\mu(E)=2m(E\cap[0,1])\), where \(m\) is Lebesgue measure. This is a finite Radon measure: it is twice the restriction of Lebesgue measure to the compact interval \([0,1]\), and regularity follows from regularity of Lebesgue measure. Its total mass is \(\mu(\mathbb{R})=2m([0,1])=2\). For \(f(x)=x(1-x)\) on \([0,1]\), extended as zero outside that interval, compute \(T_\mu(f)\).

The endpoint values of the polynomial are both zero, so the extension is continuous. Its support is contained in \([0,1]\). By the definition of \(\mu\), integration against \(\mu\) is integration against the density \(2\chi_{[0,1]}\) with respect to \(m\). Therefore

T_\mu(f)=\int_{\mathbb{R}}f\,d\mu =2\int_0^1x(1-x)\,dx =2\left(\int_0^1x\,dx-\int_0^1x^2\,dx\right) =2\left(\frac12-\frac13\right) =\frac13.

Check: the two integrals are \([x^2/2]_0^1=1/2\) and \([x^3/3]_0^1=1/3\), so their difference is \(1/6\); multiplying by the density value \(2\) gives \(1/3\). Equivalently, the symmetry \(x(1-x)=(1-x)x\) about \(1/2\) is consistent with the integral \(1/6\) before the density factor is applied.

What the Examples Show

These cases illustrate a general rule for reading the functional from the measure. A single Dirac mass samples the function at one point. A finite weighted sum of Dirac masses gives a weighted sum of point values. A measure with density \(\rho\) relative to Lebesgue measure gives \(T_\mu(f)=\int f(x)\rho(x)\,dx\) whenever that density representation is in force and the integral is finite. In every case, compact support of \(f\) and local finiteness of the measure ensure finiteness of the integral.

For example, Lebesgue measure \(m\) on \(\mathbb{R}\) has infinite total mass, since \(m(\mathbb{R})=\infty\), but it is finite on every compact set. Consequently \(f\mapsto\int f\,dm\) is still defined and finite for every \(f\in C_c(\mathbb{R})\). The important hypothesis here is local finiteness, not finite total mass.

Takeaway: A positive Radon measure finite on compact sets defines a positive linear functional on \(C_c(X)\) by integration. Compact support supplies the finiteness needed for each individual integral, while the measure determines whether the functional samples values at atoms or averages them through a density.

A Common Pitfall

A common mistake is to infer that integration against a measure of infinite total mass must give an infinite value for every nonzero test function. For example, \(m(\mathbb{R})=\infty\) does not imply \(\int f\,dm=\infty\) for nonzero \(f\in C_c(\mathbb{R})\). Consider \(f(x)=\max(1-|x|,0)\). Its support is \([-1,1]\), and it is bounded by \(1\), so the local estimate gives \(\int|f|\,dm\le 1\cdot m([-1,1])=2\). In fact, integrating the two linear pieces gives

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

The incorrect calculation would replace \(m([-1,1])\) by \(m(\mathbb{R})=\infty\) as though the function were nonzero on the whole line. The correction is to use the compact support: outside \([-1,1]\), the integrand is zero. This distinction is exactly why measures finite on compact sets suffice for the construction.

Check Your Understanding

For each question, identify the measure space and verify that the function being integrated has compact support.

For \(\mu=5\delta_1\) on \(\mathbb{R}\) and \(f(x)=\max(1-|x-1|,0)\), calculate \(T_\mu(f)\).
Suppose \(f\in C_c(X)\) is nonnegative. Which property of the Lebesgue integral implies \(T_\mu(f)\ge0\)?
Let \(K\) be compact and contain \(\operatorname{supp}f\). State the estimate for \(|T_\mu(f)|\) in terms of \(\sup_K|f|\) and \(\mu(K)\), and identify the assumption that makes its right-hand side finite.
For \(\mu(E)=3m(E\cap[0,1])\) and \(f(x)=x\) on \([0,1]\), extended by zero outside, compute \(\int f\,d\mu\).
Why does \(m(\mathbb{R})=\infty\) not prevent \(f\mapsto\int f\,dm\) from defining a finite value for every \(f\in C_c(\mathbb{R})\)?