Uniqueness from Continuous Test Functions

How Continuous Tests Determine a Measure

A measure records the sizes of sets, while a continuous test function records an average weighted by its values. The uniqueness question is whether knowing all these averages determines the original measure. On a locally compact Hausdorff space, compactly supported continuous functions can approximate the indicator of an open set from below in the precise sense needed to recover its measure. Once open sets are recovered, outer regularity determines the measure of every Borel set.

Throughout this tutorial, $X$ is a locally compact Hausdorff space, and measures are positive Borel measures on $(X,\mathcal{B}_X)$. A Radon measure here means a measure that is locally finite, inner regular on open sets, and outer regular on Borel sets. Local finiteness means every point has an open neighborhood of finite measure. Inner regularity on an open set $U$ means $\mu(U)=\sup\{\mu(K):K\subseteq U,\ K\text{ compact}\}$; outer regularity on a Borel set $E$ means $\mu(E)=\inf\{\mu(U):E\subseteq U,\ U\text{ open}\}$. The total measure $\mu(X)$ need not be finite.

We use real-valued functions in $C_c(X)$, the continuous functions on $X$ with compact support. The support of $f$ is the closure of $\{x:f(x)\ne0\}$. If $\mu$ is Radon and $f\in C_c(X)$, then $f$ is bounded and $\int f\,d\mu$ is finite: its support is compact and has finite measure, so $\int |f|\,d\mu\le \|f\|_\infty\mu(\operatorname{supp}f)<\infty$. Thus equality of integrals against every $f\in C_c(X)$ is a meaningful condition, even when $\mu(X)=\infty$.

Cutoff fact: If $K\subseteq U\subseteq X$, where $K$ is compact and $U$ is open, there is $f\in C_c(X)$ with $0\le f\le1$, $f=1$ on $K$, and $\operatorname{supp}f\subseteq U$. This is the standard compactly supported Urysohn cutoff result for locally compact Hausdorff spaces.

Recovering the Measure of an Open Set

Proposition (Open-set recovery): If $(X,\mathcal{B}_X,\mu)$ is a Radon measure space and $U\subseteq X$ is open, then $$ \mu(U)=\sup\left\{\int f\,d\mu:\ f\in C_c(X),\ 0\le f\le1,\ \operatorname{supp}f\subseteq U\right\}. $$ The equality allows the value $+\infty$. For $U=\varnothing$, the supremum is $0$.

Proof. First suppose $U$ is any open set, including the empty set. For every function $f$ in the displayed family, $f=0$ outside $U$ and $0\le f\le1$, so $0\le f\le\chi_U$. Monotonicity of the nonnegative Lebesgue integral gives $\int f\,d\mu\le\int\chi_U\,d\mu=\mu(U)$. Thus the supremum is at most $\mu(U)$. If $U=\varnothing$, the only admissible function is zero, so equality holds. Now take $U\ne\varnothing$. For each compact $K\subseteq U$, the cutoff fact gives an admissible $f$ with $f=1$ on $K$. Therefore $\int f\,d\mu\ge\int\chi_K\,d\mu=\mu(K)$. Taking the supremum over compact $K\subseteq U$ and using inner regularity on $U$ shows that the supremum over admissible $f$ is at least $\mu(U)$. This argument also applies when $\mu(U)=+\infty$: inner regularity then makes the compact-set measures unbounded. The two inequalities prove the formula. $\square$

Theorem (Uniqueness from continuous test functions): Let $(X,\mathcal{B}_X,\mu)$ and $(X,\mathcal{B}_X,\nu)$ be positive Radon measure spaces on the same locally compact Hausdorff space. Suppose $\int f\,d\mu=\int f\,d\nu$ for every real-valued $f\in C_c(X)$. Then $\mu(E)=\nu(E)$ for every $E\in\mathcal{B}_X$. Neither measure is required to be finite.

Proof. Fix an open set $U\subseteq X$. The two measures have the same integral for every $f\in C_c(X)$ satisfying $0\le f\le1$ and $\operatorname{supp}f\subseteq U$. Applying open-set recovery first to $\mu$ and then to $\nu$ gives $\mu(U)=\nu(U)$, with equality understood in $[0,\infty]$. Now fix a Borel set $E$. Outer regularity gives $$ \mu(E)=\inf_{\substack{U\supseteq E\\U\text{ open}}}\mu(U) =\inf_{\substack{U\supseteq E\\U\text{ open}}}\nu(U)=\nu(E). $$ The middle equality holds because the two measures agree on every open set. This proves equality on every Borel set, including cases where the common value is infinite. $\square$

The proof uses two different properties of a Radon measure for distinct jobs: inner regularity allows compact subsets to approximate an open set from within, and outer regularity transfers equality on open sets to all Borel sets. The cutoff functions connect the compact subsets to the continuous tests. No finiteness assumption on the whole space is needed.

Worked Examples on the Real Line

Worked Example: Recovering the Length of an Open Interval

Take $X=\mathbb{R}$ with Lebesgue measure $m$, and let $U=(0,2)$. For each integer $n\ge2$, define $$ f_n(x)=\min\{1,\ (nx-1)_+,\ (n(2-x)-1)_+\},\qquad t_+=\max\{t,0\}. $$ Each $f_n$ is continuous, takes values in $[0,1]$, and vanishes outside $[1/n,2-1/n]$. Its support is contained in $U$, so it is an admissible test in the open-set recovery formula.

On the left ramp, $f_n(x)=nx-1$ for $1/n\le x\le2/n$; its integral there is the area of a triangle with base $1/n$ and height $1$, namely $1/(2n)$. On the middle interval $[2/n,2-2/n]$, the function is $1$, and that interval has length $2-4/n$. The right ramp has integral $1/(2n)$, by the same linear calculation. Thus, including the case $n=2$, when the middle interval has length zero, $$ \int_{\mathbb{R}}f_n\,dm=\frac{1}{2n}+\left(2-\frac{4}{n}\right)+\frac{1}{2n}=2-\frac{3}{n}. $$ For example, $n=2$ gives $2-3/2=1/2$, and $n=4$ gives $2-3/4=5/4$.

For each $x\in(0,2)$, both $nx-1$ and $n(2-x)-1$ eventually exceed $1$, so $f_n(x)\to1$. For $x\notin(0,2)$, $f_n(x)=0$. Also, $f_n(x)$ increases with $n$: inside $(0,2)$, each of the two nonconstant terms in the minimum increases with $n$, and outside it the functions vanish. The Monotone Convergence Theorem therefore applies to the measurable nonnegative sequence $f_n\uparrow\chi_{(0,2)}$, giving $$ \lim_{n\to\infty}\int f_n\,dm=\int\chi_{(0,2)}\,dm=m((0,2))=2. $$ This agrees with $2-3/n\to2$, and with the usual length of the interval.

Worked Example: Detecting the Mass at a Point

On $\mathbb{R}$, consider the finite Radon measures $\mu=3\delta_{-1}+2\delta_2$ and $\nu=a\delta_{-1}+b\delta_2$, where $a,b\ge0$ and $\delta_y(E)=1$ if $y\in E$, $0$ otherwise. The functions $$ f_{-1}(x)=\max\{1-|x+1|,0\},\qquad f_2(x)=\max\{1-|x-2|,0\} $$ belong to $C_c(\mathbb{R})$. The first has support $[-2,0]$, so $f_{-1}(-1)=1$ and $f_{-1}(2)=0$. Consequently, $$ \int f_{-1}\,d\mu=3\cdot1+2\cdot0=3,\qquad \int f_{-1}\,d\nu=a\cdot1+b\cdot0=a. $$ The second has support $[1,3]$, so $f_2(-1)=0$ and $f_2(2)=1$. Hence $$ \int f_2\,d\mu=3\cdot0+2\cdot1=2,\qquad \int f_2\,d\nu=a\cdot0+b\cdot1=b. $$ If the integrals agree for every compactly supported continuous test, these two tests force $a=3$ and $b=2$. Substitution gives $3\cdot1+2\cdot0=3=a$ and $3\cdot0+2\cdot1=2=b$, as required. These tests isolate the two atoms because each function is zero at the other atom.

Worked Example: A Compactly Supported Test Distinguishes Two Measures

Let $\mu=m$ and $\nu=2m$ on the Borel sets of $\mathbb{R}$. These are Radon measures, and both are locally finite. Use the tent function $g(x)=\max\{1-|x|,0\}$, which is continuous, takes values in $[0,1]$, and has support $[-1,1]$. Its graph consists of two triangles, one on $[-1,0]$ and one on $[0,1]$, each with base $1$ and height $1$. Therefore $$ \int g\,d\mu=\frac12+\frac12=1,\qquad \int g\,d\nu=2\left(\frac12+\frac12\right)=2. $$ The two measures do not agree on all continuous tests, and the computed integrals explicitly detect that failure. As a check, $g(0)=1$ and $g(\pm1)=0$, so each triangular area is indeed $\tfrac12(1)(1)=1/2$.

A Common Pitfall: Indicators Are Not Continuous Tests

For $U=(0,2)$, it is incorrect to claim that $\chi_U\in C_c(\mathbb{R})$ and then use $\int\chi_U\,dm$ as one of the prescribed test integrals. The indicator is discontinuous at $0$ and $2$. The correct argument uses continuous cutoffs supported inside $U$, such as the $f_n$ above, and takes a supremum or a limit after computing their integrals. Their values are $2-3/n$, not exactly $2$ for any finite $n$, but their limit recovers $m(U)=2$.

Another tempting shortcut is to test the constant function $1$ to compare total masses on $\mathbb{R}$. But $1\notin C_c(\mathbb{R})$, since its support is not compact. The uniqueness theorem does not rely on such a test and does not require finite total mass. It recovers each open set locally through compactly supported cutoffs, then uses outer regularity for Borel sets. That is the essential reason the argument works on noncompact spaces.

Check Your Understanding

Use the open-set recovery formula and the uniqueness proof to organize your answers.

For a compact set $K$ contained in an open set $U$, what properties must the cutoff function have in the open-set recovery argument?
Why does $\int f\,d\mu\le\mu(U)$ hold whenever $0\le f\le1$ and $\operatorname{supp}f\subseteq U$?
Which regularity property transfers equality on open sets to equality on all Borel sets?
For the measures $3\delta_{-1}+2\delta_2$ and $a\delta_{-1}+b\delta_2$, which test functions isolate the two coefficients?
Why can the constant function $1$ not generally be used as a test in $C_c(\mathbb{R})$?

Uniqueness from Continuous Test Functions

What You'll Learn