Open-Set Contents Induced by a Functional

From a Positive Functional to Values on Open Sets

Let $X$ be a locally compact Hausdorff space. A positive linear functional on $C_c(X)$ assigns a nonnegative number to every nonnegative compactly supported continuous function. Here $C_c(X)$ denotes the real-valued continuous functions on $X$ with compact support, and the support of $f$ is the closure of $\{x:f(x)\ne0\}$. Positivity means $L(f)\ge0$ whenever $f\ge0$. The functional is finite-valued on each member of $C_c(X)$, though it need not have a finite bound on all of $C_c(X)$.

For an open set $U$, we test $L$ on functions that stay between zero and one and whose support lies inside $U$. Their functional values give a number, possibly $+\infty$, that records how much of $L$ can be detected from within $U$. Since the notation $m$ is used for Lebesgue measure, write $m_L(U)$ for this functional-induced content when $L$ is fixed. It is not, by definition alone, a measure on a sigma-algebra.

Definition: For an open set $U\subseteq X$, define $$ m_L(U)=\sup\{L(f): f\in C_c(X),\ 0\le f\le1,\ \operatorname{supp}f\subseteq U\}. $$ The supremum is in $[0,\infty]$. The zero function is admissible, so the set over which the supremum is taken is never empty.

Two details in the definition matter. First, the condition is on the support, not merely on the points where $f$ is nonzero: the closure of those points must also remain in $U$. Second, $m_L(U)$ is a supremum and need not be attained by any single admissible function. The cutoff result established in the earlier material on Urysohn cutoffs will be useful: if $K\subset U$, with $K$ compact and $U$ open, then there is $f\in C_c(X)$ with $0\le f\le1$, $f=1$ on $K$, and $\operatorname{supp}f\subset U$.

When the Functional Is Given by a Radon Measure

A Radon measure here means a Borel measure $\nu$ on $X$ that is finite on compact sets and inner regular on open sets: for every open $U$, $\nu(U)=\sup\{\nu(K):K\subset U,\ K\text{ compact}\}$. This regularity is precisely what allows compactly supported tests inside $U$ to recover the full value $\nu(U)$, even when that value is infinite.

Theorem: Let $X$ be locally compact Hausdorff, let $\nu$ be a positive Radon measure on its Borel sets, and define $L(f)=\int_X f\,d\nu$ for $f\in C_c(X)$. Then $L$ is a positive linear functional and, for every open $U\subseteq X$, $m_L(U)=\nu(U)$, including when $\nu(U)=+\infty$.

Proof. Each $f\in C_c(X)$ is bounded and vanishes outside a compact set. Since $\nu$ is finite on compact sets, $\int_X |f|\,d\nu<\infty$, so $L(f)$ is defined and finite. Linearity follows from linearity of the integral, and $f\ge0$ implies $L(f)\ge0$. Now fix an open $U$. If $f$ is admissible in the definition of $m_L(U)$, then $0\le f\le1$ and $f=0$ on $X\setminus U$. Therefore $L(f)\le\nu(U)$, giving $m_L(U)\le\nu(U)$. In the other direction, let $K\subset U$ be compact. The cutoff result gives an admissible $f$ equal to one on $K$; thus $L(f)=\int_X f\,d\nu\ge\nu(K)$, so $m_L(U)\ge\nu(K)$. Taking the supremum over compact $K\subset U$ and using inner regularity gives $m_L(U)\ge\nu(U)$. These inequalities also imply equality when $\nu(U)$ is infinite: in that case the compact-set values are unbounded, so $m_L(U)=+\infty$. If $U=\varnothing$, only the zero function is admissible and both sides equal zero. Hence $m_L(U)=\nu(U)$ for every open $U$. $\square$

This result is a calculation from a functional whose measure is already known. It does not yet assert that every positive functional has a representing measure; that construction requires further steps. The theorem does show why the supremum is a natural candidate for the value a representing measure should assign to an open set.

Worked Example: Lebesgue Measure of an Open Interval

Take $X=\mathbb R$, let $\nu=m$ be Lebesgue measure, and define $L(f)=\int_{\mathbb R}f\,dm$ for real-valued $f\in C_c(\mathbb R)$. For $U=(0,1)$, the theorem gives $m_L((0,1))=m((0,1))=1$. To see explicitly how admissible functions approach this value, for each integer $n\ge2$ define a continuous function $f_n$ that is zero outside $[1/(2n),1-1/(2n)]$, rises linearly from zero to one on $[1/(2n),1/n]$, equals one on $[1/n,1-1/n]$, and falls linearly to zero on $[1-1/n,1-1/(2n)]$.

Each $f_n$ has compact support contained in $(0,1)$ and satisfies $0\le f_n\le1$. Its plateau has length $1-2/n$. Each linear ramp has base $1/(2n)$ and height one, so its area is $\tfrac12\cdot\tfrac1{2n}\cdot1=\tfrac1{4n}$. Consequently,

L(f_n)=\int_{\mathbb R} f_n\,dm =1-\frac2n+\frac1{4n}+\frac1{4n} =1-\frac{3}{2n}.

For $n=2$, this is $1-\tfrac34=\tfrac14$; for $n=3$, it is $1-\tfrac12=\tfrac12$; and for $n=6$, it is $1-\tfrac14=\tfrac34$. These values approach one, so the supremum is at least one. Conversely every admissible $f$ satisfies $0\le f\le\chi_{(0,1)}$, and hence $\int f\,dm\le m((0,1))=1$. Thus the supremum is exactly one. Check: the theorem gives $m((0,1))=1$, agreeing with the explicit upper bound and approximating values.

Worked Example: A Weighted Functional on an Open Interval

On $X=\mathbb R$, let $\nu(A)=\int_A e^{-|x|}\,dm(x)$ for Borel sets $A$, and set $L(f)=\int_{\mathbb R}f(x)e^{-|x|}\,dm(x)$. This is a positive functional on $C_c(\mathbb R)$, induced by the Radon measure with density $e^{-|x|}$. Let $U=(-\ln2,\ln2)$, which contains zero. Since $e^{-|x|}$ is even and $\ln2>0$,

m_L(U)=\nu(U) =2\int_0^{\ln2}e^{-x}\,dx =2\bigl(1-e^{-\ln2}\bigr) =2\left(1-\frac12\right) =1.

The value at either endpoint does not affect the integral because each singleton has Lebesgue measure zero. Check: the integral over the positive half of $U$ is $1-\tfrac12=\tfrac12$; doubling it gives one, as in the displayed calculation.

Worked Example: A Functional Supported at Two Points

On $X=\mathbb R$, define $L(f)=3f(0)+2f(2)$. This is positive and linear on $C_c(\mathbb R)$. It is integration against the finite Radon measure $\nu=3\delta_0+2\delta_2$, where $\delta_a$ assigns mass one to a set containing $a$ and zero otherwise. Consider $U=(-1,1)$. It contains $0$ but not $2$, so $\nu(U)=3$, and the theorem gives $m_L(U)=3$.

The supremum can also be checked directly. Any admissible $f$ has $0\le f(0)\le1$. Since $2\notin U$ and $\operatorname{supp}f\subset U$, $f(2)=0$. Thus $L(f)=3f(0)\le3$. The cutoff result supplies an admissible $f$ equal to one at $0$, for which $L(f)=3$. Hence the supremum is exactly three. Check: substituting $f(0)=1$ and $f(2)=0$ into the functional gives $3\cdot1+2\cdot0=3$.

A Common Mistake: Using the Indicator as a Test Function

For $U=(0,1)$ in the Lebesgue example, a tempting but invalid argument is to choose $f=\chi_{(0,1)}$ in the defining supremum. The calculation $\int\chi_{(0,1)}\,dm=1$ is correct, but this function is not continuous, so it does not belong to $C_c(\mathbb R)$. Moreover, its support is $[0,1]$, which is not contained in $(0,1)$. It violates both admissibility requirements.

The correction is to use actual continuous cutoffs such as the $f_n$ above. They have support strictly inside $U$, and their integrals $1-\tfrac{3}{2n}$ approach one. The upper bound $\int f\,dm\le1$ applies to every admissible function, so the supremum is one without pretending it is attained by the indicator.

What the Calculation Establishes

The definition uses only $L$, the topology of $X$, and the open set $U$. In the examples, it returns familiar quantities because the functionals are already given by integration against Radon measures. In particular, compact cutoffs capture all the mass of an open set: they test compact subsets from within, while regularity says those compact subsets account for the full value.

Keep the distinction between the objects in view. $L$ is a functional on functions, while $m_L$ is at this stage defined only on open sets. Its values may be infinite, even though $L(f)$ is finite for every individual $f\in C_c(X)$. Establishing additional properties of $m_L$, and constructing a measure from it, are separate steps.

Check Your Understanding

In each question, use the support condition as well as the pointwise bounds in the definition.

For a positive Radon measure $\nu$ and $L(f)=\int f\,d\nu$, explain why every admissible $f$ gives $L(f)\le\nu(U)$.
Let $L(f)=5f(1)$ on $\mathbb R$. What are $m_L((0,2))$ and $m_L((2,3))$? Justify both values.
For $L(f)=\int_{\mathbb R}f\,dm$, why is $\chi_{(0,1)}$ not an admissible function for computing $m_L((0,1))$?
For the density $e^{-|x|}$, calculate $\nu((-a,a))$ for $a>0$ by integrating on the positive and negative halves.
Why can $m_L(U)$ be infinite even though $L(f)$ is finite for every $f\in C_c(X)$?

Open-Set Contents Induced by a Functional

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