Urysohn Cutoffs and Compactly Supported Functions

Separating Compact Sets with Continuous Functions

In the Riesz representation theory of positive linear functionals, continuous functions with compact support serve as test functions. A useful construction is a cutoff: a continuous function that is one on a chosen compact set and vanishes outside a slightly larger region. The topological hypotheses matter. Local compactness supplies compact neighborhoods, while the Hausdorff property lets us separate closed sets by open neighborhoods. Together these facts yield cutoffs without assuming a metric or a formula for distance.

All functions in this tutorial are real-valued and continuous unless stated otherwise. For a topological space \(X\), the support of a function \(f:X\to\mathbb{R}\) is \(\operatorname{supp} f=\overline{\{x\in X:f(x)\ne0\}}\), where the closure is taken in \(X\). The space \(C_c(X)\) consists of continuous real-valued functions on \(X\) whose support is compact. No measure or almost-everywhere qualification is involved in these definitions.

The main construction rests on the compact Hausdorff Urysohn lemma. We prove it first, including the separation property needed for its dyadic construction.

The Urysohn Lemma on Compact Hausdorff Spaces

Proof. First establish the separation property used below. If \(x\in Y\) and \(F\subseteq Y\) is closed with \(x\notin F\), then \(F\) is compact. For every \(y\in F\), the Hausdorff property gives disjoint open sets \(G_y,H_y\) containing \(x,y\), respectively. Finitely many of the \(H_y\) cover \(F\), say \(H_{y_1},\ldots,H_{y_n}\). The open set \(G=\bigcap_{j=1}^nG_{y_j}\) contains \(x\) and is disjoint from the open set \(H=\bigcup_{j=1}^nH_{y_j}\), which contains \(F\). Since \(H\) is open and disjoint from \(G\), it is also disjoint from \(\overline G\). Thus \(\overline G\subseteq Y\setminus H\subseteq Y\setminus F\).

It follows that any two disjoint closed sets \(A,B\) have disjoint open neighborhoods. For each \(x\in A\), apply the preceding property to \(F=B\), obtaining an open \(G_x\) containing \(x\) with \(\overline{G_x}\cap B=\varnothing\). Compactness of \(A\) gives finitely many such sets covering \(A\). Their union \(G\) is open, contains \(A\), and its closure is contained in the finite union of their closures, hence misses \(B\). The open sets \(G\) and \(Y\setminus\overline G\) are disjoint neighborhoods of \(A\) and \(B\). This also covers an empty closed set by taking its neighborhood to be empty.

Now construct open sets \(U_r\) indexed by the dyadic rationals \(r\in[0,1]\), meaning numbers of the form \(k/2^n\). Set \(U_1=Y\setminus B\). By the separation property, choose an open \(U_0\) such that \(A\subseteq U_0\) and \(\overline{U_0}\subseteq U_1\). Inductively, whenever \(r<s\) are adjacent dyadics already constructed and \(\overline{U_r}\subseteq U_s\), apply the separation property to the closed set \(\overline{U_r}\) and the closed set \(Y\setminus U_s\). It gives an open set \(U_{(r+s)/2}\) with \(\overline{U_r}\subseteq U_{(r+s)/2}\subseteq\overline{U_{(r+s)/2}}\subseteq U_s\). Repeating at each dyadic level yields \(\overline{U_r}\subseteq U_s\) whenever \(r<s\) are dyadic.

For \(y\in Y\), define \(u(y)=\inf\bigl(\{r\in\mathbb{Q}: r\text{ is dyadic},\ 0\le r<1,\ y\in U_r\}\cup\{1\}\bigr)\). Then \(0\le u\le1\). Every point of \(A\) belongs to \(U_0\), so \(u=0\) on \(A\). No point of \(B\) belongs to any \(U_r\) with \(r<1\), since \(U_r\subseteq U_1=Y\setminus B\); thus \(u=1\) on \(B\).

To verify continuity, for each real \(a\) the strict sublevel set \(\{y:u(y)<a\}\) is the union of the \(U_r\) over dyadics \(r<a\) in \([0,1)\); for \(a\le0\) it is empty, and for \(a>1\) it is all of \(Y\). Indeed, membership in such a \(U_r\) implies \(u(y)\le r<a\), while an infimum strictly below \(a\) has some member of its defining set strictly below \(a\). For \(0\le a<1\), the strict superlevel set is \(\{y:u(y)>a\}=\bigcup_{a<r<1,\ r\text{ dyadic}}(Y\setminus\overline{U_r})\). If \(y\notin\overline{U_r}\), then \(y\) belongs to no \(U_q\) with \(q\le r\), by the nesting property, so \(u(y)\ge r>a\). Conversely, if \(u(y)>a\), choose dyadics \(a<r<s<u(y)\). If \(y\in\overline{U_r}\), nesting would give \(y\in U_s\) and hence \(u(y)\le s\), a contradiction. The displayed level sets are open; the cases \(a<0\) and \(a\ge1\) give either the empty set or all of \(Y\) as appropriate. Therefore \(u\) is continuous. This proves the theorem, including the cases of empty \(A\) or \(B\). \(\square\)

Finding a Relatively Compact Neighborhood

To apply the compact result in a locally compact space, first make room between the given compact set and the prescribed open set. A set is relatively compact here if its closure is compact.

Proof. If \(K=\varnothing\), take \(V=\varnothing\). Otherwise, fix \(x\in K\). Choose a compact neighborhood \(C\) of \(x\), so \(x\in\operatorname{int}C\). A compact subset of a Hausdorff space is closed, so \(C\) is a compact Hausdorff space. The set \(O=U\cap\operatorname{int}C\) is an open neighborhood of \(x\) in \(X\); it is also relatively open in \(C\). The point-versus-closed-set separation argument in the theorem's proof, applied inside \(C\), gives a relatively open neighborhood \(W\) of \(x\) whose closure in \(C\) lies in \(O\). Because \(W\) is relatively open in \(C\), write \(W=G\cap C\) for some open \(G\subseteq X\), and put \(W'=G\cap\operatorname{int}C\). Then \(W'\) is open in \(X\), contains \(x\), and lies in \(W\). Its closure in \(X\) lies in \(C\), since \(C\) is closed, and in the closure of \(W\) relative to \(C\), since \(W'\subseteq W\subseteq C\). Consequently \(\overline{W'}\) is compact and contained in \(O\subseteq U\). For each \(x\in K\), choose such a \(W'_x\). Finitely many cover \(K\); let \(V\) be their union. Its closure is contained in the finite union of the compact sets \(\overline{W'_x}\), so it is compact and contained in \(U\). \(\square\)

The Compactly Supported Cutoff Theorem

Proof. The empty-set case holds because the zero function is continuous and its support is empty. Suppose \(K\ne\varnothing\), and take \(V\) from the lemma, so \(K\subseteq V\) and the compact set \(C=\overline V\) lies in \(U\). In the compact Hausdorff space \(C\), the sets \(K\) and \(C\setminus V\) are disjoint and closed. Apply the Urysohn lemma to obtain a continuous \(g:C\to[0,1]\) with \(g=1\) on \(K\) and \(g=0\) on \(C\setminus V\). Define \(f(x)=g(x)\) for \(x\in C\) and \(f(x)=0\) for \(x\in X\setminus C\). The set \(C\) is closed, and \(f\) vanishes on its relative boundary: indeed, \(C\setminus V\) contains \(C\setminus\operatorname{int}C\), because \(V\subseteq C\) and \(\operatorname{int}C\subseteq V\) need not hold, but any point of \(C\setminus V\) is already a zero of \(g\). To verify continuity at a point of \(C\setminus V\), use that \(g\) is continuous on \(C\) and equals zero there; points outside \(C\) have value zero as well. More formally, for any \(\varepsilon>0\), the relative open set \(\{z\in C:g(z)<\varepsilon\}\) contains \(C\setminus V\); its union with \(X\setminus C\) is open in \(X\) and is a neighborhood of each point of \(C\setminus V\) on which \(f<\varepsilon\). At points in \(V\), continuity follows locally from \(f=g\), since \(V\subseteq\operatorname{int}C\) is not required: instead \(V\) is open in \(X\) and contained in \(C\), so \(V\subseteq\operatorname{int}C\). Thus all points are covered, and \(f\) is continuous. Its nonzero set lies in \(V\), since \(g=0\) on \(C\setminus V\); hence \(\operatorname{supp}f\subseteq\overline V=C\subseteq U\). The support is closed in the compact set \(C\), so it is compact. Thus \(f\in C_c(X)\), as required. \(\square\)

The continuity check in the extension step can also be seen directly from the two open sets \(V\) and \(X\setminus\overline V\), together with their boundary. On \(V\), the function is the restriction of \(g\) to the open subspace \(V\); on \(X\setminus\overline V\), it is identically zero. At a boundary point of \(V\), which lies in \(C\setminus V\), continuity of \(g\) on \(C\) and \(g=0\) there ensure that values on \(C\) approach zero, while values outside \(C\) are zero. This is exactly why the function was required to vanish on \(C\setminus V\).

Worked Cutoffs

Why Support, Not Just Vanishing, Matters

A frequent mistake is to check only that a function vanishes at every point outside \(U\). That shows \(\{f\ne0\}\subseteq U\), but the cutoff theorem requires the stronger condition \(\operatorname{supp}f=\overline{\{f\ne0\}}\subseteq U\). For example, let \(U=(-1,1)\) and \(f(x)=\max\{0,1-|x|\}\) on \(\mathbb{R}\). It vanishes for every \(x\notin U\), but \(\{f\ne0\}=(-1,1)\), whose closure is \([-1,1]\); the support is not contained in \(U\). The correction is to make the nonzero region smaller, for example \(g(x)=\max\{0,1-2|x|\}\). Then \(\{g\ne0\}=(-1/2,1/2)\) and \(\operatorname{supp}g=[-1/2,1/2]\subset U\), while \(g(0)=1\). The theorem's intermediate neighborhood \(V\) provides precisely this extra margin.

On a discrete space, the same principle has a simple form. In a discrete space every function is continuous and every finite set is compact. If \(K\) is finite and \(K\subseteq U\), then \(\chi_K\) is a cutoff: its support is \(K\), which is contained in \(U\). In a general locally compact Hausdorff space, the theorem replaces this indicator—which need not be continuous—with a continuous function while retaining compact support inside the neighborhood. This is why cutoffs are useful when approximating or testing with functions in \(C_c(X)\).