Countable Dense Simple Functions from Countable Generators

From a Countable Algebra to a Countable Dense Family

Density says that every element of a function space can be approximated as accurately as desired by elements of a specified family. Separability is the stronger-looking but often more useful conclusion that one can choose that family to be countable. On a finite measure space, a countable algebra that generates the measurable sets modulo null sets supplies a concrete countable dense family in every \(L^p\) with \(0<p<\infty\).

Fix a finite measure space \((X,\mathcal{M},\mu)\), so \(\mu(X)<\infty\), and let \(\mathcal{A}\subseteq\mathcal{M}\) be a countable algebra of sets. An algebra contains \(X\) and is closed under complements and finite unions. Assume it generates \(\mathcal{M}\) modulo null sets in the following precise sense: for every \(E\in\mathcal{M}\), there is a \(B\in\sigma(\mathcal{A})\) such that \(\mu(E\triangle B)=0\). The space need not be complete. Here \(L^p(\mu)\) consists of equivalence classes, under equality \(\mu\)-a.e., of measurable real- or complex-valued functions \(f\) for which \(\int |f|^p\,d\mu<\infty\). When \(0<p<1\), the usual \(L^p\) quantity is a quasi-norm rather than a norm; the density assertion remains valid.

Overlaps among the sets \(A_j\) are allowed. A finite sum of this form is still a simple function, takes finitely many rational (or rational-complex) values, and is measurable. Since \(\mathcal{A}\) is countable, there are only countably many choices of each finite list of sets and coefficients. Taking the union over all finite lengths shows that \(\mathcal{D}_{\mathcal{A}}\) is countable. Finiteness of \(\mu(X)\) also ensures every member belongs to \(L^p(\mu)\): each is bounded and \(\int |g|^p\,d\mu\le \|g\|_\infty^p\mu(X)<\infty\).

Approximating Measurable Sets in Measure

The first step is a set approximation fact. It is important that it gives approximation in measure, not exact equality with an element of the algebra.

Proof. Let \(\mathcal{C}\) be the collection of sets \(E\in\sigma(\mathcal{A})\) for which, for every \(\varepsilon>0\), some \(A\in\mathcal{A}\) satisfies \(\mu(E\triangle A)<\varepsilon\). Every \(A\in\mathcal{A}\) belongs to \(\mathcal{C}\), by using \(A\) itself. If \(E\in\mathcal{C}\), choose \(A\in\mathcal{A}\) with \(\mu(E\triangle A)<\varepsilon\). Since \(A^c\in\mathcal{A}\) and \(E^c\triangle A^c=E\triangle A\), the complement \(E^c\) is in \(\mathcal{C}\).

Now let \(E_1,E_2,\ldots\in\mathcal{C}\), and put \(E=\bigcup_{j=1}^{\infty}E_j\). Because \(\mu(X)<\infty\), continuity from below gives \(\mu(E\setminus\bigcup_{j=1}^N E_j)\to0\). Given \(\varepsilon>0\), choose \(N\) so this measure is less than \(\varepsilon/2\). For each \(j=1,\ldots,N\), choose \(A_j\in\mathcal{A}\) with \(\mu(E_j\triangle A_j)<\varepsilon/(2N)\), and set \(A=\bigcup_{j=1}^N A_j\), which is in \(\mathcal{A}\). The symmetric difference between the two finite unions is contained in \(\bigcup_{j=1}^N(E_j\triangle A_j)\). Therefore, by subadditivity,

Thus \(\mathcal{C}\) contains \(\mathcal{A}\) and is closed under complements and countable unions. It is a sigma-algebra, so it contains \(\sigma(\mathcal{A})\). This proves the lemma. \(\square\)

Under the modulo-null-sets assumption, this lemma also approximates every \(E\in\mathcal{M}\): choose \(B\in\sigma(\mathcal{A})\) with \(\mu(E\triangle B)=0\), then approximate \(B\) by \(A\in\mathcal{A}\). The inclusion \(E\triangle A\subseteq(E\triangle B)\cup(B\triangle A)\) gives \(\mu(E\triangle A)=\mu(B\triangle A)\), which can be made arbitrarily small.

Density of the Countable Family

Proof. Fix \(f\in L^p(\mu)\) and a desired error \(\varepsilon>0\). The earlier density result for simple functions in finite-\(p\) \(L^p\) gives a finite-valued measurable simple function \(s\) with \(\|f-s\|_p<\varepsilon/2\). Write \(s=\sum_{j=1}^k c_j\chi_{E_j}\), where \(E_j\in\mathcal{M}\) are disjoint and the coefficients \(c_j\) are real or complex. If \(s=0\), the zero member of \(\mathcal{D}_{\mathcal{A}}\) already has zero distance from \(s\). Otherwise, terms with coefficient zero may be discarded, so take \(k\ge1\) and at least one \(c_j\ne0\).

By set approximation, for any \(\delta>0\) choose \(A_j\in\mathcal{A}\) with \(\mu(E_j\triangle A_j)<\delta\). Put \(h=\sum_{j=1}^k c_j\chi_{A_j}\) and \(U=\bigcup_{j=1}^k(E_j\triangle A_j)\). Outside \(U\), \(s=h\). Everywhere, \(|s-h|\le C\), where \(C=\sum_{j=1}^k|c_j|\). Hence

Choose \(\delta\) small enough that \(C^p k\delta<\varepsilon^p/(2K_p)\), where \(K_p=1\) if \(0<p\le1\), and \(K_p=2^{p-1}\) if \(p>1\). Next choose rational (or rational-complex) \(q_j\) so that \(\sum_{j=1}^k|q_j-c_j|<\eta\), where \(\eta>0\) is small enough that \(\eta^p\mu(X)<\varepsilon^p/(2K_p)\). Such coefficients exist because the rationals are dense in the reals, and \(\mathbb{Q}_{\mathbb{C}}\) is dense in the complex numbers. If \(\mu(X)=0\), all \(L^p\) classes are zero and the theorem is immediate; otherwise this choice of \(\eta\) is possible.

Set \(g=\sum_{j=1}^k q_j\chi_{A_j}\in\mathcal{D}_{\mathcal{A}}\). Pointwise, \(|h-g|\le\sum_j|c_j-q_j|<\eta\), so \(\int|h-g|^p\,d\mu\le\eta^p\mu(X)\). For any complex or real numbers \(u,v\), \(|u+v|^p\le K_p(|u|^p+|v|^p)\): for \(p\le1\) this follows from subadditivity of the \(p\)th power on nonnegative numbers, and for \(p>1\) it follows from convexity applied to \((|u|+|v|)/2\). Applying this inequality to \(s-g=(s-h)+(h-g)\) yields

Finally, if \(p\ge1\), Minkowski's inequality gives \(\|f-g\|_p\le\|f-s\|_p+\|s-g\|_p<\varepsilon\). If \(0<p<1\), integrate \(|f-g|^p\le|f-s|^p+|s-g|^p\); the two terms are each less than \((\varepsilon/2)^p\), so \(\|f-g\|_p^p<2(\varepsilon/2)^p\), which alone need not be less than \(\varepsilon^p\). To ensure the desired bound in this case, start instead with both errors less than \(\varepsilon/4\): simple-function density supplies \(\|f-s\|_p<\varepsilon/4\), and the choices above can make \(\|s-g\|_p<\varepsilon/4\). Then \(\|f-g\|_p^p<2(\varepsilon/4)^p\le\varepsilon^p\), since \(2/4^p<1\) is not true for every positive \(p\). More generally choose both errors less than \(\varepsilon\,2^{-1/p-1}\); then \(2(\varepsilon\,2^{-1/p-1})^p=\varepsilon^p/2^p<\varepsilon^p\). Simple-function density and the construction allow this choice. Thus \(\|f-g\|_p<\varepsilon\) also for \(0<p<1\), proving density. \(\square\)

Why “Generated” Does Not Mean “Already in the Algebra”

A common mistake is to infer that every measurable set belongs to the original algebra because that algebra generates the sigma-algebra. Generation means that countable set operations produce \(\sigma(\mathcal{A})\); it does not say that each resulting set is already an element of \(\mathcal{A}\). In the dyadic example, \(E=(1/3,2/3)\) is not a finite union of dyadic intervals, so using \(\chi_E\) as if it were one of the algebra indicators is unjustified. The correction is to use \(B_2=[1/4,3/4)\), for which the exact symmetric-difference measure is \(1/6\), and then use the sequence \(B_n\) whose errors tend to zero. For indicators, the relevant relation is exact: \(\|\chi_E-\chi_A\|_p^p=\mu(E\triangle A)\). Thus approximation in measure of sets is precisely what converts into \(L^p\) approximation of their indicators.

The finite-measure assumption enters the set lemma when continuity from below is used to discard the tail of a countable union. It also ensures bounded simple functions lie in \(L^p\). The theorem therefore does not claim, without additional hypotheses or localization, that the same construction applies on an arbitrary infinite-measure space.