Nonseparability from Uncountably Many Disjoint Sets

Disjoint Sets as Separated \(L^p\) Vectors

A metric space is separable if it has a countable dense subset. For a finite-p \(L^p\) space, one way to prove nonseparability is to find uncountably many elements that stay a fixed positive distance apart. Disjoint measurable sets provide a direct construction: their normalized indicators have norm one, and disjointness makes the distance between any two of them exactly \(2^{1/p}\).

Let \((X,\mathcal{M},\mu)\) be a measure space, with no assumption that \(\mu\) is finite, sigma-finite, or complete. Fix \(p\) with \(1\le p<\infty\). We use real- or complex-valued measurable functions. The space \(L^p(\mu)\) consists of equivalence classes modulo equality \(\mu\)-a.e. of functions \(f\) for which \(\int_X |f|^p\,d\mu<\infty\), with norm \(\|f\|_p=(\int_X |f|^p\,d\mu)^{1/p}\). Its metric is \(d(f,g)=\|f-g\|_p\). The argument below concerns this metric, not pointwise separation of chosen representatives.

Theorem (Disjoint-set criterion for nonseparability): Suppose \((X,\mathcal{M},\mu)\) contains an uncountable family \((A_\alpha)_{\alpha\in I}\) of pairwise disjoint measurable sets such that \(0<\mu(A_\alpha)<\infty\) for every \(\alpha\in I\). Then \(L^p(\mu)\) is nonseparable for every \(1\le p<\infty\).

Proof. For each \(\alpha\), define \(u_\alpha=\mu(A_\alpha)^{-1/p}\chi_{A_\alpha}\). The assumptions \(0<\mu(A_\alpha)<\infty\) make this coefficient finite and positive. Moreover,

\|u_\alpha\|_p^p =\int_X \mu(A_\alpha)^{-1}\chi_{A_\alpha}\,d\mu =\mu(A_\alpha)^{-1}\mu(A_\alpha) =1.

For distinct \(\alpha,\beta\), the sets are disjoint, so \(|u_\alpha-u_\beta|^p\) equals \(\mu(A_\alpha)^{-1}\) on \(A_\alpha\), equals \(\mu(A_\beta)^{-1}\) on \(A_\beta\), and is zero elsewhere. Consequently,

\|u_\alpha-u_\beta\|_p^p =\mu(A_\alpha)^{-1}\mu(A_\alpha) +\mu(A_\beta)^{-1}\mu(A_\beta) =2, \qquad \|u_\alpha-u_\beta\|_p=2^{1/p}.

In particular these \(L^p\) classes are distinct, and their pairwise distances are the same positive number \(\delta=2^{1/p}\). Suppose, to obtain a contradiction, that \(D\) is a countable dense subset of \(L^p(\mu)\). For each \(\alpha\), density gives a \(d_\alpha\in D\) with \(\|d_\alpha-u_\alpha\|_p<\delta/3\). If \(\alpha\ne\beta\) but \(d_\alpha=d_\beta\), the triangle inequality would give \(\|u_\alpha-u_\beta\|_p<2\delta/3\), contradicting \(\|u_\alpha-u_\beta\|_p=\delta\). Thus \(\alpha\mapsto d_\alpha\) is an injection from the uncountable set \(I\) into the countable set \(D\), which is impossible. Hence \(L^p(\mu)\) is nonseparable. \(\square\)

Working with the Criterion

The finite, positive measure condition has two jobs. It ensures that \(\chi_{A_\alpha}\) belongs to \(L^p(\mu)\), and it allows the indicator to be normalized to norm one. Pairwise disjointness then gives the same distance for every distinct pair. The argument does not require the whole space to have finite measure, nor does it require the family to cover \(X\).

Worked Example: Normalizing Disjoint Sets of Different Sizes

Take \(X=[0,4)\) with Lebesgue measure \(m\), and let \(A_1=[0,1)\), \(A_2=[1,3)\), and \(A_3=[3,4)\). These sets are measurable and pairwise disjoint, with measures \(1,2,1\), respectively. For \(p=2\), their normalized indicators are \(u_1=\chi_{A_1}\), \(u_2=2^{-1/2}\chi_{A_2}\), and \(u_3=\chi_{A_3}\). Directly,

\|u_1\|_2^2=1^2\cdot1=1,\qquad \|u_2\|_2^2=(2^{-1/2})^2\cdot2=\frac12\cdot2=1,\qquad \|u_3\|_2^2=1^2\cdot1=1.

For instance, \(u_1\) and \(u_2\) have disjoint supports, so

\|u_1-u_2\|_2^2 =\int_{A_1}1\,dm+\int_{A_2}\left|2^{-1/2}\right|^2\,dm =1+\frac12\cdot2=2, \qquad \|u_1-u_2\|_2=\sqrt2.

The same computation for \(u_1,u_3\) gives \(1+1=2\), and for \(u_2,u_3\) gives \(\frac12\cdot2+1=2\). Thus all three pairwise distances are \(\sqrt2=2^{1/2}\), as the theorem predicts. This finite family illustrates the calculation; nonseparability requires an uncountable family.

Uncountable Counting Measure

For an arbitrary set \(I\), counting measure on \((I,\mathcal{P}(I))\) assigns to a set its number of elements when finite, and \(+\infty\) when it is infinite. Here every singleton has measure one. If \(I\) is uncountable, the singletons form an uncountable pairwise disjoint family of measurable sets, each of finite positive measure. The theorem therefore applies.

Worked Example: Counting Measure on an Uncountable Set

Let \(I=\mathbb{R}\), equipped with counting measure \(\#\) on its power-set sigma-algebra, and fix \(1\le p<\infty\). For every \(i\in I\), set \(A_i=\{i\}\). Then \(\#(A_i)=1\), and the normalized indicator is \(u_i=\chi_{\{i\}}\). Its norm and the distance between two distinct such vectors are

\|u_i\|_p^p=\sum_{j\in I}|\chi_{\{i\}}(j)|^p=1, \qquad \|u_i-u_k\|_p^p =\sum_{j\in I}|\chi_{\{i\}}(j)-\chi_{\{k\}}(j)|^p =1+1=2 \quad (i\ne k).

Thus \(\|u_i-u_k\|_p=2^{1/p}\) for all distinct \(i,k\). Since \(\mathbb{R}\) is uncountable, the theorem proves that \(L^p(\#)\) is nonseparable. In fact, the definition of the integral for counting measure gives \(\int_I |f|^p\,d\#=\sum_{i\in I}|f(i)|^p\), where the sum of nonnegative terms is the supremum of its finite partial sums. The theorem needs only the singleton vectors; this description also identifies the familiar sequence-space form of this \(L^p\) space when the index set is uncountable.

Check: substituting \(\#(A_i)=1\) into the normalization gives coefficient \(1^{-1/p}=1\); substituting \(\#(A_i)=\#(A_k)=1\) into the distance formula gives \(1+1=2\). These agree with the direct sums above.

The same reasoning applies when the disjoint sets are not singletons. For example, in a product space, each set in an uncountable collection of fibers may have measure one. This makes clear that the criterion concerns the measure and geometry of the sets, not the names or cardinalities of their individual points.

Worked Example: Uncountably Many Unit Fibers

Let \(I\) be any uncountable set with counting measure, let \(Y=[0,1]\) with Lebesgue measure, and use the product measure on \(X=I\times[0,1]\) with the product sigma-algebra. For each \(i\in I\), let \(A_i=\{i\}\times[0,1]\). These measurable fibers are pairwise disjoint. The product-measure rectangle formula gives \(\mu(A_i)=\#(\{i\})m([0,1])=1\cdot1=1\). Hence \(u_i=\chi_{A_i}\) has \(L^3(\mu)\) norm satisfying \(\|u_i\|_3^3=1^3\cdot1=1\).

For distinct \(i,k\), the difference is \(1\) on \(A_i\), \(-1\) on \(A_k\), and zero elsewhere. Therefore,

\|u_i-u_k\|_3^3 =\int_{A_i}1\,d\mu+\int_{A_k}1\,d\mu =1+1=2, \qquad \|u_i-u_k\|_3=2^{1/3}.

There are uncountably many such vectors, so \(L^3(\mu)\) is nonseparable by the theorem. Substitution into the general formula gives the same check: the two measures are both \(1\), and \(1^{-1}\cdot1+1^{-1}\cdot1=2\).

Why Sigma-Finiteness Changes the Picture

There is no contradiction with separability results for sigma-finite spaces. In fact, sigma-finiteness rules out an uncountable family of pairwise disjoint measurable sets of positive measure. Recall that \((X,\mathcal{M},\mu)\) is sigma-finite if \(X=\bigcup_{k=1}^{\infty}X_k\) for measurable sets \(X_k\) with \(\mu(X_k)<\infty\).

Proposition: In a sigma-finite measure space \((X,\mathcal{M},\mu)\), every pairwise disjoint family of measurable sets of positive measure is at most countable.

Proof. Let \((A_\alpha)_{\alpha\in I}\) be pairwise disjoint, with \(\mu(A_\alpha)>0\) for every \(\alpha\), and choose a measurable cover \(X=\bigcup_{k\ge1}X_k\) with \(\mu(X_k)<\infty\). For each \(\alpha\), at least one intersection \(A_\alpha\cap X_k\) has positive measure; otherwise countable subadditivity would give \(\mu(A_\alpha)=\mu(\bigcup_k(A_\alpha\cap X_k))=0\). For such a \(k\), there is an integer \(n\ge1\) with \(\mu(A_\alpha\cap X_k)>1/n\), since a positive number exceeds \(1/n\) for some \(n\). Thus every \(\alpha\) belongs to at least one class indexed by \((k,n)\), where that class consists of indices satisfying \(\mu(A_\alpha\cap X_k)>1/n\).

Fix \(k,n\). Any finite collection of \(r\) distinct sets from this class has pairwise disjoint intersections with \(X_k\). Finite additivity and monotonicity give \(r/n<\sum_{\alpha}\mu(A_\alpha\cap X_k)\le\mu(X_k)\), so \(r<n\mu(X_k)\). In particular this class cannot be infinite: if it were infinite, it would contain finite subcollections of arbitrarily large size, contradicting this fixed finite bound. Each class is therefore finite. There are countably many pairs \((k,n)\), so their union is countable, and it contains every index \(\alpha\). Hence \(I\) is at most countable. \(\square\)

The proposition also shows why the finite positive-measure hypothesis in the nonseparability criterion is natural in sigma-finite settings: an uncountable disjoint family of positive-measure sets cannot occur there. The theorem is a sufficient criterion, not a characterization of all nonseparable \(L^p\) spaces.

A Pitfall: Positive Measure Is Not Enough

If disjoint sets have infinite measure, their indicators need not belong to \(L^p\), and there may be no finite positive-measure subsets to normalize. Consequently, an uncountable family of disjoint positive-measure sets by itself does not imply nonseparability. The finiteness condition is not a technical detail that can simply be omitted.

Worked Example: Infinite-Measure Singletons Do Not Suffice

Let \(X\) be an uncountable set, let \(\mathcal{M}=\mathcal{P}(X)\), and define \(\mu(\varnothing)=0\) and \(\mu(E)=+\infty\) for every nonempty \(E\). This is a measure: for any pairwise disjoint sequence, if all sets are empty then both the measure of the union and the sum of their measures are zero; if any set is nonempty, the union is nonempty and both are \(+\infty\). The singletons are an uncountable disjoint family of positive measure, but each has measure \(+\infty\).

For any measurable \(f\) with \(f(x)\ne0\) at some \(x\), set \(c=|f(x)|^p>0\). The function \(|f|^p\) is at least \(c\chi_{\{x\}}\), so monotonicity of the nonnegative integral gives \(\int_X|f|^p\,d\mu\ge c\mu(\{x\})=+\infty\). Thus every function in \(L^p(\mu)\) is zero everywhere, and \(L^p(\mu)\) consists only of the zero class. It is separable. The failed calculation would be to set \(\mu(\{x\})^{-1/p}\chi_{\{x\}}=0\) and call it a unit vector: the coefficient is not a finite positive real number, and the singleton indicator is not in \(L^p(\mu)\). The theorem does not apply.

When using the criterion, check the hypotheses before normalizing: the sets must be measurable, pairwise disjoint, and each must have measure strictly greater than zero and finite. Once those conditions hold for an uncountable family, the normalized indicators supply a uniform separation in the \(L^p\) metric, and the countable-dense-set contradiction completes the argument.

Check Your Understanding

For each question, identify which part of the separation argument or its hypotheses is involved.

If \(0<\mu(A)<\infty\), compute the \(L^p\) norm of \(\mu(A)^{-1/p}\chi_A\).
For two disjoint sets \(A,B\) of finite positive measure, show that the distance between their normalized indicators is \(2^{1/p}\).
Why must the balls of radius \(2^{1/p}/3\) around the normalized indicators be disjoint?
Apply the criterion to counting measure on an uncountable set, and state the distance between distinct singleton indicators.
Explain why an uncountable family of disjoint positive-measure sets cannot occur in a sigma-finite measure space.
Which hypothesis fails for the singleton family in the measure space where every nonempty set has infinite measure?