Constructing Measures From Positive Lp Functionals

From a Functional to a Measure

A positive bounded linear functional on \(L^p\) assigns a nonnegative number to every nonnegative \(L^p\) function. In particular, it can be tested on indicator functions—but only when those indicators belong to \(L^p\). On a set of finite measure, they do. The resulting values behave like a measure, and sigma-finiteness lets us assemble these local measures across the entire space.

Throughout, let \((X,\mathcal{M},\mu)\) be a sigma-finite measure space, and fix \(1\le p<\infty\). Functions are real-valued, with \(L^p(\mu)\) understood as equivalence classes modulo equality \(\mu\)-a.e. A bounded linear functional \(\Lambda:L^p(\mu)\to\mathbb{R}\) has norm \(\|\Lambda\|=\sup_{\|f\|_p\le1}|\Lambda(f)|<\infty\). It is positive if \(\Lambda(f)\ge0\) whenever \(f\ge0\) \(\mu\)-a.e. We use the norm bound \(|\Lambda(f)|\le\|\Lambda\|\|f\|_p\).

The first issue is countable additivity. If \(E_1,E_2,\ldots\) are disjoint and their union \(E\) has finite measure, then \(\chi_{\cup_{k=1}^n E_k}\) converges to \(\chi_E\) in \(L^p\): the \(p\)-th power of the norm of the difference is \(\mu(E\setminus\cup_{k=1}^nE_k)\), which tends to zero by continuity from below of \(\mu\). Boundedness of \(\Lambda\) therefore implies convergence of the functional values. This proves countable additivity whenever the union remains in the finite-measure domain. To obtain a measure on all of \(X\), we apply this argument separately on finite-measure pieces.

Local Measures and Their Assembly

Proof. Since \(\mu\) is sigma-finite, choose measurable sets \(F_j\) of finite \(\mu\)-measure whose union is \(X\). Set \(X_1=F_1\), and for \(j\ge2\) set \(X_j=F_j\setminus\cup_{i<j}F_i\). These sets are pairwise disjoint, have finite measure, and their union is \(X\); empty pieces are permitted.

For each \(j\), define \(\nu_j(A)=\Lambda(\chi_{A\cap X_j})\) for every \(A\in\mathcal{M}\). This is finite and nonnegative, since \[ 0\le\nu_j(A)\le\|\Lambda\|\mu(A\cap X_j)^{1/p} \le\|\Lambda\|\mu(X_j)^{1/p}<\infty. \] Also, \(\nu_j(\varnothing)=0\).

To prove countable additivity, let \(A_1,A_2,\ldots\) be pairwise disjoint measurable sets, and write \(A=\cup_{k=1}^{\infty}A_k\). For each \(n\), the indicator of \((\cup_{k=1}^nA_k)\cap X_j\) converges in \(L^p(\mu)\) to \(\chi_{A\cap X_j}\). Indeed, the sets \((\cup_{k=1}^n A_k)\cap X_j\) increase to \(A\cap X_j\), which has finite measure, and the \(p\)-th power of the norm of the indicator difference is the measure of the remaining set. Continuity from below makes that measure tend to zero. Applying the continuous functional \(\Lambda\), and using finite additivity on the disjoint indicators, gives \[ \nu_j(A)=\lim_{n\to\infty}\nu_j\!\left(\bigcup_{k=1}^n A_k\right) =\lim_{n\to\infty}\sum_{k=1}^n\nu_j(A_k) =\sum_{k=1}^{\infty}\nu_j(A_k). \] Thus each \(\nu_j\) is a finite measure on \((X,\mathcal{M})\).

Define \(\nu(A)=\sum_{j=1}^{\infty}\nu_j(A)\), allowing the value \(+\infty\). For disjoint \(A_k\), countable additivity of each \(\nu_j\) gives \(\nu(\cup_k A_k)=\sum_j\sum_k\nu_j(A_k)\). The two orders of summation of a nonnegative array agree: both equal the supremum of its finite rectangular partial sums. Hence this is \(\sum_k\sum_j\nu_j(A_k)=\sum_k\nu(A_k)\). Also \(\nu(\varnothing)=0\), so \(\nu\) is a measure. Since \(\nu(X_j)=\nu_j(X_j)<\infty\) and the \(X_j\) cover \(X\), \(\nu\) is sigma-finite.

If \(\mu(A)=0\), then \(\mu(A\cap X_j)=0\), so \(\chi_{A\cap X_j}\) is the zero element of \(L^p(\mu)\). Consequently \(\nu_j(A)=0\) for every \(j\), and \(\nu(A)=0\). This proves \(\nu\ll\mu\).

Finally, let \(E\in\mathcal{M}\) have finite \(\mu\)-measure. The sets \(E\cap\cup_{j=1}^nX_j\) increase to \(E\), and their indicators converge to \(\chi_E\) in \(L^p(\mu)\), since the measure of the omitted part tends to zero. Continuity of \(\Lambda\) therefore gives \[ \Lambda(\chi_E)=\lim_{n\to\infty}\Lambda\!\left(\chi_{E\cap\cup_{j=1}^nX_j}\right) =\lim_{n\to\infty}\sum_{j=1}^n\nu_j(E)=\nu(E). \] The stated bound now follows from \(|\Lambda(\chi_E)|\le\|\Lambda\|\|\chi_E\|_p=\|\Lambda\|\mu(E)^{1/p}\), together with positivity. \(\square\)

Worked Constructions

Why the Finite-Measure Restriction Matters

The assignment \(E\mapsto\Lambda(\chi_E)\) is initially defined only when \(\mu(E)<\infty\). If \(\mu(E)=\infty\), then \(\|\chi_E\|_p=\infty\), so \(\chi_E\) does not represent an element of \(L^p(\mu)\) and \(\Lambda(\chi_E)\) has no meaning. For instance, on \(\mathbb{R}\) with Lebesgue measure and \(p=2\), the constant function \(\chi_{\mathbb{R}}=1\) is not in \(L^2(m)\), because its squared integral is \(m(\mathbb{R})=\infty\).

A related error is to infer convergence of functional values merely from pointwise convergence of indicators. For disjoint sets with an infinite-measure union, the partial-union indicators may converge pointwise, but that alone does not imply convergence in \(L^p\), which is what continuity of \(\Lambda\) requires. The proof instead works on each finite-measure cell, where the omitted measure tends to zero and hence the indicator convergence is in \(L^p\). Sigma-finiteness then provides a countable assembly of those valid local constructions.

The resulting measure records exactly how the positive functional acts on indicators of finite-measure sets. Absolute continuity says that null sets for the original measure remain null for the constructed measure. These are the key structural facts needed before a density can be sought to represent the functional by integration.