There are two definitions of $L^p(S, \Sigma,\mu)$ in the literature. (Here $S$ is a set, $\Sigma$ is a $\sigma$-algebra of subsets of $S$ and $\mu$ is a positive measure.) The two definitions are the same in some (most?) cases (such as if $\mu$ is $\sigma$-finite) but are not always the same. One seems to be standard in most graduate textbooks and the other is from Dunford and Schwartz's Linear Operators, Part I: General Theory (abbreviated DSI). Can anyone recount the history and present day usage of the two definitions? For easier reference the two definitions are recalled here.

The first definition is taken from: Michael E Taylor, Measure Theory and Integration, Graduate Studies in Mathematics, volume 76, Amer Math Soc. 2006, page 43. It is:

Definition 1. $L^p(S, \Sigma,\mu)$ is the set of all (equivalence classes) of $\Sigma$-measurable functions $f$ so that $$ \|f\|_{p}^p:= \int_S \lvert f\rvert^p d \mu < \infty \text{ if $1\le p <\infty$} $$ or the essential supremum of $\lvert f\rvert$ is finite if $p=\infty$. (The equivalence classes are for the relation $f \equiv g$ if $f=g$ $\mu$-almost everywhere.)

Here $\Sigma$-measurable is (according to DSI page 240) defined as: A function $f:S \to \mathbb{C}$ is $\Sigma$-measurable if $f^{-1}(B) \in \Sigma$ for all Borel subsets $B$ of the complex plane $\mathbb{C}$.

The second definition is from DSI, page 119:

Definition 2. It is exactly the same as Definition 1 except that “$\Sigma$--measurable” is replaced by “$\mu$-measurable”.

Let us recall the definition of $\mu$-measurable from DSI. First they introduce (DSI page 101) the topology associated with convergence in $\mu$ measure.

A function $f:S \to \mathbb{C}$ is said (page 106) to be totally $\mu$-measurable if it belongs to the closure of the set of all $\Sigma$-measurable simple (complex valued) functions.

A function $f:S \to \mathbb{C}$ is said to be $\mu$-measurable if $\chi_E f$ is totally $\mu$-measurable whenever $E \in \Sigma$ has finite $\mu$ measure. (Here $\chi_E$ denotes the characteristic function of $E$.)

In general $\Sigma$-measurable implies $\mu$-measurable but not conversely. Hence Definition 1 defines a smaller set $L^p$ than Definition 2. Is it strictly smaller? Not if $\mu$ is $\sigma$-finite. When then is there a difference?

Following DSI, page 296 we define $\Sigma_1$ as follows. We suppose (without loss of generality) that $(S,\Sigma,\mu)$ is complete (so that $\Sigma$ contains all subsets of sets in $\Sigma$ of measure zero). Define $\Sigma_1$ to be the family of all subsets $E$ of $S$ so that $A \cap E \in \Sigma$ whenever $A \in \Sigma$ has finite measure. Certainly $\Sigma_1 \supseteq \Sigma$ and if the containment is proper and if $E \in \Sigma_1\setminus \Sigma$ then $\chi_E$ belongs to $L^\infty(S,\Sigma, \mu)$ as defined in DSI (second definition), but is not in $L^\infty(S,\Sigma, \mu)$ as defined in the first definition. Constructing an example where $\Sigma_1 \neq \Sigma$ seems to be straightforward.

Surely this is known and noted somewhere in the literature. Can anyone direct me?

  • $\begingroup$ When you quote the definition of $L^p$ from Taylor, you still refer to DSI for the definition of the subterm "$\Sigma$-measurable". Is their definition the same as Taylor's? $\endgroup$
    – LSpice
    Commented Jun 7, 2020 at 18:55
  • $\begingroup$ Taylor calls $\Sigma$--measurable, simply measurable like many authors. $\endgroup$ Commented Jun 7, 2020 at 20:17
  • $\begingroup$ Re: Whatever Taylor calls the notion, is the definition the same as the one you cite from DSI? (I don't have access to either book right now, so I can't check.) $\endgroup$
    – LSpice
    Commented Jun 7, 2020 at 20:19
  • $\begingroup$ Yes. (I had to choose one author's terminology) Denis $\endgroup$ Commented Jun 8, 2020 at 18:58

1 Answer 1


The property Σ=Σ_1 amounts to (X,Ε,μ) being locally determined.

A measure space (X,Σ,μ) is locally determined if μ is semifinite and A∈Σ if and only if A∩F∈Σ for all F∈Σ such that μ(F) is finite. See Fremlin, Measure Theory, Definition 211H.

Almost all measurable spaces that arise in practice (e.g., from Radon measures) are strictly localizable and therefore locally determined by Theorem 211L(d) in op. cit.

A measure space (X,Σ,μ) is strictly localizable if it can be partitioned into a disjoint family of measurable subsets of finite measure such that (X,Σ,μ) is the disjoint union (coproduct) of the resulting measure spaces.

The closely related (but slightly weaker) property of being localizable amounts to saying that μ is semifinite and the Boolean algebra Σ/N is complete, where N is the σ-ideal of sets of μ-measure 0.

Localizability is equivalent to the following results in measure theory: the Hahn–Jordan decomposition theorem for signed measures that are absolutely continuous with respect to μ, the Riesz representation theorem, the Radon–Nikodym theorem, and the fact that bounded measurable functions form a von Neumann algebra. Thus, once you move outside of the domain of localizable measure spaces, all of measure theory falls apart and you have much bigger problems than L^p-spaces.

As for the remaining case of localizable measure spaces that are not strictly localizable, the known examples are quite pathological, but in any case, they can be improved to a complete locally determined measure space (see Proposition 213D in op. cit.) and by Proposition 213H(d) there is no practical difference between the two spaces.

In fact, for any localizable measure space we can find a strictly localizable measure space with an isomorphic Boolean algebra Σ/N, which for all practical purposes can be used instead of the original space.

  • $\begingroup$ The Hahn-Jordan decomposition theorem is stated on Wikipedia without any preconditions $\endgroup$
    – wlad
    Commented Jun 8, 2020 at 6:47
  • 1
    $\begingroup$ @ogogmad: It's a different Hahn–Jordan decomposition theorem (one measure instead of two). I added a clarification. $\endgroup$ Commented Jun 8, 2020 at 15:32
  • $\begingroup$ Sounds like the Lebesgue decomposition theorem, no? $\endgroup$
    – wlad
    Commented Jun 8, 2020 at 17:32
  • 1
    $\begingroup$ @ogogmad: First of all, the Lebesgue decomposition theorem does not require σ-finiteness or any other conditions (see the nLab). Secondly, as I already explained, it has nothing to do with the Hahn–Jordan decomposition, which decomposes a signed measure into its positive and negative components. $\endgroup$ Commented Jun 8, 2020 at 17:58
  • 1
    $\begingroup$ @ogogmad: This is Theorem 2 in Kelley's paper that is linked in the answer. $\endgroup$ Commented Jun 8, 2020 at 18:17

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