Let $X$ be a locally compact Hausdorff space and $\mu$ a Radon measure on $X$.


A subset $E\subseteq X$ is called locally Borel if $F \cap E$ is Borel for every Borel set $F\subseteq X$ with $\mu(F) < \infty.$ The locally Borel subset $E$ is called locally null if $\mu(E \cap F)= 0$ for every $F\subseteq X$ with $\mu(F) < \infty$. We say that a property of the points of $X$ holds locally almost everywhere (= a.e.) if the set of points where the property fails is locally null. A function $f: X \to \mathbb{C}$ is called locally measurable if $f^{-1}(A)$ is locally Borel for every Borel subset $A \subseteq \mathbb{C}$, i.e. $f$ is measurable w.r.t. the larger $\sigma$-algebra of locally Borel sets on the domain.

Construction $1$: [As e.g. defined in Folland's book "A course in abstract harmonic analysis", p50.]

Define $\mathscr{L}_1^\infty$ to be the set of locally measurable functions $f: X \to \mathbb{C}$ with the following property: there exists $\alpha \in (0, \infty)$ such that $$\{x \in X: \lvert f(x)\rvert > \alpha\}$$ is a locally null set. The vector space $\mathscr{L}_1^\infty$ has the seminorm $$\lVert f\rVert_\infty:= \inf\{\alpha \ge 0: \{x \in X: \lvert f(x)\rvert> \alpha\} \text{ is locally null}\}.$$

Let $\mathscr{N}_1$ be the subspace of locally measurable functions that are $0$ locally a.e.. We then define the quotient vector space $$L_1^\infty:= \mathscr{L}_1^\infty/\mathscr{N}_1$$ on which the seminorm $\lVert\cdot\rVert_\infty$ becomes a norm.

The standard argument proves that this is a Banach space.

Construction 2: [As e.g. defined in "Abstract harmonic analysis" by Hewitt and Ross, p151, or as e.g. defined in "Measure theory" by Cohn, section 3.3.]

Let $\mathscr{L}_2^\infty$ be the set of bounded measurable functions (with the usual supremum norm) and let $\mathscr{N}_2$ be the set of measurable bounded functions that are $0$ locally almost everywhere. Then we define the quotient vector space $$L_2^\infty:= \mathscr{L}_2^\infty/\mathscr{N}_2$$ (together with the quotient norm).

Question: Do these two constructions give the same Banach space, i.e. is there a canonical isometric isomorphism $$\Phi: L_2^\infty \to L_1^\infty?$$

There is an obvious well-defined linear isometry $$\Phi: L_2^\infty \to L_1^\infty: f + \mathscr{N}_2 \mapsto f + \mathscr{N}_1.$$

However, is it surjective? I.e. can we choose a representative of an element of $L_1^\infty$ to be measurable, instead of only locally measurable?

There is a possibility that this doesn't need any topological assumptions and that the above is true for general measure spaces and measures.

Context question: I'm reading the book "A course in abstract harmonic analysis" by Folland. On p50, it is claimed that the canonical map $L^\infty_1 \to (L^1)^*$ is an isometric isomorphism, but for the surjectivity Folland refers to Hewitt and Ross, where the other definition is used, so I'm trying to unify them. In this context, it would be desirable that the map $\Phi$ above is surjective.

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    $\begingroup$ What you are missing is a key measurability criterion for Radon measures (Theorem 11.42 in Hewitt & Ross) , which basically says that locally measurable is the same as measurable for these measures (and slightly more). More conceptually, locally measurable and measurable coincide for the so-called "decomposable" measures (see e.g. Hewitt & Stromberg, Real and Abstract Analysis, 19.25--19.30), which includes Radon measures and $\sigma$-finite ones, the core is basically Theorem 11.39 in Hewitt & Ross, which in turn relies on the measurability of locally-null sets for such measures. $\endgroup$
    – Hua Wang
    Feb 1, 2022 at 23:58
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    $\begingroup$ A direct "unification" you are looking for is exactly the theory of decomposable measures (see Hewitt & Stromberg for examples), with one of key results being the Radon-Nikodym theorem without the restriction on $\sigma$-finiteness. It is clear that one needs to look at locally null sets instead of null sets in these contexts, which are the "correct" notion in view of the $L^1$-$L^\infty$ duality. $\endgroup$
    – Hua Wang
    Feb 2, 2022 at 0:14
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    $\begingroup$ You might also want to look for localizable measures, but seeing your interest in operator algebras, perhaps it is more interesting to point out these $L^\infty$ spaces are exactly abelian von Neumann algebras up to isomorphism, no more, no less. Seeing that von Neumann algebras are the noncommutative analogue of $L^\infty$ with the predual as the analogue of $L^1$, and it is known that there are various Radon-Nikodym type theorems for von Neumann algebras, how about that for the unification you are looking for ? :) $\endgroup$
    – Hua Wang
    Feb 2, 2022 at 0:17
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    $\begingroup$ @HuaWang I think there might be a subtlety here: Hewitt and Ross define measurability with respect to the Carathéodory outer measure and locally null sets are defined with respect to such measurable sets (this is for example the case in the theorem where they show that locally null sets are measurable), whereas for me measurable set is a synonym for Borel subset. Thus, I don't see how the results of Hewitt and Ross imply that every locally measurable function (in the sense of Folland) is automatically measurable, i.e. that the preinverse of a Borelset of $\mathbb{C}$ is a Borel in $X$. $\endgroup$
    – Andromeda
    Feb 2, 2022 at 8:09
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    $\begingroup$ If I understand correctly, you want to know if locally Borel functions can be modified on a locally null set to get a Borel function? If so, I don't have a good answer, but I strongly suspect the answer is negative. On the flip side, there are many good reasons to use the measurable sets coming from Caratheodory constructions instead of merely Borel ones, e.g. in order to deal with many more subtleties for the measurability when defining convolution in general locally compact groups. I am not very familiar with Folland's book, but in this regard, I think Hewitt & Ross is invaluable. $\endgroup$
    – Hua Wang
    Feb 3, 2022 at 2:23

1 Answer 1


I think one does need to be careful (c.f. the comments). There are Radon measures for which locally null and null are different. The canonical example is to take $X=\mathbb R^2$ with the topology that $U$ is open if and only if $U_x = \{ y : (x,y)\in U \}$ is open, for each $x$. Then define a Radon measure by the functional $$ C_{00}(X)\rightarrow\mathbb C; \quad f\mapsto \sum_x \int f(x,y) \ dy $$ where we use Lebesgue measure. Then $\{(x,0)\}$ is locally null, but not null. See exercises 3.3.6 and 7.2.4 in Cohn's book.

I think the resolution to the problem asked is to realise that locally compact groups are not entirely arbitrary locally compact spaces. The "trick" which Folland uses in his Harmonic Analysis book is to realise that for any $G$ we can always find an open and closed, $\sigma$-compact subgroup $H$. Then $G/H$ has the discrete topology, and being $\sigma$-compact, everything works fine on $H$. Then you can reconstruct the results you want by working piecewise on each coset of $H$. (See page 51 of Folland).

This resolution works quite adequately with "construction 2". Indeed, see Cohn, Theorem 9.4.8 which shows exactly that $L^\infty$ is the dual of $L^1$ for any regular Borel measure on $G$. However, beware of the subtle point in the proof, that you cannot exactly just work on each coset of $H$: to get a Borel function, some trick using Lusin's theorem (to approximate by continuous functions) is needed.

(An alternative approach to "fix" the duality issue is to work with a larger $\sigma$-algebra than the Borel sets; compare Cohn Section 7.5. If I understand things right, this is the same notion of "measurability" which Hewitt+Ross uses. Here, Exercise 7.5.5 in Cohn is interesting: $L^p$ for this $\sigma$-algebra, or $L^p$ for Borel sets, are isometrically isomorphic, so long as $p$ is finite.)

In conclusion, the two constructions asked about are in general different. But in this special case (locally compact groups) they agree.

A meta-question is: which construction to use? I guess I don't know (and it doesn't matter, if you believe this answer!) From my experience, the literature in abstract harmonic analysis almost always uses construction 1, and the "locally" language. Furthermore, in most cases, it doesn't really matter exactly which $\sigma$-algebra one works with.

More pragmatically, the vast majority of examples are $\sigma$-compact, and so there is relatively little to be lost by just assuming that $G$ is $\sigma$-compact. Then it's clear that both constructions agree. (Related to the often seen "In this paper we assume that all Hilbert spaces are separable" caveats).

  • $\begingroup$ Thanks again for the extremely useful answer. Would you mind offering me some advice? Will I run into trouble anywhere while reading Folland if I just use the definition of $L^\infty$ that Cohn uses (I think your answer suggests that it doesn't matter since we work in the context of locally groups anyway)? My goal is not to become an expert in harmonic analysis or perform research in it, but to become proficient enough in it and to build enough intuition to understand locally compact quantum groups (which is why I am also reading the Takesaki's, but it is a slow process...) $\endgroup$
    – Andromeda
    Feb 2, 2022 at 21:41
  • $\begingroup$ @Andromeda: See my update. I guess, if I was forced to give a recommendation in your case, I think the "assume everything is $\sigma$-compact" seems about right. $\endgroup$ Feb 2, 2022 at 21:53
  • $\begingroup$ Thanks! But now I am wondering: is Folland's claim true that for construction 1, we get the desired duality $(L_1)^* = L^\infty$ (for a Radon measure on an arbitrary locally compact space)? He refers to Hewitt-Ross, but I'm not sure the results there are applicable. $\endgroup$
    – Andromeda
    Feb 2, 2022 at 22:15
  • $\begingroup$ Also, in your answer you claim that both constructions are different: but if I have to believe Folland, it will turn out that both spaces are canonically isomorphic to $(L^1)^*$, so that they do coincide. Could you clarify what you mean? $\endgroup$
    – Andromeda
    Feb 2, 2022 at 22:19
  • $\begingroup$ Basically what I want to know: Is Folland correct that using his construction we get the desired $(L^1)^* = L^\infty$ duality, because I don't see how the results that he refers to apply to his situation. Let me know if I should ask another question for this. $\endgroup$
    – Andromeda
    Feb 2, 2022 at 22:21

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