TITLE: Unifying two definitions of $L^\infty$
QUESTION [5 upvotes]: Let $X$ be a locally compact Hausdorff space and $\mu$ a Radon measure on $X$.

Definitions:
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.

REPLY [3 votes]: 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).