The dual space to the Banach space $L^1(\mu)$ for a sigma-finite measure $\mu$ is $L^\infty(\mu)$, given by the correspondence

$\phi \in L^\infty(\mu) \mapsto I_\phi$, where $I_\phi(f) = \int f \cdot \phi \,d\mu$ for $f \in L^1(\mu)$.

Now, we consider this for more general measures $\mu$. If $\mu$ is not semifinite, then there is a measurable subset $A$ with $\mu(A) = \infty$ but $\mu(B) = 0$ for every subset $B$ of $A$ with $\mu(B) < \infty$. Any $\phi$ supported on $A$ will then map to the zero functional, so the correspondence is not one-to-one.

If $\mu$ is semifinite, the correspondence is one-to-one, but is it onto?

Using the Radon-Nikodym Thorem, any member of $L^1(B)^*$ for $\mu(B) < \infty$ can be represented by an $L^\infty(B)$ function class $[\phi_B]$. Unlike in the $\sigma$-finite measure case, however, I do not see an easy way to form a "patchwork" $\phi$ with $\phi|_B = \phi_B$ a.e.$[\mu]$. I suspect this may have an easy answer, but it eludes me. Or perhaps there is a well-known counterexample?

Davide Giraudo
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Ben Passer
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2 Answers2


To answer the question in the title, there is the following general theorem:

The natural map $I \colon L^\infty(X) \to (L^1(X))^\ast$ is

  1. an isometric injection if and only if $(X,\Sigma,\mu)$ is semifinite.
  2. an isometric isomorphism if and only if $(X,\Sigma,\mu)$ is localizable.

You already covered point 1 in your question (the isometry property being straightforward). The second point is more delicate.

A measure space $(X,\Sigma,\mu)$ is called localizable if it is semifinite and in addition the following condition holds:

For every family $\mathcal{F} \subseteq \Sigma$ there is $H \in \Sigma$ such that

  1. $F \setminus H$ is a null set for all $F \in \mathcal{F}$.
  2. If $G \in \Sigma$ is such that $F \setminus G$ is a null set for all $F \in \mathcal{F}$ then $H \setminus G$ is a null set.

Loosely speaking, this property asserts that every family $\mathcal{F}$ of measurable sets has a smallest measurable envelope $H$ (up to null sets).

The definition of localizability implies via a slightly technical argument that one can glue together measurable functions. More precisely:

Let $\mathscr{F}$ be a family of functions such that each $f \in \mathscr{F}$ is defined and measurable on a measurable subset $D_f$ of the localizable measure space $(X,\Sigma,\mu)$. Suppose in addition that $f_1 = f_2$ a.e. on $D_{f_1} \cap D_{f_2}$ whenever $f_1,f_2 \in \mathscr{F}$. Then there is a measurable function $g \colon X \to \mathbb{R}$ whose restriction to $D_f$ satisfies $g|_{D_f} = f$ a.e. for all $f \in \mathscr{F}$.

Using this result one can patch together the Radon-Nikodym derivatives one obtains from restricting a continuous linear functional $\varphi \colon L^1(X) \to \mathbb{R}$ to the subspaces $L^1(F) \subseteq L^1(X)$ where $F$ runs through the subsets of finite measure of $X$.

The proof of the converse direction proceeds by a direct verification of the localizability property of $(X,\Sigma,\mu)$ the fact that $I$ is isometric implies semi-finiteness by point 1. of the theorem and the "envelope condition" uses surjectivity of $I$.

Details can be found in 243G on page 153 of this PDF. The gluing property is proved in 213N on page 28.

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    Thanks! I had not previously heard of localizability as a property of measure spaces, and that gluing result is exactly the kind of property needed. – Ben Passer May 29 '13 at 16:42

Let $X$ be an uncountable set, equipped with the $\sigma$-algebra $\mathcal{F}$ consisting of all countable sets and their complements, or any other $\sigma$-algebra that contains all singletons but is not $2^X$. Let $\mu$ be counting measure, which is semifinite. Note that any function in $L^1(\mu)$ is supported on a countable set. Let $E$ be a subset of $X$ which is not in $\mathcal{F}$, and for $f \in L^1(\mu)$ set $I(f) = \sum_{x \in E} f(x)$. Clearly this is a continuous linear functional on $L^1(\mu)$ (with norm 1). However, if $I(f) = \int fg\,d\mu$ it's easy to see we would have to have $g = 1_E$, but this is not in $L^\infty(\mu)$ because $E$ is not measurable.

But $\sigma$-finiteness is not necessary, either: repeat this example with $\mathcal{F} = 2^X$.

Nate Eldredge
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  • Thanks a bunch! I'm kicking myself for not thinking of this. I'd upvote it, but I don't have enough "reputation" yet. – Ben Passer May 29 '13 at 05:34