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If $X$ is an arbitrary measure space, I already know with proof that $L^p(X)$ and $L^q(X)$ are mutually duals as Banach spaces, when $1<p$ and $p$, $q$ are dual indices. I also know a different proof that only works when $X$ is sigma finite, but then it establishes also that the dual of $L^1(X)$ is $L^\infty(X)$. Is this still true when $X$ is not sigma finite?

Please let me know how to prove your answer.

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How do you define $L^\infty$ if $X$ is not $\sigma$-finite? Bounded functions modulo null sets or modulo locally null sets? In the former case the duality holds if and only if the measure space is localizable. Consider the countable-cocountable $\sigma$-algebra on an uncountable set with the measure $\mu(S) = \infty$ if $S$ has countable complement and $\mu(S) = 0$ if $S$ is countable. $L^1(\mu) = 0$ while $L^\infty(\mu) \neq 0$. –  t.b. May 27 '12 at 23:27
    
I was not even aware that sometimes people make a special case for $L^\infty$. –  Jeff May 28 '12 at 1:56
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No, this is not true if $X$ is not $\sigma-$finite. Let $X = [0,1]$, $S = \{A \subset X| A$ or $A^c$ is countable$\}$ and $\mu$ denote the counting measure on $X$. (Note that $\mu$ is not $\sigma-$ finite!) Then one can check that a function $f$ is in $L^1(\mu)$ iff $f$ vanishes on all but countably many points $c_1,c_2,\dots$ and $\sum_{i=1}^{\infty} f(c_i) < \infty$. Define $T: L^1(\mu) \rightarrow \mathbb{R}$ by $T(f) = \sum_{x \in [0,1]} xf(x)$. Then we have $|T(f)| \leq ||f||_1$ but there is no $g \in L^{\infty} [0,1]$ such that $Tf = \int fg d \mu$. (Indeed, if such $g$ existed, then $g(x) = x$ for all x, but this function is not measurable!) The statement $L^1(X)^* = L^{\infty}(X)$ can be extended to spaces that are more general than $\sigma-$ finite spaces, the so-called decomposable measure spaces, but I don't know much about them.

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As I mentioned in a comment, localizability is a necessary and sufficient condition for the duality to hold, see e.g. Fremlin, measure theory volume II, Theorem 242 G b), page 153. Decomposability (by that people usually mean what Fremlin calls strict localizability) is slightly stronger. –  t.b. May 28 '12 at 8:47
    
@t.b. Thank you for your comment and for the link. I appreciate it. –  algebra_fan May 28 '12 at 18:11
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