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Today I thought about this for the first time and I really cannot see what is going on. I think it is a very stupid question but I really cannot see it.

Consider the space $L^{\infty}(\mathbb{R})$ with the Lebesgue measure. According to this: The Duals of $l^\infty$ and $L^{\infty}$, an element in the dual of this space is a finite signed measure $v$ which is absolutely continuous with respect to the Lebesgue measure.

By Radon - Nikodym theorem we obtain: $dv = fd\mu$ and then the bounded total variation property is equivalent to $f \in L^1$.

Thus we may thus construct an isometry between $(L^{\infty})^*$ and $L^1$ in an obvious way to get reflexivity of $L^1$ which is absurd. So my question is: where is the mistake?

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Nick Kolliopoulos
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    This is not an answer to your question but rather a general comment : having a isomorphism between $X$ and $X^{**}$ does not suffice to have reflexivity. It must be the canonical embedding $x \mapsto \widehat{x}$ that is isomorphic. – M.G Jun 05 '15 at 19:49
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    You left out "finitely additive" from the characterization of the dual of $L^\infty$. Such measures that are not countably additive correspond to element of the dual that do not come from $L^1$. – GEdgar Jun 05 '15 at 19:52
  • Thank you for your answer. Now it is clear. About M.G 's comment: Yes I'm aware of this, but here the problem is somewhere else and provided that we could apply radon-nikodym we would have the canonical isometry. – Nick Kolliopoulos Jun 05 '15 at 20:11

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The dual space of $L^\infty$ is the space of "finitely addditive" signed measures and not the space of finite signed measures. The usual Radon-Nikodym theorem does not apply.

Steve652015
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