# When are $L^1(X,{\mathcal X},\mu)$ and $L^\infty(X,{\mathcal X},\mu)$ reflexive?

Terence Tao's notes on "Duality and the Hahn Banach theorem want me to show:

Exercise 17: Show that any closed subspace of a reflexive space is again reflexive. Also show that a Banach space $X$ is reflexive if and only if its dual is reflexive. Conclude that if $(X,{\mathcal X}, \mu)$ is a measure space which contains a countably infinite sequence of disjoint sets of positive measure, then $L^1(X,{\mathcal X},\mu)$ and $L^\infty(X,{\mathcal X},\mu)$ are not reflexive. (Hint: Reduce to the $\sigma$-finite case. $L^\infty$ will contain an isometric copy of $\ell^\infty({\Bbb N})$.)

My question is about the last part of the exercise. Actually, I have tried to construct a counterexample. Consider the real numbers with the borel sigma algebra. Endow the space with the measure $\mu$ such that $\mu(E)$ is zero if the $E$ does not contain an integer and is $\infty$ is $E$ contains an integer. We can see that this space contains countable many sets of positive measure, namely each singleton $\{n\}, n\in \Bbb Z$ has positive measure. Now, we consider $L^1$ on this space. Any $f\in L^1$ has to be identically zero on $\Bbb Z$ to remain integrable. Since, all the measure is concentrated on $\Bbb Z$, $f$ has to be zero almost everywhere. Hence, the only element of $L^1$ is the identically zero function (Functions equal almost everywhere represent the same element). Since $L^1=\{0\}$, $(L^1)^*=\{0\}$ and $(L^1)^{**}=\{0\}$. But now, $L^1$ is reflexive. What is the mistake here?

• There is no mistake, but I'm sure the intended interpretation of "disjoint sets of positive measure" is "disjoint sets of finite positive measure". – Michael Greinecker Oct 12 '17 at 15:00