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Let $X=\{a,b\}$,and $\mu(\{a\} )=1$, and $\mu(\{b\} )=\mu(X)=+\infty$ and $\mu(\emptyset)=0$. Is it truth that $L^\infty(\mu)$ is the dual space of $L^1(\mu)$. Whether $L^\infty(\mu)=L^1(\mu)^\ast$? If no, why?

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Thank you very much, you help me a lot! – elizabeta gigova Mar 30 '13 at 15:47
If you find the answer useful you can "accept" it. – Siméon Apr 3 '13 at 13:28

1 Answer 1

Here we have

$$ \|f\|_{L^1} = \begin{cases}|f(a)| & \text{if } f(b) = 0\\ +\infty & \text{else}\end{cases} $$ so that $L^1(\mu) = \{f \in\mathcal{F}(\{a,b\},\Bbb R) \mid f(b) = 0\}\simeq \Bbb R$ and $L^\infty(\mu) = \mathcal{F}(\{a,b\},\Bbb R)\simeq \Bbb R^2$.

In particular, $\dim L^1(\mu) = 1$ and $\dim L^\infty (\mu) = 2$ ...

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Derive details from the proof of L∞(μ) of the following claim: claim : If {E_α} is an arbitrary family of measurable sets on [0,1], then there is E⊂[0,1] such that: E contains almost all E_α; If F is measurable, than F contains almost all E_α and F⊇E almost everywhere. – elizabeta gigova Mar 30 '13 at 19:28
It is better to ask this as a new question. – Siméon Apr 3 '13 at 13:29

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