# Lebesgue space - $L^p$ spaces

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

$$\|f\|_{L^1} = \begin{cases}|f(a)| & \text{if } f(b) = 0\\ +\infty & \text{else}\end{cases}$$ 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 $\dim L^\infty (\mu) = 2$ ...