Dual of $L^1$ when measure is the counting measure

Question

Let $X$ be an uncountable set, $\mu$ the counting measure on $X$ and $\mathcal{M}$ the $\sigma$- algebra of countable or co-countable sets. How can I prove that the dual of $L^1(\mu|\mathcal{M})$ is $L^\infty(\mu)$?

Answer 1

On one hand, given an $y \in L^\infty(\mu) = \ell^\infty(X)$, we may define an functional $Ty \in L^1(\mu|_{\def\M{\mathcal M}\M})^*$ by $$ (Ty)(x) = \int_X xy\,d\mu = \sum_{i \in X} x(i)y(i) $$ Then $T \colon L^\infty(\mu) \to L^1(\mu|_\M)^*$ is one-to-one, as $$ (Ty)(e_i) = y_i, \quad i \in X $$ where $e_i \in L^1(\mu|_\M)$ is the function with $e_i(j) = \delta_{ij}$, $j \in X$.

$\def\L{L^\infty(\mu)}\def\K{L^1(\mu|_\M)}$ For $y \in \L$, $x\in \K$ we have \begin{align*} \def\abs#1{\left|#1\right|}\def\norm#1{\left\|#1\right\|} \abs{(Ty)(x)} &= \abs{\sum_i x(i)y(i)}\\ &\le \sum_i \abs{x(i)}\abs{y(i)}\\ &\le \norm{y}_\infty\norm x_1 \end{align*} Hence $\norm{Ty}_{\K^*} \le \norm{y}_\infty$, so $T$ is continuous.

Now, given $x^*\in \K^*$, define $y \colon X \to \mathbb K$ by $y(i) := x^*(e_i)$. Then $\abs{y(i)}\le \norm{x^*}_{\K^*}\norm{e_i}_1 = \norm{x^*}$, hence $y \in \L$. Now $Ty = x^*$, as $x^*$ and $Ty$ agree on $\{e_i \mid i \in X\}$, which span is dense. Moreover, by the above $$ \norm{y}_\infty = \sup_i \abs{y(i)} \le \norm{x^*}_{\K^*} = \norm{Ty} \le \norm y_\infty $$ Altogether, $T \colon \L \to \K$ is an isometric isomorphism.