This is Exercise 6.12 from Rudin's RCA.
Let $\mathscr{M}$ be the collection of all sets $E$ in the unit interval $[0,1]$ such that either $E$ or its complement is at most countable. Let $\mu$ be the counting measure on this $\sigma$-algebra $\mathscr{M}$. If $g(x)=x$ for $0\le x \le 1$ show that $g$ is not $\mathscr{M}$-measurable,although the mapping
$$f \mapsto \sum xf(x) = \int fg d\mu$$ makes sense for every $f\in L^1(\mu)$ and defines a bounded linear functional on $L^1(\mu)$. Thus $(L^1)^* \neq L^\infty$ in this situation.
It is easy to prove the assertions. Why does this lead to $(L^1)^* \neq L^\infty$? We have only shown the existence of a bounded linear functional of the form $\int fg d\mu$ where $g \notin L^\infty$. Thus, the canonical map $g \mapsto (f \mapsto \int fg d\mu)$ is not isomorphic, but why is this enough to conclude that $(L^1)^* \neq L^\infty$? Is it not possible that there may be other isomorphism that maps an element of $L^\infty$ to the map $f \mapsto \int fg d\mu$?
Best Answer
If $\int fg \, d\mu = \int x f \, d\mu$ for all $f\in L^1(\mu)$, then necessarily $g(x) = x$. To see this, fix $y \in [0,1]$ and choose $$ f(x) = \begin{cases} 1, & x=y, \\ 0, & \text{otherwise}. \end{cases} $$ Then $g(y) = \int f g \, d\mu = \int x f \, d\mu = y$. But $y$ was arbitrary, so $g(x) = x$ for all $x\in [0,1]$.