TITLE: Example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$
QUESTION [7 upvotes]: Let $m$ be the Lebesgue measure on $[a,b]$.
What would be an example of a bounded linear functional $T:L^ \infty \to \mathbb{R}$ which cannot be expressed by an integral $\int_{[a,b]}fg$ for an integrable $f$?
I know that, if $1 \leq p < \infty$, then any bounded linear functional $T:L^p \to \mathbb{R}$ can be expressed by an integral $\int_{[a,b]}fg$ for an $f \in L^q$. (Where $q$ is the conjugate of $p$.) But the proof I am familiar with relies on the fact that $p$ is finite.

REPLY [3 votes]: You may take a sequence of linear functionals, $T_n\in(L^\infty)'$, which for $n>\frac{1}{b-a}$  and $f\in L^\infty$ is defined by:
$$ T_n f = n\int_a^{a+1/n} f(x) \, dx.$$ 
Then $|T_n f| \leq \|f\|_\infty$ so by weak-* compactness (Banach-Alaoglu), the sequence has a weak-* accumulation point $S\in (L^\infty)'$. 
Being in a non-separable case there need not be a convergent subsequence and in view of the abstract nature of the construction (Axiom of Choice being involved), it is often quite difficult to do any calculations with such an S. In the present case we may, however, extract sufficient information.
First, the interpretation of being a weak-* accumulation point is that to any given $f\in L^\infty$ there is a subsequence $(n_k)$, which in general will depend upon $f$, so that $\lim_k T_{n_k} f = S f$. 

If we take $f=1$ we see that $T_n 1 = 1$ so that
$S1 = \lim T_{n_k} 1 = 1$. In particular, $S$ is non-zero in $(L^\infty)'$. 
If $f$ has support in $[a+\varepsilon,b]$ for $\varepsilon>0$  we see that $T_nf=0$ for $n$ large enough, implying that $Sf=0$. 

Suppose now, that $Sf=\int_a^b gf \, dx$ for some $g\in L^1$. The last part implies that $g\equiv 0$ (Lebsegue a.e.) so $S=0$, thus contradicting the first part. So $S$ may not be described in this way.