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 $f$
My answer is very similar to the solution described in Folland's real analysis book.
Consider the map $L : C([a, b]) \to \mathbb{R}$ given by $$L(f) = f(a).$$ Note that, here $C([a,b])$ is the space of continuous functions on $[a,b]$. We note that $C([a,b])$ can be viewed as a vector subspace of $L^\infty([a,b])$.
Remark: Technically, $C([a,b])$ is not a subspace of $L^\infty([a,b])$ since $C([a,b])$ is a collection of functions whereas $L^\infty([a,b])$ is a collection of equivalence classes. On the other hand, one can map each function $f\in C([a,b])$ to it's equivalence class in $L^\infty([a,b])$. Thus, we see that $C([a,b])$ corresponds to a vector subspace $Y$ of $L^\infty([a,b])$. We are choosing to view $C([a,b])$ as this subspace $Y$. Then $L$ is defined on $Y$ by the map $L([f]) = f(a)$ where $f$ is the continuous representative of the equivalence class $[f]$.
Returning to our problem, we note that $$ L(f) \leq \lVert f\rVert_{\infty}. $$ By the Hahn-Banach Theorem, there exists a linear functional $$T:L^\infty([a,b]) \to \mathbb{R}$$ extending $L$ such that $T(f) \leq \lVert f\rVert_\infty$ for every $f\in L^\infty([a,b])$. That is, $T : L^\infty([a,b]) \to \mathbb{R}$ is a continuous linear functional such that $$ T\mid_{C([a,b])} = L. $$ Suppose for a contradiction that there exists a function $g\in L^1([a,b])$ such that $$ T(f) = \int_{[a,b]} fg $$ for all $f\in L^\infty([a,b])$. Consider the sequence of functions $$ f_n(x) = \max\left(0, 1 - n(x-a)\right). $$ By definition, we have $$T(f_n) = f_n(a) = 1$$ for each $n\in\mathbb{N}$. On the other hand, it follows from the dominated convergence theorem that $$ \lim_{n\to\infty} T(f_n) = \int_{[a,b]} f_n g = 0, $$ which is absurd.