Using the axiom of choice and a Hamel basis for a normed space, one can prove the existence of everywhere defined discontinuous linear functionals.
My question: Does there exist a discontinuous everywhere defined linear functional on $L^p(\mathbb{R}^n)$, $1\leq p <+\infty$, representable by a measurable function $g : \mathbb{R}^n \to \mathbb{R}$?
In other words, the question is whether the following false: If $g$ is measurable and $$ \forall f\in L^p, \quad \left| \int f g \right| < +\infty, $$ then $g \in L^{p*}$?
I have tried to argue by counterpositive with no luck: The idea is to assume that $g\notin L^{p*}$ and construct an $f\in L^p$ such that $\big|\int fg \big|= +\infty$.
If $1 \leqslant p < +\infty$, and $g \colon \mathbb{R}^n \to \mathbb{R}$ (or $\mathbb{R}^n \to \mathbb{C}$) is a measurable function such that for all $f \in L^p(\mathbb{R}^n)$ the integral
$$\int_{\mathbb{R}^n} fg \,d \lambda$$
exists, then it follows that $g \in L^{p^{\ast}}(\mathbb{R}^n)$, where $p^{\ast}$ is the conjugate exponent to $p$, hence the linear functional $T \colon f \mapsto \int fg\,d\lambda$ is continuous on $L^p(\mathbb{R}^n)$.
We can prove that using the Banach-Steinhaus theorem: For $n \in \mathbb{N}$ define
$$g_n(x) = \begin{cases} \qquad 0 &, \lVert x\rVert > n \\ \quad\;\; g(x) &, \lVert x\rVert \leqslant n \text{ and } \lvert g(x)\rvert \leqslant n \\ \frac{n}{\lvert g(x)\rvert}\cdot g(x) &, \lVert x\rVert \leqslant n < \lvert g(x)\rvert.\end{cases}$$
Then $g_n \in L^1(\mathbb{R}^n) \cap L^{\infty}(\mathbb{R}^n) \subset L^{p^{\ast}}$, and we have $\lim\limits_{n \to +\infty} g_n(x) = g(x)$ for all $x\in \mathbb{R}^n$. The family of continuous linear functionals $T_n \colon f \mapsto \int fg_n\,d\lambda$ is pointwise bounded, since
$$\int_{\mathbb{R}^n} f(x)g_n(x)\,d\lambda \xrightarrow{n\to +\infty} \int_{\mathbb{R}^n} f(x) g(x)\,d\lambda$$
by the dominated convergence theorem. The Banach-Steinhaus theorem asserts that the family $\{ T_n : n \in \mathbb{N}\}$ is equicontinuous, i.e. $\sup\limits_{n\in \mathbb{N}} \lVert T_n\rVert < +\infty$. It then follows that $T$ is continuous and
$$\lVert T\rVert = \lVert g\rVert_{L^{p^{\ast}}(\mathbb{R}^n)} \leqslant \sup_{n\in \mathbb{N}} \lVert T_n\rVert.$$
An application of the monotone convergence theorem shows that in fact
$$\lVert T\rVert = \lVert g\rVert_{L^{p^{\ast}}(\mathbb{R}^n)} = \sup_{n\in \mathbb{N}} \lVert T_n\rVert = \lim_{n\to +\infty} \lVert g_n\rVert_{L^{p^{\ast}}(\mathbb{R}^n)},$$
since $\lvert g_n(x)\rvert^{p^{\ast}}$ converges monotonically to $\lvert g(x)\rvert^{p^{\ast}}$.
The same result with basically the same proof works for an arbitrary $\sigma$-finite measure space, one just replaces the balls of radius $n$ with centre $0$ used here by a sequence of sets with finite measure exhausting the space.