I was thinking about a comment on this question, and how some functions are not Riemann integrable but are Lebesgue integrable. But are there any functions which are defined everywhere in an interval of $\mathbb{R}$ but are not Lebesgue integrable? In other words, does it make sense for a function to have no antiderivative by Lebesgue integration?
For example, is there an $f(t)$ that satisfies the properties of the third section below:
$$\begin{aligned} &\quad\quad\text{$\sin(t)$ defined everywhere ✓}\\ \int_0^x \sin(t) \;\mathrm{d}t :&\quad\quad\text{Riemann integrable ✓} \\ &\quad\quad\text{Lebesgue integrable ✓} \\ \\ \hline \\ &\quad\quad\text{$I_\mathbb{Q}(t)$ defined everywhere in interval ✓}\\ \int_0^x I_\mathbb{Q}(t) \;\mathrm{d}t :&\quad\quad\text{Riemann integrable ✗} \\ &\quad\quad\text{Lebesgue integrable ✓} \\ \\ \hline \\ &\quad\quad\text{$f(t)$ defined everywhere in interval ✓}\\ \int_0^x f(t) \;\mathrm{d}t :&\quad\quad\text{Riemann integrable ✗} \\ &\quad\quad\text{Lebesgue integrable ✗} \\ \end{aligned}$$
Where $I_\mathbb{Q}(t)$ is the Dirichlet function.
Yes, let $A\subset[0,1]$ be a set which is not measurable with respect to the Lebesgue measure. Let $f(t)=I_A(t)$. Then $f$ is defined everywhere in the interval $[0,1]$, but is neither Riemann nor Lebesgue integrable.