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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.

Jam
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    I suspect an approach like this could work: partition $[0, 1]$ into countably many translations $V_1, V_2, \ldots$ of the immeasurable Vitali set (https://math.stackexchange.com/questions/137949/the-construction-of-a-vitali-set), and then define $f(x) = n$ wherever $x \in V_n$. – Connor Harris May 24 '18 at 14:03
  • Do you want a function that is not Lebesgue integrable (if so, over what set?) or do you want a function that is nowhere Lebesgue integrable? If the latter, then you may want to say what you mean by "nowhere Lebesgue integrable" (i.e. not Lebesgue integrable over any open interval, not Lebesgue integrable over any nonempty open set, not Lebesgue integrable over any Borel set with positive Lebesgue measure, not Lebesgue integrable over any Lebesgue measurable set with positive Lebesgue measure, etc.). Yes, I know the various equivalences, but is this also part of your question? – Dave L. Renfro May 24 '18 at 14:57
  • Hi, @DaveL.Renfro. I wanted a function that wasn't Lebesgue integrable over some interval on the real line. I hadn't considered your points about what set the integral's over so I'll have to read up on that and get back to you. Thanks – Jam May 24 '18 at 16:00
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    @Jam, then my example would work. This function is not integrable over $[0,1]$. – Oleg May 24 '18 at 16:58
  • @Oleg I'll give you the bounty when the minimum time's elapsed. – Jam May 24 '18 at 17:11
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    the characteristic function of a Bernstein set would be "nowhere Lebesgue integrable". https://dantopology.wordpress.com/tag/bernstein-set/ and https://en.wikipedia.org/wiki/Bernstein_set and https://math.stackexchange.com/questions/169714/whats-application-of-bernstein-set – Mirko May 26 '18 at 23:26

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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.

Oleg
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