# Does an example of a function exist where the Lebesgue Integral doesn't work?

I am studying the Lebesgue Integral as part of a project and I am convinced the Lebesgue Integral is a more powerful method of integration than that of Bernhard Riemann's method. An example of where this is established to be true is the Dirichlet Function.

However I am wondering if there is an example of a function where the Lebesgue Integral also struggles to define an accurate area under a function?

QUESTION EDITED - 00:22 - 04/04/19

• There is not, since the Lebesgue integral generalizes and extends that of Riemann's – JustDroppedIn Apr 3 '19 at 23:15
• If you are asking for an example of a function that is neither Riemann integrable nor Lebesgue integrable there are plenty of them,. Even non-measurable functions exist. – Kavi Rama Murthy Apr 3 '19 at 23:18
• That is what I am asking. What is an example please? – Keighleyite Apr 3 '19 at 23:18
• $f(x)=1$ for all real numbers $x$. – Kavi Rama Murthy Apr 3 '19 at 23:19
• $\frac{1}{x}\sin\left(\frac{1}{x^3}\right)$ is not Lebesgue integrable - an example from wiki entry of Henstock-Kurzweil integral. – achille hui Apr 3 '19 at 23:33

Let $$A$$ be a non measurable set then the characteristic function of $$A$$ is not Lebesgue integrable.
For a more wild example define $$f:[0,1]\to [0,1]$$ as follows: If after some term every second number in the decimal expansion of $$x$$ is recurring (for exampe $$0.375389192919593...$$) then $$f(x)=0.$$(the numbers between the recurring digits.) in our case $$f(0.375389192919593...)=0.12153....$$ If there is no recurrence set $$f(x)=0$$