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

  • $\begingroup$ There is not, since the Lebesgue integral generalizes and extends that of Riemann's $\endgroup$ – JustDroppedIn Apr 3 '19 at 23:15
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    $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy Apr 3 '19 at 23:18
  • $\begingroup$ That is what I am asking. What is an example please? $\endgroup$ – Keighleyite Apr 3 '19 at 23:18
  • $\begingroup$ $f(x)=1$ for all real numbers $x$. $\endgroup$ – Kavi Rama Murthy Apr 3 '19 at 23:19
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    $\begingroup$ $\frac{1}{x}\sin\left(\frac{1}{x^3}\right)$ is not Lebesgue integrable - an example from wiki entry of Henstock-Kurzweil integral. $\endgroup$ – achille hui Apr 3 '19 at 23:33

There are lots and lots of non Lebesgue integrable functions. Here are two examples:

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 $f(x)=0$


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