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What is an example that involves a fuction on an interval of the real numbers where the Lebesgue integral is better than the Riemann integral.

By better, it probably means that the Lebesgue intregral is defined while the Riemann integral is not.

By an example, I mean an example that has importance in mathematics for reasons other than just showing that the Lebesgue integral is more general than the Riemann integral. If the importance of the example is not clear, then please explain why it is important.

A possible answer could be an example where the use of the dominated convergence theorem plays a role.

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    $\begingroup$ I am not sure what you are looking for exactly, but I would offer that the Lebesgue spaces that can be defined are complete, metric, vector spaces. Something which is not possible with the Riemann integral. The emphasis is on completeness, which is always a very usesful thing to have. $\endgroup$ – Snake707 Jan 8 at 16:02
  • $\begingroup$ You may also interested in other integral constructions as the gauge integral (also known as Henstock Kurzweil integral), see Wikipedia for more details. $\endgroup$ – p4sch Jan 13 at 21:30
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Take the Dirichlet function (restricted to, say, $[0,1]$), for instance:

  • it is Lebesgue-integrable;
  • it is not Riemann integrable;
  • it was defined by Dirichlet before Lebesgue was even born.
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In fact the intuition behind Lebesgue integrability is that how likely the surface under a non-negative function is defined and how much is its surface (note that Lebesgue integrability is defined for non-negative functions). For example define the surface of a $1\times 1$ solid square (I mean the perimeter and inside the square). If any time we remove vertical lines from the solid square (that is, removing a line from a surface) the intuition says as if "nothing" has been removed from the solid square and the area must be preserved. The intuition is show below:

enter image description here

As you can see, the white lines are step by step removed from the dark shaded solid square, leaving a "reduced" square which is "almost" as of the primary square (by "almost" we mean a set whose Lebesgue measure is equal to that of the primary square i.e. each time a zero-set has been subtracted from the square). Now let a function be defined as $1$ on the domain of the "reduced" square. For example let$$f_n(x)=\begin{cases}1&,\quad x={1\over k}\ \ ,\ \ \text{for some }1\le k\le n\\0 &\quad \text{elsewhere}\end{cases}$$whose underneath surface is as follows:

enter image description here

Surely, $f_{\infty}(x)$ exists and is Reimann un-integrable in $[0,1]$ but this is unfair under the generalization to Lebesgue integrability since the surface is "almost" the same except some zero-set infinitely many distinct lines counted and subtracted from the surface of the function. Probably, this intuition was the main lead Lebesgue to establish probably his most prominent theory on measure-sets.

An interesting interpretation is that Reimann integral counts the surface beneath a function vertically while Lebesgue integral counts it horizontally.

Hope I could have clarified enough so far why such a generalization exists. You are highly encouraged to check out Lebesgue integration on Wikipedia.

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