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Given a function $f:[a,b]\to \mathbb R$ it could make sense, depending on the situation, to define the integral of $f$ in either of the following ways:

  1. Standard Riemann integral, assuming it exists.
  2. Limit as $\varepsilon\to 0$ of the Riemann integral over $[a+\varepsilon, b]$, assuming the latter integral exists for sufficiently small $\varepsilon$.

Does anyone know an example of a function $f$ for which these two do not coincide? For instance a function for which the second method is well-defined but the Riemann integral does not exist?

Inzinity
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1 Answers1

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Consider the function $f(x)= n$ for $x=1/n$ for $n$ a natural number and $f(x)=0$ otherwise.

For any $\varepsilon > 0$ on the interval $[\varepsilon, 1]$ this function is bounded and zero up to finitely many points. Hence the Riemann integral exists and is zero. So using definition 2. the limit $\varepsilon \rightarrow 0$ exists and is equal to $0$.

However, on the interval $[0,1]$ this function has infinitely many discontinuities, is unbounded and is not Riemann integrable according to definition 1.

quarague
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  • On the other hand, if $f$ is bounded, then 1 and 2 are the same. See e.g. https://math.stackexchange.com/q/311395, https://math.stackexchange.com/q/1301170. – Calum Gilhooley Jun 28 '23 at 12:54
  • The failure of Riemann integrability (your last paragraph) is not due to infinite number of discontinuities, but more due to the fact that Riemann integration applies only to bounded functions. – Paramanand Singh Sep 21 '23 at 11:16
  • @ParamanandSingh The wording was misleading, I adjusted it. – quarague Sep 21 '23 at 12:47