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Let $f_n:[0,1]\to\mathbb{R}$ be a function defined as:

$$f_n(x)=\begin{cases}1\ \text{if}\ x=r_n\\0\ \text{otherwise}\end{cases}$$

where $r_n$ is an enumeration of rationals in $[0,1]$. Then, is the limit function $f=\lim_{n\to\infty}f_n$ Riemann integrable/Lebesgue integrable?

I think the function is Lebesgue integrable, as its absolute value is bounded, since $$f(x)=\begin{cases}1\ \text{if}\ x\in\mathbb{Q}\\0\ \text{otherwise}\end{cases}$$. The function is easily seen to be non-Riemann integrable because of nowhere continuity. Am I right? Thanks beforehand.

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  • $\begingroup$ It's $1$ when $x\in\Bbb Q$. It is zero almost everywhere, but continuous nowhere. $\endgroup$ – Angina Seng Jan 25 '19 at 7:15
  • $\begingroup$ @LordSharktheUnknown but the sequence of rationals converges to irrationals right, since the set of rationals is not complete? $\endgroup$ – vidyarthi Jan 25 '19 at 7:17
  • $\begingroup$ When you say easily seen... you're applying a non-trivial Lebesgue theorem about integrability of Riemann functions. $\endgroup$ – mathcounterexamples.net Jan 25 '19 at 7:18
  • $\begingroup$ @vidyarthi The sequence cannot converge as it enumerates all the rationals. $\endgroup$ – mathcounterexamples.net Jan 25 '19 at 7:19
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    $\begingroup$ @vidyarthi Right. And the best argument is the one of Maksim answer below. No need to bring sophisticated Lebesgue Riemann theorem here. $\endgroup$ – mathcounterexamples.net Jan 25 '19 at 7:48
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$f$ is not Riemann integrable, that is true, but the reason is not of missing any continuity. It is just that the upper Darboux sum will always be 1 and the lower Darboux sum will always be zero given any partition of $ [0,1]$ and therefore the function is not Riemann integrable.

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  • $\begingroup$ so do you, as @mathcounterexamples.net pointed out, mean the pointwise limit of the sequence is the zero function? $\endgroup$ – vidyarthi Jan 25 '19 at 7:44

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