I have been given the example $$ f_n(x)= \begin{cases} 1, & \text{if}\; x\in\{(x_k)_{k=1}^{n}\}, \\ 0, & \text{otherwise.} \end{cases} $$ Here the sequence $(x_k)_{k=1}^{\infty}$ is the sequence that "enumerates" elements of $\mathbb{Q}$. The sequence of functions converges pointwise to $$ f(x)= \begin{cases} 1, & x\in\mathbb{Q}, \\ 0, & x\notin\mathbb{Q}, \end{cases} $$ which isn't Riemann integrable. However, I am not quite convinced that $f_n$ is integrable. I would like some help in seeing how $f_n$ is Riemann integrable. Also, I would like more examples of sequences of Riemann integrable functions $f_n$ that converge pointwise to a function $f$ that is not Riemann integrable.

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    $\begingroup$ With $f_n(x)=1$ for $x=x_1,\ldots, x_n$, and $f_n(x)=0$ otherwise, each $f_n$ is continuous at all but finitely many points; thus, integrable. $\endgroup$ Dec 18, 2013 at 20:14
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    $\begingroup$ Integrable means "Riemann Integrable"? $\endgroup$
    – Igor Rivin
    Dec 18, 2013 at 20:15
  • $\begingroup$ I see. So the integral of $(f_n)$ is 0? $\endgroup$ Dec 18, 2013 at 20:15
  • $\begingroup$ Yes, I always forget. Sorry, I'll correct it $\endgroup$ Dec 18, 2013 at 20:15
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    $\begingroup$ In your definition of $f_n$, I believe you should have $n$ as the upper bound on the sequence. $\endgroup$
    – user61527
    Dec 18, 2013 at 20:17

1 Answer 1


For the $f_n(x)$, it comes down to the fact that the Riemann sum $1dx \rightarrow 0$ when the partition width is small-enough. Then you do this finitely-many times, and you get a total of $0$ for any $f_n(x)$. You can show that this result is true no matter how small you make the partition width $||P||$, i.e., you can show $|\Sigma f(x)dx-0 |<\epsilon$ for any partition width.

  • $\begingroup$ I am not sure whether making the partition width small enough would suffice since $\mathbb{Q}$ is dense in $\mathbb{R}$. Could you elaborate? $\endgroup$ Dec 18, 2013 at 20:24
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    $\begingroup$ But remember that $f_n(x)$ is the integral where only $x_1,x_2,..,x_n$ are non-zero. So you don't need to worry about all terms in $\mathbb Q$ , nor density. $\endgroup$
    – user99680
    Dec 18, 2013 at 20:26

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