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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.

Nathanael Skrepek
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CoffeeIsLife
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1 Answers1

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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.

user99680
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  • I am not sure whether making the partition width small enough would suffice since $\mathbb{Q}$ is dense in $\mathbb{R}$. Could you elaborate? – CoffeeIsLife Dec 18 '13 at 20:24
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    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. – user99680 Dec 18 '13 at 20:26