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I want to find a sequence of functions $f_n$ converging to $f$ pointwise, where all of the $f_n$'s are bounded in magnitude (say by $1$) and integrable on $[−1, 1]$, but $f$ is not integrable on $[−1, 1]$. I was thinking about using $f(x) = \sin\left(\frac{1}{x - 1}\right)$ and $f_n(x) = f\left(x - \frac{1}{n}\right)$, but I'm not sure whether or not $f(x)$ is actually integrable or not on $[-1,1]$, and also $f_n$ only converges pointwise to $f$ on $[-1,1)$ since $f(1)$ is undefined. Is there a better example?

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    Integrable in the sense of Lebesgue, or in the sense of Riemann? – 2017-02-28
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    @Starfall Riemann – 2017-02-28

2 Answers 2

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The characteristic function of the rational numbers in $ [-1, 1] $ is not a Riemann integrable function, and is clearly the pointwise limit of bounded integrable functions.

In your example, $ f(x) $ is not defined at $ 1 $; and no matter how you define it at $ 1 $, it will only be discontinuous at $ 1 $, thus it will be integrable.

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    I thought about using that as $f$ as well, but how is it the pointwise limit of bounded intebrable functions? – 2017-02-28
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    @ihmth Pick an enumeration of the rationals in the interval, say $ q_1, q_2, \ldots $; and define $ f_1 $ to be $ 1 $ at $ q_1 $ and $ 0 $ elsewhere, define $ f_2 $ to be $ 1 $ at $ q_1, q_2 $ and $ 0 $ elsewhere... What is the limit of this sequence? – 2017-02-28
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Let $q_1, q_2, \dots$ be the rationals in $[-1,1].$ For $n\in \mathbb N,$ let $f_n=\mathbb 1_{\{q_1,\dots , q_n\}}.$ Then each $f_n$ is Riemann integrable on $[-1,1],$ each $f_n$ is bounded by $1$ in magnitude, and $f_n \to \mathbb 1_{\mathbb Q\cap [-1,1]}$ pointwise everywhere in $[-1,1].$ However, $\mathbb 1_{\mathbb Q\cap [-1,1]}$ is not Riemann integrable on $[-1,1].$