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I am wondering of a proof strategy to show. That a sequence of Riemann integrable functions which converges point wise to a function may not actually uniformly converge to it. If it makes the argument simpler i also know that the function $f$ the sequence of functions are converging point wise to, is not Riemann integrable. Although on a second thought is it possible to build such an argument without the knowledge of $f$ being non-Riemann integrable ?

Any help would be much appreciated.

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2 Answers 2

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For $n\in\mathbb N$, let $f_n$ be the indicator function of the set $\{n,n+1,n+2,\ldots\}$, i.e. we have $$f_n= \begin{cases} 1 &\text{if } x\in \{n,n+1,n+2,\ldots\}\\ 0 &\text{otherwise}\end{cases}$$ and note that each $f_n$ is Riemann-integrable (with integral $0$), and that $f_n\to 0$ pointwise but not uniformly.

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$$\text{Try}\ f_n=\mathbf 1_{(0,1/n)}.$$

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