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I am trying to find an example of a sequence of functions $(f_n)$ on $\mathbb{R}$ such that: Each $f_n$ is integrable, $f_n\rightarrow f$ uniformly; the limit function $f$ is not integrable. Integrable means here normal Riemann integrable. Any help will be appreciated.

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    $\begingroup$ Hint: $f_n=\sum_{k=1}^n k^{-1} \mathbb 1_{[k-1, k)}$ where $\mathbb 1_A$ is the indicator function of $A$. $\endgroup$ – user251257 Aug 13 '16 at 14:17
  • $\begingroup$ @user251257 Can you please explain this example little bit. $\endgroup$ – Mr. MBB Aug 14 '16 at 3:29
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Let $\mathbb 1_A$ denote the indicator function of the subset $A\subseteq \mathbb R$, that is $\mathbb 1_A(x)$ equals $1$ if $x\in A$ and $0$ otherwise.

Further for $n\in\mathbb N$ let $$ f_n = \sum_{k=1}^n k^{-1}(\mathbb 1_{[k-1,k)} - \mathbb 1_{[-k, -k+1)}) $$ and $$ f = \lim_{n\to \infty} f_n. $$

Then, $f_n$ is Riemann integrable. But, for any $a\in \mathbb R$ we have $$ \int_a^\infty f = \infty $$ and $$ \int_{-\infty}^a f = -\infty. $$ Hence, their sum is not defined and $f$ is not (improper) integrable.

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