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If $f_n \rightarrow f$ uniformly and $f_n$ are measureable, integrable functions with $\mu$ $\sigma$-finite. Then $f$ has not to be integrable. Note that $f_n: \Omega \rightarrow \mathbb{R}$. I am new to measure-theory and got few problems with the idea and the right writing. I am using the definition $$\int_{\Omega} f d\mu=\sum_{n=1}^\infty a_n \mu(A_n).$$

My idea was to construct $f_n$ such that for $f$ we have $\Omega=\{x\in \mathbb{R}|x\geq 1\}$, $a_n=1/n$ and $A_n=[n,n+1]$ so that $\mu([n,n+1])=1$ and the integral would equal the harmonic series which diverges. How can I construct such $f_n$? And how can I determine those $a_n$ in general? thanks for your help.

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With $f=\displaystyle\sum_{n}\dfrac{1}{n}\chi_{[n,n+1)}$, $f_{n}=\displaystyle\sum_{1\leq k\leq n}\dfrac{1}{k}\chi_{[k,k+1)}$, then $|f_{n}-f|\leq\dfrac{1}{n+1}$ and $\|f\|_{L^{1}([1,\infty))}=\displaystyle\sum_{n}\dfrac{1}{n}=\infty$.

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  • $\begingroup$ Thanks for your answer. In another question ($f$ assumed integrable) I was asked to show that in general : $lim_{n \to \infty} \int f_n d\mu \neq \int f d\mu$. I used $f_n=\frac{1}{n} \chi_{[-n,n]}$. So $\frac{1}{n}$ gives a example for both questions with only a different characteristic function? $\endgroup$ – Deavor Nov 3 '17 at 23:49
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    $\begingroup$ If $f_{n}$ are such constructed, then $f=0$ which is integrable. $\endgroup$ – user284331 Nov 3 '17 at 23:52
  • $\begingroup$ In the other task $f$ was allowed to be integrable. If I got it right the indicator function is "responsible" in this case whether $f$ is integrable or not? $\endgroup$ – Deavor Nov 3 '17 at 23:54
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    $\begingroup$ Not sure what you are asking. If the question needs $f$ to be integrable but not allowing limit and integrals swiping, then your $f_{n}=1/n\chi_{[-n,n]}$ works. $\endgroup$ – user284331 Nov 3 '17 at 23:56
  • $\begingroup$ Yes that's it. Thank you. $\endgroup$ – Deavor Nov 3 '17 at 23:59

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