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Again while preparing to calculus I found another interesting question:

Prove or give counterexample that if $f_n\to f$ uniformally $[0,\infty]$ and $\forall n\in\mathbb {N}\int_0^\infty|f_n|dx\le M$, then $\int_0^\infty|f(x)|dx<\infty$

It seems incorrect. I'm looking for a function that tends to a constant $c$ uniformly and its integrals are bounded. If the convergence was in $[2,b] (b\in\mathbb R)$, I could have taken $f_n(x)=\frac {x^n}{x^n+1}$ which indeed is a counterexample. How can I find a counterexample?

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  • $\begingroup$ There are no counterexamples of this kind. If $f_n \to c$ uniformly, then for some $n$ $|f_n| > c/2$ on $[0, \infty)$, and the integral of $|f_n|$ is infinite. $\endgroup$ – Dan Shved Jul 6 '13 at 9:49
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Fix $A>0$. Then $$\int_0^A|f(x)|\mathrm dx=\lim_{n\to +\infty}\int_0^A|f_n(x)|\mathrm dx\leqslant \liminf_{n\to +\infty}\int_0^{+\infty}|f_n(x)|\mathrm dx\leqslant M.$$ As $A$ was arbitrary, $\int_0^{+\infty}|f(x)|\mathrm dx\leqslant M$.

It actually works when we only have pointwise convergence, and it is called Fatou's lemma, but it involves more advanced tools.

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  • $\begingroup$ why $\lim\int_0^A|f_n|\ge lim\inf\int_0^{+\infty}\le M$? $\endgroup$ – user65985 Jul 6 '13 at 10:57
  • $\begingroup$ For each $n$, $\int_0^A|f_n|dx\leqslant \int_0^{+\infty}|f_n|dx$, now take the $\liminf$ on both sides. $\endgroup$ – Davide Giraudo Jul 6 '13 at 11:29

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