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Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of locally summable non-negative functions on $\mathbb{R}$.

Suppose that $\int_{a}^{b}x^2\,f_n(x)\,dx \xrightarrow[n\to\infty]{}0$ for every $a,b\in\mathbb{R}$ ,

1) can I conclude that $\int_{a}^{b}f_n(x)\,dx \xrightarrow[n\to\infty]{}0$ for every $a,b\in\mathbb{R}$ ?

2) can I conclude that $f_n(x) \xrightarrow[n\to\infty]{}0$ for Lebesgue-almost every $x\in\mathbb{R}$ ?

I think the answer to 1) should be yes, I tried to prove it using a change of variable ($y=x^3$) but I didn't menage.

Edit:I added the hypothesis $f_n\geq0$ later.

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Neither are true. For example, $f_n (x) = \frac {1}{nx}$ on the interval $[0,1]$. This should give you a hint about what you failed to consider.

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    $\begingroup$ Not a counterexample to 2) though. $\endgroup$
    – mrf
    Dec 29 '12 at 13:13
  • $\begingroup$ It was meant to only be the counter example to 1, since he was so certain the answer is yes. 2) is not true either, and you need another counter example. I do not want to add it, as that will completely give it away. $\endgroup$
    – Calvin Lin
    Dec 29 '12 at 13:24
  • $\begingroup$ But $\frac{1}{nx}$ is not integrable on $[0,1]$.. I edit the question writing "summable" instead of "integrable" to be clearer. $\endgroup$
    – qwertyuio
    Dec 29 '12 at 13:25
  • $\begingroup$ @qwertyuio Yes, which is why I intentionally chose it. Now, you should be able to create a similar sequence that is actually integrable and the integral tends to infinity. $\endgroup$
    – Calvin Lin
    Dec 29 '12 at 13:27
  • $\begingroup$ I tried with $f_n(x)=1/(n x^{1+1/n})$, but it is not a counter-example. Instead $f_n(x)=1/(n x^{1+1/n^2})$ is a counter-example, right? $\endgroup$
    – qwertyuio
    Dec 29 '12 at 13:40

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