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I have the following question: Give a non-negative lebesgue measurable function $f$ on $[0,1]$ such that

\begin{equation} \lim_{t \to \infty} t\mu(f > t) = 0 \end{equation}

but $f \not \in L^1$

In a previous part of a problem, I was able to show that $f \in L^1$ is a sufficient condition for the limit to be $0$. I have tried stuff like $1/x$ which is not in $L^1$ but the limit is $1$ in this case. I have tried other variations like $1/x^n$ where $n \geq 0$ but none of these work.

I think that $[0,1]$ having finite measure is likely relevant, but don't know how.

Mike
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  • From basic Probability Theory there exists $f$ such that $\mu (f>t)=\frac 1 {t \ln t}$ for $t \geq 2$. This $f$ satisfies your requirements. – Kavi Rama Murthy Nov 18 '20 at 00:14
  • HInt: the line between integrable and non-integrable for $1/x^p$ is $p=1.$ So you need to squeeze a function in between. So, maybe a $\log$ of some kind? Well, just saw that the other comment gave it away. I will leave mine though because it may indicate the thinking on how the function is obtained. – Matematleta Nov 18 '20 at 00:16

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