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Trying to solve the Exercise 8.5 of Bass's Analysis book, I want to prove that

$$\lim_{t\to\infty}t\,m(\{x:f(x)\geq t\})=0,$$

where $m$ is the Lebesgue measure and $f(x)=\dfrac{1}{x\log(1/x)}$ defined in $[0,1]$.

It is difficult to get the intervals in which $f(x)\geq t$, I think it envolves W-Lambert function, I am trying to get some approximation, but I am stucked. Could you help me? Thank you in advance.

Mittens
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Quiet_waters
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  • @OliverDiaz it's defined in $[0,1]$, sorry, edited. – Quiet_waters Feb 06 '22 at 00:57
  • @Randall it's defined in $[0,1]$, sorry, edited. – Quiet_waters Feb 06 '22 at 00:57
  • Well surely you cannot have $f(x)$ defined like this for $x=0$ as that would cause division by zero, so the domain of $f$ must be wrong – Lorago Feb 06 '22 at 01:26
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    @OliverDiaz indeed, however then the domain should technically be $(0,1)$ and not $[0,1]$, no? – Lorago Feb 06 '22 at 01:29
  • But please be sure to read the comments, because there is a glitch in the posted answer. – Matematleta Feb 06 '22 at 02:10
  • @OliverDiaz but the calculation $m{x, f(x)\geqslant t}=x_t=-\frac 1{t\log(x_t)}.$ is not correct, is it? The function has vertical asymptotes at $0$ and $1$ and has a local minimum in $(0,1).$ – Matematleta Feb 06 '22 at 02:40
  • @OliverDiaz but $f$ is not strictly increasing on $(0,1)$ so how can $m{x, f(x)\geqslant t}=x_t?$. The claim is true but you need to split up the interval, from $0$ to the local min and from there to $1,$ – Matematleta Feb 06 '22 at 02:58
  • Yes, ok. But my point is the original answer does not split the interval up, which I guess you have to do to make the proof go through. – Matematleta Feb 06 '22 at 03:02

1 Answers1

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This is rather long comment that fits better in this section.

The problem in the aformentioned book in the OP states

Exercise 8.5 Find a non-negative function $f$ on $[0, 1]$ such that $\lim_{t\rightarrow\infty} t\,m({x:f(x)\geq t})=0$, but $f$ is not integrable, where $m$ is Lebesgue measure.

I gather the OP had a hint and tried to work it out. A similar or related question has beed asked before in MSE (see here. In his solution, Davide Giraudo considered the function $f(x)=\frac{1}{x|\log x|}$ for $x\in (0,1)$. This is not quite the function that solves question 8.5. however, the modification $$\phi(x):=\frac{1}{x|\log x|}\mathbb{1}_{(0,e^{-1}]}(x) + e\,\mathbb{1}_{(e^{-1},1]}(x)$$ works exactly as explained by Davide.

Quiet_waters
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Mittens
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  • Thank you so much! You're right. Indeed, at the version 4.2, there is the function $f(x)=\dfrac{1}{x\log(1/x)}$ as an answer in the book. My question could been solved by this function $\phi$, but I am thinking about the book, what do you think? Thank you! – Quiet_waters Feb 06 '22 at 16:43
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    @Quiet_waters: I added a comment in the answer section showing that the function $f$ you propose does not quite work. – Mittens Feb 06 '22 at 19:35