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How can I find a function $f$ such that $f \notin L^{p} (\mathbb{R})$ for all $p$ but you can find a constant $c>0$ for it with

$m(x \in \mathbb{R} \, s.t. |f(x)|>t) \leq \frac{c}{t}$ for $\forall t$

I tried $f(x)={1 \over x^2}$ since $f \notin L^{p} (\mathbb{R})$ for all $p$, but it looks like this is not the right answer. I guess we could use the Chevyshev's inequality on $f(x)$ , the measure theoretic one (see Section Measure-theoretic statement in https://en.wikipedia.org/wiki/Chebyshev%27s_inequality)

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    $\begingroup$ What is $p$? ${}{}{}$ $\endgroup$ – copper.hat Oct 16 '15 at 15:54
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    $\begingroup$ Do you mean $| f(x) | > t$ instead of $|f (x) > t|$? $\endgroup$ – Hetebrij Oct 16 '15 at 15:56
  • $\begingroup$ Perhaps you can tell us what you tried for $f.$ $\endgroup$ – zhw. Oct 16 '15 at 16:05
  • $\begingroup$ @copper.hat for all $p$ $\endgroup$ – Susan_Math123 Oct 16 '15 at 16:19
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    $\begingroup$ You didn't try $f(x) = 1/x$? $\endgroup$ – zhw. Oct 16 '15 at 16:22
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Let $f(x) = 1/x, x \ne 0.$ Then $f\not \in L^p(\mathbb {R}), 0 < p \le \infty,$ but $m(\{x: |f(x)| > t\}) = 2/t$ for all $t>0.$

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