Almost sure boundedness is a strong condition which implies the finiteness of moment at any order. Now let $f:\mathbb R \to \mathbb R$ be square-integrable (w.r.t. Lebesgue measure), i.e., $\int f^2(x) \mathrm d x < \infty$. Could you provide an example of such $f$ that is not almost surely bounded (w.r.t. Lebesgue measure)?

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    $\begingroup$ $f(x)=\frac{1}{x^{1/4}}\boldsymbol 1_{(0,1]}(x)$. $\endgroup$
    – Surb
    May 10, 2022 at 16:16
  • $\begingroup$ @Surb Could you transfer your comment to an answer so that i can accept it? $\endgroup$
    – Akira
    May 10, 2022 at 16:18
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    $\begingroup$ Ah!! Surb beat me to it.. I was typing my answer lol $\endgroup$ May 10, 2022 at 16:19

1 Answer 1


$\frac{1}{\sqrt{x}}$ in $(0,1)$ with the Lebesgue measure is a classic example in showing that $L^{1}$ need not be a subset of $L^{\infty}$ . (in fact $L^{1}$ need not be a subset of $L^{p}$ for any $p>1$.

So just take the square root (i.e. $\frac{1}{x^{\frac{1}{4}}})$ and you'll get an $L^{2}$ function that is not $L^{\infty}$


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