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Let $X$ be a measure space of finite total measure and suppose that $f \in L^1(X)$. I know that since $\mu(X) < \infty$, $f\in L^p$ implies that $f \in L^q$ for all $q \leq p$. I am interested in $p = \sup\{r\in [1,\infty) : f\in L^r\}$. In particular, I am wondering if it is possible for $f$ to actually be in $L^p$. Clearly its possible for $f$ to fail to be in $L^p$ (cf. $1/x^{1/p}$ on $[0,1]$. It seems somewhat implausible that the integrals of $| f|^p$ would all be uniformly bounded up to some point $p$ after which they are infinite, but I am not sure how to prove this, or even if its true.

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excuse me, I'm working just on an affirmation that you did here, that $f(x)=\dfrac{1}{x^{\dfrac{1}{p}}}$ with $x\in [0,1]$ fails to be in $L_p$, is it because $\int_{[0,1]}\mid f\mid ^p d\mu = \mid x\mid ^{-\dfrac{p}{p}}.\mu([0,1])=\mid \dfrac{1}{x}\mid $, and this goes to $\infty$ when $x\mapsto 0$? –  Irene May 14 at 14:25

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up vote 4 down vote accepted

Consider $\displaystyle{\frac{1}{x(\log x)^2}}$ on $\left(0,\frac{1}{e}\right)$, where $p=1$ and the function is in $L^1$. By taking powers of this function, you can get any other finite value of $p$. See also this related question. Of course, the case $p=\infty$ is covered by any bounded function, and there are also examples of unbounded functions for which $p=\infty$, like $\log$ on $(0,1)$.

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