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Following this question: Limit of $L^p$ norm

Given a finite measure space $(X, \mathcal{M}, \mu)$ and a measurable function $f:X\to\mathbb{R}$ in $L^\infty$, $\displaystyle\lim_{p\to\infty}\|f\|_p=\|f\|_\infty$

Can we find a counterexample when $\mu$ is not a finite sigma measure?

Hermi
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  • This statement is true in general. You can patch the proof together from the answers to the exact question that you linked. – stomfaig Mar 10 '24 at 21:12
  • See if this works: as is pointed out in Davide's comment, the statement is true for $\sigma-$finite measure space, by using Levi's theorem. On the other hand, if $p\ge 1$ and $f\in L^p$, then ${x: f(x)\ne 0}$ is $\sigma-$finite, since ${x: f(x)\ne 0}={x:|f(x)|^p\ne 0}=\bigcup_{n\in\mathbb{Z}_+}{|f(x)|^p>\frac{1}{n}}$. – Asigan Mar 11 '24 at 06:33
  • Well, to be honest, Davide said it is true for $\sigma-$finite space, but I has a little doubt. But as I said in my comment, if the statement holds in $\sigma-$finite space, then it should be true in any space. – Asigan Mar 12 '24 at 05:32

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There is no counterexample, since this statement is true in general. I am going to extend this answer, so that the argument holds for the non-$\sigma$-finite case too. I am using the same notation as used there.

When proving that $\liminf_{p \to \infty} \|f\|_p \leq \|f\|_\infty$ the same proof goes through, even if your space is not $\sigma$-finite, so there is nothing to do.
When proving that $\liminf_{p \to \infty} \|f\|_p \geq \|f\|_\infty$, the only thing that needs discussion is that $\mu(S_\delta)$ is finite. Now, by Tchebychev’s inequality, $$\mu(S_\delta) \leq \frac{\|f\|_q^q}{(\|f\|_\infty-\delta)^q}$$ where the right hand side is definitely finite, thus so is the left hand side.

stomfaig
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    @Hermi your $f$ does not lie in any $L^q$ with $q\in [1,\infty)$, so it doesn’t satisfy the hypothesis of the theorem, i.e not a counterexample. – peek-a-boo Mar 12 '24 at 21:17
  • If you drop that assumption, then your counterexample works. – peek-a-boo Mar 12 '24 at 21:22
  • But note that if you drop that $f \in L^q$ for some $q$, then the statement doesn't hold even for $\sigma$-finite measures, as shown by the counterexample you gave. – stomfaig Mar 12 '24 at 21:29