I am thinking about the proof that the usual $L^p$ spaces are complete. So, let $(X,\mathcal{F},\mu)$ be a measure space and let $p\in[1,+\infty)$. Important: by a measure I mean a nonnegative $\sigma$-additive, rather than just finitely additive, set function. Let $\tilde{L}^p(X,\mathcal{F},\mu)$ be the space of $\mathcal{F}$-measurable functions $f:X\to\mathbb{R}$ such that $\int_X |f|^p d\mu<+\infty$. If $f,g\in\tilde{L}^p(X,\mathcal{F},\mu)$, then $f+g$ and $\alpha f$ also belong to this space for every real $\alpha$. So $\tilde{L}^p(X,\mathcal{F},\mu)$ is a vector space. Define the seminorm $s_p$ on $\tilde{L}^p(X,\mathcal{F},\mu)$ by $s_p(f)=\left(\int_X |f|^p d\mu\right)^{1/p}$, $f\in\tilde{L}^p(X,\mathcal{F},\mu)$. The kernel of this seminorm is the subspace $N$ of $\mathcal{F}$-measurable functions $f\in\tilde{L}^p(X,\mathcal{F},\mu)$ such that $s_p(f)=0$, i.e., $N$ is the subspace of functions which are equal to zero $\mu$-almost everywhere. Then $L^p(X,\mathcal{F},\mu)=\tilde{L}^p(X,\mathcal{F},\mu)/N$ with the norm $\|F\|_p=s_p(f)$, $f\in F$, is the usual $L^p$ space.

I inspection the proof that $L^p(X,\mathcal{F},\mu)$ is complete, i.e., it is a Banach space. So, we need to show that every Cauchy sequence $\{F_n\,|\,n\geqslant 1\}$ in $L^p(X,\mathcal{F},\mu)$ is convergent to some element of the space. If I understand things right, we need first to choose functions $f_n\in F_n$, $n\geqslant 1$, and after this we can use standard arguments to construct an $\mathcal{F}$-measurable function $f:X\to\mathbb{R}$ such that $f\in\tilde{L}^p(X,\mathcal{F},\mu)$ and $s_p(f_n-f)\to 0$ as $n\to\infty$, which means that $\|F_n-(f+N)\|_p\to 0$ as $n\to\infty$. After this we are done: $F_n\to f+N$ as $n\to\infty$ in $L^p(X,\mathcal{F},\mu)$.

So, if I understand things right, in the first step of the proof we need the Axiom of Countable Choice.

Question 1: am I right?

Question 2: is it possible for a space $L^p(X,\mathcal{F},\mu)$ to be non-complete without the Axiom of Countable Choice?

  • $\begingroup$ Is it clear that the proof $s_p(f) = 0 \implies \text{$f = 0$ a.e.}$ does not use any form of choice? (I have no reason to think it would, but I'm not accustomed to wondering about such things.) $\endgroup$
    – LSpice
    Feb 4, 2022 at 0:13
  • $\begingroup$ Also, as very much not an expert, my impression is that choice axioms are mainly about making choices simultaneously (i.e., showing that a product is inhabited). Maybe you can get away with choosing the representatives inductively? If so, would that still use ACC implicitly? $\endgroup$
    – LSpice
    Feb 4, 2022 at 0:15
  • 2
    $\begingroup$ @LSpice I can't comment about this specific situation, but yes, making choices inductively often requires choice axioms. In fact often, if the choices at different steps are to depend on earlier choices, you may need to use axiom of dependent choice, which is strictly stronger than countable choice. (edit: briefly looking at it, it doesn't seem like this argument requires dependent choice. I wouldn't be surprised if it did require countable choice) $\endgroup$
    – Wojowu
    Feb 4, 2022 at 1:14
  • 6
    $\begingroup$ My suspicion is that countable choice will creep into the argument in many ways, not just in this one relatively obvious place. Many fundamental facts in measure theory require countable choice. For instance, Lebesgue measure may fail to be countably additive due to the fact $\mathbb R$ might be a countable union of countable (hence measure zero) sets. This also implies monotone and dominated convergence theorems may fail. $\endgroup$
    – Wojowu
    Feb 4, 2022 at 1:26
  • 1
    $\begingroup$ I am not sure whether the axiom of choice is necessary for this particular step (although as others commented above, the choice might be necessary in the foundation), since it might be possible to invoke Lebesgue differentiation theorem to pick a canonical choice (take the average on balls of radius $1$). $\endgroup$
    – Z. M
    Feb 5, 2022 at 4:49


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