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?

simultaneously(i.e., showing that a product is inhabited). Maybe you can get away with choosing the representativesinductively? If so, would that still use ACC implicitly? $\endgroup$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$5more comments