[proofplan]
We use the criterion that a normed space is complete if and only if every absolutely convergent series converges. The case $p \in [1, \infty)$ uses the monotone convergence theorem to control the partial sums of $|f_n|$ and the dominated convergence theorem to upgrade pointwise convergence of $\sum f_k$ to $L^p$-convergence. The case $p = \infty$ is handled directly: we replace the essentially bounded representatives by everywhere bounded ones on the complement of a single null set, where the sequence becomes uniformly Cauchy in the sup norm. The bulk of the work lies in producing an integrable dominator and extracting an a.e. pointwise limit.
[/proofplan]
[step:Reduce completeness to convergence of absolutely convergent series]
A normed space $X$ is complete if and only if every absolutely convergent series converges in norm. Specifically: $X$ is complete iff whenever $(x_n) \subset X$ satisfies $\sum_{n=1}^\infty \|x_n\|_X < \infty$, the series $\sum_{n=1}^\infty x_n$ converges in $X$.
We accept this criterion as known (it follows by extracting a fast-Cauchy subsequence from any Cauchy sequence). Hence to prove $L^p(\mu)$ is complete, it suffices to take any $(f_n) \subset L^p(\mu)$ with $\sum_{n=1}^\infty \|f_n\|_p < \infty$ and exhibit a limit $f \in L^p(\mu)$ with $\|\sum_{k=1}^n f_k - f\|_p \to 0$.
[/step]
[step:For $p \in [1, \infty)$, build the dominator $s = \sum_{k=1}^\infty |f_k|$ and bound $\|s\|_p$]
Fix $p \in [1, \infty)$ and let $(f_n) \subset L^p(\mu)$ with $M := \sum_{n=1}^\infty \|f_n\|_p < \infty$. Define
\begin{align*}
s_n: \Omega &\to [0, \infty) \\
\omega &\mapsto \sum_{k=1}^n |f_k(\omega)|, \qquad s := \lim_{n \to \infty} s_n = \sum_{k=1}^\infty |f_k|: \Omega \to [0, \infty].
\end{align*}
Each $|f_k|$ is measurable and finite-valued $\mu$-a.e., so $s_n$ is measurable and $s = \lim_n s_n$ is measurable as a pointwise limit of measurable functions (with values in $[0, \infty]$).
By the triangle inequality for the $L^p$ norm, for each $n$
\begin{align*}
\|s_n\|_p = \Big\|\sum_{k=1}^n |f_k|\Big\|_p \le \sum_{k=1}^n \||f_k|\|_p = \sum_{k=1}^n \|f_k\|_p \le M.
\end{align*}
The sequence $s_n^p$ is non-negative, measurable, and increases pointwise to $s^p$ (since $s_n \uparrow s$ pointwise and $t \mapsto t^p$ is monotone on $[0, \infty]$). The monotone convergence theorem applied to $(s_n^p)$ gives
\begin{align*}
\int_\Omega s^p \, d\mu = \lim_{n \to \infty} \int_\Omega s_n^p \, d\mu = \lim_{n \to \infty} \|s_n\|_p^p \le M^p.
\end{align*}
In particular, $\int_\Omega s^p \, d\mu \le M^p < \infty$, so $s^p \in L^1(\mu)$ and consequently $s < \infty$ $\mu$-a.e.: indeed, $\mu(\{s = \infty\}) > 0$ would force $\int s^p \, d\mu = \infty$.
[guided]
Why is the monotone convergence theorem the right tool here? The hypotheses are: a non-negative sequence increasing pointwise. We have $s_n^p \ge 0$ and $s_n^p \uparrow s^p$ (because the partial sums $s_n = |f_1| + \cdots + |f_n|$ are increasing in $n$ at each $\omega$, and $t \mapsto t^p$ is monotone on $[0, \infty]$). Both hypotheses are met, and MCT gives us the swap $\int \lim = \lim \int$ that we need.
The bound $\|s_n\|_p \le M$ comes from the triangle inequality applied $n-1$ times, then bounded uniformly by $\sum_{k=1}^\infty \|f_k\|_p = M$. After taking the limit, $\int s^p \, d\mu \le M^p$, so $s^p$ is integrable. The set where $s = \infty$ must therefore be $\mu$-null: any positive measure of $\{s = \infty\}$ would contribute $+\infty$ to the integral $\int s^p \, d\mu$ (since $s^p = \infty$ there), contradicting integrability.
[/guided]
[/step]
[step:Define the candidate limit $f$ and verify $f \in L^p(\mu)$]
Let $N := \{\omega \in \Omega : s(\omega) = \infty\}$, a $\mu$-null measurable set by the previous step. On $\Omega \setminus N$ the series $\sum_{k=1}^\infty f_k(\omega)$ is absolutely convergent in $\mathbb{F}$ (its absolute value is bounded by $s(\omega) < \infty$), hence convergent. Define
\begin{align*}
f: \Omega &\to \mathbb{F} \\
\omega &\mapsto \begin{cases} \sum_{k=1}^\infty f_k(\omega) & \omega \in \Omega \setminus N, \\ 0 & \omega \in N. \end{cases}
\end{align*}
This is measurable (limit of $\mathbb{1}_{\Omega \setminus N} \sum_{k=1}^n f_k$ pointwise) and $|f(\omega)| \le s(\omega)$ for all $\omega \in \Omega$ (with $|f| = 0 \le \infty = s$ on $N$). Therefore $|f|^p \le s^p \in L^1(\mu)$, so $f \in L^p(\mu)$ with $\|f\|_p \le \|s\|_p \le M$.
[/step]
[step:Apply dominated convergence to obtain $\|\sum_{k=1}^n f_k - f\|_p \to 0$]
Set $S_n := \sum_{k=1}^n f_k$. On $\Omega \setminus N$ we have $S_n(\omega) \to f(\omega)$ by definition of $f$, so $S_n \to f$ $\mu$-a.e. Also,
\begin{align*}
|S_n - f|^p \le (|S_n| + |f|)^p \le (s + s)^p = (2s)^p \quad \text{on } \Omega \setminus N,
\end{align*}
with $\int (2s)^p \, d\mu = 2^p \int s^p \, d\mu \le 2^p M^p < \infty$. So $(2s)^p \in L^1(\mu)$ is an integrable dominator for $|S_n - f|^p$.
The dominated convergence theorem applied to $|S_n - f|^p \to 0$ $\mu$-a.e., dominated by $(2s)^p \in L^1(\mu)$, gives
\begin{align*}
\|S_n - f\|_p^p = \int_\Omega |S_n - f|^p \, d\mu \to 0,
\end{align*}
so $\|S_n - f\|_p \to 0$. Hence the absolutely convergent series $\sum f_n$ converges to $f \in L^p(\mu)$, and $L^p(\mu)$ is complete in the case $p \in [1, \infty)$.
[guided]
The dominated convergence theorem requires three ingredients: a sequence of measurable functions, pointwise a.e. convergence to a limit, and an integrable dominator for the absolute values. We verify all three explicitly:
1. **Measurable sequence**: Each $|S_n - f|^p$ is measurable (composition of $|\cdot|^p$ with the difference of measurable functions).
2. **Pointwise a.e. convergence**: On $\Omega \setminus N$, $S_n(\omega) \to f(\omega)$ by construction of $f$ as the sum of the convergent series. So $|S_n - f|^p \to 0$ on $\Omega \setminus N$, and $\mu(N) = 0$.
3. **Integrable dominator**: $|S_n| \le s$ pointwise on $\Omega \setminus N$ (triangle inequality applied to the partial sum) and $|f| \le s$ pointwise on $\Omega$. Hence $|S_n - f|^p \le (s + s)^p = 2^p s^p$ on $\Omega \setminus N$. Since $\int s^p \, d\mu \le M^p < \infty$ from Step 2, we have $2^p s^p \in L^1(\mu)$.
DCT then gives the swap $\lim \int |S_n - f|^p \, d\mu = \int \lim |S_n - f|^p \, d\mu = 0$, i.e. $\|S_n - f\|_p \to 0$.
[/guided]
[/step]
[step:For $p = \infty$, lift to genuinely bounded representatives by removing a single null set]
Now consider $p = \infty$. Let $(f_n) \subset L^\infty(\mu)$ be Cauchy in $L^\infty(\mu)$. For each $n$, choose a representative (still denoted $f_n$) and a $\mu$-null set $N_n \subset \Omega$ such that $|f_n(\omega)| \le \|f_n\|_\infty$ for all $\omega \in \Omega \setminus N_n$ (the essential supremum is attained as a true supremum off some null set). For each pair $(i, j)$, choose a $\mu$-null set $N_{ij} \subset \Omega$ such that $|f_i(\omega) - f_j(\omega)| \le \|f_i - f_j\|_\infty$ for all $\omega \in \Omega \setminus N_{ij}$.
Let $N := \bigcup_n N_n \cup \bigcup_{i, j} N_{ij}$. As a countable union of $\mu$-null sets, $\mu(N) = 0$. On $\Omega \setminus N$ we have, simultaneously for all $n$ and all $(i, j)$:
\begin{align*}
|f_n(\omega)| \le \|f_n\|_\infty, \qquad |f_i(\omega) - f_j(\omega)| \le \|f_i - f_j\|_\infty.
\end{align*}
[/step]
[step:Extract a uniform pointwise limit $f$ on $\Omega \setminus N$]
On $\Omega \setminus N$ the sequence $(f_n)$ is uniformly Cauchy: for every $\varepsilon > 0$ there exists $N_\varepsilon \in \mathbb{N}$ with $\|f_i - f_j\|_\infty < \varepsilon$ for all $i, j \ge N_\varepsilon$ (by Cauchy-ness in $L^\infty$), so
\begin{align*}
\sup_{\omega \in \Omega \setminus N} |f_i(\omega) - f_j(\omega)| < \varepsilon \quad \text{for } i, j \ge N_\varepsilon.
\end{align*}
Uniform Cauchyness in the sup norm on $\Omega \setminus N$ gives a function $f: \Omega \setminus N \to \mathbb{F}$ to which $(f_n)$ converges uniformly: define $f(\omega) := \lim_{n \to \infty} f_n(\omega)$ pointwise on $\Omega \setminus N$ (the limit exists by the Cauchy property in $\mathbb{F}$), and the convergence is uniform by the standard $\varepsilon/3$ argument. Extend $f$ to $\Omega$ by setting $f := 0$ on $N$.
[/step]
[step:Conclude $f \in L^\infty(\mu)$ and $\|f_n - f\|_\infty \to 0$]
Since $\|f_n\|_\infty$ is bounded (a Cauchy sequence in any normed space is bounded), let $K := \sup_n \|f_n\|_\infty < \infty$. Then $|f_n(\omega)| \le K$ on $\Omega \setminus N$ for all $n$, so passing to the limit, $|f(\omega)| \le K$ on $\Omega \setminus N$. Combined with $f = 0 \le K$ on $N$, we get $|f| \le K$ everywhere, hence $\operatorname{ess\,sup}_\Omega |f| \le K$, so $f \in L^\infty(\mu)$.
For convergence: given $\varepsilon > 0$, choose $N_\varepsilon$ as in the uniform Cauchy bound. For $n \ge N_\varepsilon$ and $\omega \in \Omega \setminus N$, $|f_n(\omega) - f(\omega)| = \lim_{j \to \infty} |f_n(\omega) - f_j(\omega)| \le \varepsilon$. Therefore $|f_n - f| \le \varepsilon$ on $\Omega \setminus N$, so $\|f_n - f\|_\infty \le \varepsilon$ for $n \ge N_\varepsilon$. This proves $\|f_n - f\|_\infty \to 0$.
Combining the cases $p \in [1, \infty)$ and $p = \infty$, $L^p(\mu)$ is complete for all $p \in [1, \infty]$, hence is a Banach space.
[guided]
The technical heart of the $p = \infty$ case is upgrading "essentially bounded" estimates to "everywhere bounded" estimates by removing a single null set $N$. There are two classes of statements that fail off some null set:
- For each $n$, $|f_n| \le \|f_n\|_\infty$ off $N_n$.
- For each pair $(i, j)$, $|f_i - f_j| \le \|f_i - f_j\|_\infty$ off $N_{ij}$.
Both classes are countable: $\{N_n : n \in \mathbb{N}\}$ and $\{N_{ij} : (i, j) \in \mathbb{N} \times \mathbb{N}\}$ are countable. So the union $N$ of all of them is a countable union of $\mu$-null sets, hence $\mu$-null. On $\Omega \setminus N$ all the bounds hold simultaneously.
This is the standard pattern: when working with $L^\infty$ representatives and finitely or countably many a.e. inequalities, gather all the exceptional null sets into one big union; on its complement, the representatives satisfy everywhere what they originally satisfied a.e. From there, the rest is sup-norm Cauchyness in $B(\Omega \setminus N)$ (the space of bounded functions), which is the standard completeness of bounded sequences with the sup norm, and yields a uniform limit $f$. Extending $f$ by $0$ on $N$ does not affect the $L^\infty$ norm (since $\mu(N) = 0$).
[/guided]
[/step]
