I have a proof of $L^1$ being complete (no need to read all of it since it is the traditional proof):
Let $\{f_n\}_{n=1}^\infty$ be a Cauchy sequence in $L^1(X)$. It is sufficient to find a convergent subsequence. By definition, for all $k>0$, there is some $N_k$ such that $||f_n-f_m||<2^{-k}$ for all $n,m\geq N_k$. Set $g_k:=f_{N_k}$. Then $||g_k-g_{k+1}||\leq2^{-k}$ as we can pick $N_{k+1}> N_k$.
Now, set $h_1(x)=|g_1(x)|$ and $$h_{n+1}(x)=|g_1(x)|+\sum_{k=1}^{n}|g_{k+1}(x)-g_k(x)|$$ Then $\{h_n\}_{n=1}^\infty$ is nonnegative and nondecreasing. We also have by the linearity of the integral $$\int h_n=||{g_1}||+\sum_{k=1}^n||{g_{k+1}-g_k}||\leq ||{g_1}||+\sum_{k=1}^n{2^{-k}}\leq||{g_1}||+1$$ Thus $\lim \int h_n\leq ||{g_1}||+1$. Applying the monotone convergence theorem to $\{h_n\}_{n=1}^\infty$ gives us that the limit $$|g_1(x)|+\sum_{n=1}^\infty |g_{n+1}(x)-g_n(x)|$$ is integrable since it is a non-negative function that has a finite integral. This also implies that it is finite a.e. Writing $$g_n(x)=g(x)+\sum_{k=1}^n[g_{k+1}(x)-g_k(x)]$$ gives us that $f(x)=\lim g_n(x)$ exists (it is the sum of an absolutely convergent series) for a.e $x$ (so $f\in L^1$). Finally, to show convergence in the $L^1$-norm, we note that $$|f(x)-g_n(x)|\leq\left\lvert\sum_{k=n}^\infty [g_{k+1}(x)-g_k(x)]\right\rvert\leq \sum_{k=n}^\infty |g_{k+1}(x)-g_k(x)|$$ so then we can apply the monotone convergence theorem to the partial sums we get $$||{f(x)-g_n(x)}||\leq \sum_{k=n}^\infty ||{g_{k+1}(x)-g_k(x)}||\leq 2^{1-n}$$ and thus $f$ is the desired limit.
I have a few questions:
- Is the reason why we only need to show that $\lim g_n(x)$ exists for a.e. $x$ because the $L^1$ space does not care about null sets? It seems to me that $f$ does not equal $\lim g_n(x)$ on all $x$ because $\lim g_n(x)$ could take on values of infinity, which is not in $\mathbb{R}$ (but $f$ maps to $\mathbb{R}$). Would $f$ technically be a function that agrees with $\lim g_n(x)$ pointwise everywhere except when $\lim g_n(x)=\infty$ (taking any value in $\mathbb{R}$ on those points)? If $f$ is not precisely $\lim g_n(x)$, does that break the last part of the proof?
- Building off of that idea, if I had a sequence of functions $f_n\rightarrow f$ for a.e. $x$, would I be able to say $f = \liminf f_n$ even if the convergence is not strictly pointwise? I guess in general, what does "$\leq$" mean in $L^1$?