[Math] On a proof of Riesz-Fischer Theorem

Question

Questions : [See below for context.]

$\rm\color{#c00}{a)}$ First, is the proof presented below $100$ % correct ?

$\rm\color{#c00}{b)}$ How would one justify the LHS of $(2)$ ? Are my thoughts correct ?

Thoughts : One way I see it is in three times : first we note that $(\cdot)^{1/p}$ is continuous, then we apply the Monotone Convergence Theorem (MON) and finally we note that $|\cdot|^p$ is continuous, so that all in all we can bring the limit all the way in (even if $\displaystyle\sum_{k\geq1} |f_k|$ is infinite in this case).

Another way I see it is by noting that every norm is (Lipschitz) continuous, so that we can effectively bring the limit inside (even if it is infinite in this case).

$\rm\color{#c00}{c)}$ Why must we take the absolute value of $f_k$ when investigating $\displaystyle\sum|f_k|$ ? Are my thoughts correct ?

Thoughts : I think that MON would not apply for $\rm\color{#c00}{b)}$ and I think that if we did not take the absolute value, then we could not speak, a priori, of the limit, that is, $\displaystyle\sum_{k\geq1} f_k$, in $(2)$. Taking the absolute value gives a series with positive terms and such a series either converges or diverges to $\infty$. A series with negative terms could diverge by oscillating.

$\rm\color{#c00}{d)}$ In $(4)$, defining $F$ by $0$ is purely arbitrary, right ? We could define $F$ to be anything on the null set, right ? Also, $\displaystyle\sum_{k\geq1} f_k$ does exist on the given set since absolute convergence implies convergence for series of numbers, right ?

$\rm\color{#c00}{e)}$ The equality $(5)$ is true since
$$
F-\sum_{k=1}^{n}f_k=\sum_{k=n+1}^{\infty}f_k
$$
almost everywhere, right ?

$\rm\color{#c00}{f)}$ Would it be fair to say that what was sufficient to prove was $\displaystyle\sum_{k\geq1}|f_k|<\infty$ a.e., [added] and that the reason the proof is not trivial is that $L^p(\mu)$ contains, by definition, finite-valued functions ? I mean, in measure theory we work with functions with range in the extended real line, $[-\infty,\infty]$, and if we allowed the functions in $L^p(\mu)$ to take infinite values, then in fact every series would be convergent (trivially) in $L^p(\mu)$. Is this correct ?

Def. : Let $(X,\cal{A},\mu)$ be a measured space. Then for $1\leq p<\infty$ we define
$$
L^p(\mu):=\{f:X\to\mathbb{R}\text{ measurable and s.t. }\|f\|_p<\infty\},
$$
where $\|f\|_p:=\left(\int|f|^pd\mu\right)^{1/p}$. As usual we really take equivalence classes of functions differing only on a null set.

Thm (Riesz-Fischer) : $(L^p(\mu),\|\cdot\|_p)$ is complete for $1\leq p<\infty$.

Dem. : We know it suffices to show that every absolutely convergent series converges.

Let $(f_k)_{k\geq1}\subset L^p(\mu)$ be a sequence such that
$$
\sum_{k=1}^{\infty}\|f_k\|_p<\infty.\tag{0}
$$
Since, by Minkowski's inequality, one has
$$
\left\|\sum_{k=1}^n|f_k|\right\|_p\leq\sum_{k=1}^n\|f_k\|_p,\tag{1}
$$
letting $n\to\infty$ gives
$$
\left\|\sum_{k=1}^{\infty}|f_k|\right\|_p\leq\sum_{k=1}^{\infty}\|f_k\|_p<\infty,\tag{2}
$$
thus
$$
\sum_{k=1}^{\infty}|f_k|<\infty\quad\text{a.e.}\tag{3}
$$
Now define
$$
F:=\begin{cases}\displaystyle\sum_{k\geq1}f_k&\text{on }\left\{\displaystyle\sum_{k\geq1}|f_k|<\infty\right\},\\0&\text{otherwise}.\end{cases}\tag{4}
$$
Then $F$ is in $L^p(\mu)$ since $|F|\leq\displaystyle\sum_{k\geq1}|f_k|$ and by $(2)$. Also, $F$ is such that
$$
\sum_{k=1}^{\infty}f_k=F,
$$
because
\begin{align}
\left\|F-\sum_{k=1}^{n}f_k\right\|_p&=\left\|\sum_{k=n+1}^{\infty}f_k\right\|_p\tag{5}\\
&\leq\left\|\sum_{k=n+1}^{\infty}|f_k|\right\|_p\\
&\stackrel{(2)}{\leq}\sum_{k=n+1}^{\infty}\|f_k\|_p\xrightarrow[(n\to\infty)]{(0)}0.\quad\text{Q.E.D.}
\end{align}

Answer 1

1. Yes.
2. This has nothing to do with continuity; only with the monotone convergence theorem. Note that we cannot use that continuity of the $L^p$-norm: In order to deduce $$\|g_n\|_p \to \|g\|_p$$ from the continuity of the norm, we already have to know that $g_n \to g$ in $L^p$ (recall that the norm is continuous as a mapping $\|\cdot\|: (L^p,\|\cdot\|_p) \to (\mathbb{R},|\cdot|)$. The argumentation goes as follows: By Minkowski's inequality, $$\left\| \sum_{k=1}^n |f_k| \right\|_p \leq \sum_{k=1}^n \|f_k\|_p \leq \sum_{k=1}^{\infty} \|f_k\|_p. \tag{1}$$ Since $|f_k| \geq 0$, we know $$\sum_{k=1}^n |f_k| \uparrow \sum_{k=1}^{\infty} |f_k| \qquad \text{as} \,\, n \to \infty. \tag{2}$$ Therefore, it follows from the monotone convergence theorem that $$\left\| \sum_{k=1}^{\infty} |f_k| \right\|_p = \sup_{n \in \mathbb{N}} \left\| \sum_{k=1}^n |f_k| \right\|_p \stackrel{(1)}{\leq} \sum_{k=1}^{\infty} \|f_k\|_p < \infty.$$
3. Yes, that's correct. The series $\sum_{k=1}^{\infty} |f_k| \in [0,\infty]$ is well-defined since each addend is non-negative. Moreover, the non-negativity is crucial in $(2)$ (otherwise we cannot apply monotone convergence theorem).
4. Yes.
5. Yes.
6. No, it's not that easy. If I understand you correctly, you would like to set $$F := \sum_{k=1}^{\infty} f_k$$ (as a pointwise limit) which might take values in $[-\infty,\infty]$. The problem is that the series $$\sum_{k=1}^{\infty} f_k$$ is not necessarily convergent, consider e.g. $f_k = (-1)^k$. This means that $F$ is not well-defined on some exceptional set and we have to ensure that this exceptional set is small (in $L^p$-sense). That's exactly what the proof above is about: We show that the exceptional set has measure $0$.