I got lost when reading the proof of $L^{\infty}$ is complete. The book proceed the proof as follows:
We show that each absolutely convergent series in $L^{\infty}(X,\mathscr{A},\mu)$ is convergent. We do this by considering functions (instead of equivalence classes) in $\mathscr{L}^{\infty}(X,\mathscr{A},\mu)$.
Let $\{f_k\}$ be a sequence of functions that belong to $\mathscr{L}^{\infty}(X,\mathscr{A},\mu)$ and satisfy $\sum_k\|f_k\|<+\infty$. For each positive integer $k$, let $N_k=\{x\in X:|f_k(x)|>\|f_k\|_{\infty}\}$. Then the series $\sum_kf_k(x)$ converges at each $x$ outside $\bigcup_kN_k$, and the function $f$ defined by
\begin{align*}
f(x) =
\begin{cases}
\sum_kf_k(x)\quad&\text{if $x\notin\bigcup_kN_k$},\\
\\
0\quad&\text{if $x\in\bigcup_kN_k$}
\end{cases}
\end{align*}
is bounded and $\mathscr{A}$-measurable. Since $\bigcup_kN_k$ is locally $\mu$-null, the inequality
\begin{align*}
\left\|f-\sum_{k=1}^nf_k\right\|_{\infty} \leq \sum_{k=n+1}^{\infty}\|f_k\|_{\infty}\tag1
\end{align*}
holds for each $n$, and so
\begin{align*}
\lim_{n\to\infty}\left\|f-\sum_{k=1}^nf_k\right\|_{\infty} \leq \lim_{n\to\infty}\sum_{k=n+1}^{\infty}\|f_k\|_{\infty} = 0.\tag2
\end{align*}
Thus $L^{\infty}(X,\mathscr{A},\mu)$ is complete.
I have a couple of questions about this proof.
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The definition of $L^p(X,\mathscr{A},\mu)$ says that the elements of $L^p(X,\mathscr{A},\mu)$ are equivalence classes of functions. Why is it legit to proceed the proof by considering functions in $\mathscr{L}^p(X,\mathscr{A},\mu)$?
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In the proof, it says "the series $\sum_kf_k(x)$ converges at each $x$ outside $\bigcup_kN_k$". Here is how I understand this step, and I want to know if it is correct?
If $x\notin\bigcup_{k=1}^{\infty}N_k$, then $|f_k(x)|\leq\|f_k\|_{\infty}$ for all $k\in\mathbb{N}$, and thus the convergence of $\sum_{k=1}^{\infty}\|f_k\|_{\infty}$ implies that $\sum_{k=1}^{\infty}f_k(x)$ converges by the comparison test.
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The proof claims that $f$ is $\mathscr{A}$-measurable. I couldn't see why this is true. I want to show that for each $t\in\mathbb{R}$ the set $\{x\in X:f(x)<t\}\in\mathscr{A}$. But I honestly don't know how to do this. Can someone please help me out?
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I got complete lost by inequality (1) and (2). Why does $\bigcup_kN_k$ being locally $\mu$-null imply that the inequality (1) holds for each $n$? Why can we just take limit for both side without proving the limits exist?
Please please help! I really appreciate it!
Note:$\quad$ The book I am reading defines $\|f\|_{\infty}$ to be the infimum of those nonnegative numbers $M$ such that $\{x\in X:|f(x)>M|\}$ is locally $\mu$-null.
Reference:$\quad$ Theorem 3.4.1 from Measure Theory by Donald Cohn
Best Answer
Indeed, let $A_n = \{f_n \neq g_n\} \ \forall n, Cf = \{\lim_n f_n \neq f\}$ and set $$Cg = \left(\bigcup_{n \in \mathbb{N}} A_n \right) \cup Cf$$
Then $I$ has zero measure, and on $I^c$, the complement of $I$, $f_n = g_n \ \forall n, f_n \rightarrow f$. A simple argument will tell that $\lim_n g_n(x) = f(x) \ \forall x \in I^c$. Thus, $g_n \rightarrow f$ a.e
With this statement, the convergence property of a sequence in $L^p$ doesn't depend on the choice of equivalent classes of functions. If one choice of a sequence of functions that converges, every other choice that is equivalent to this sequence will also converge.
The function $g_N$ is the sum of $N$ measurable functions so it is measurable. The limit of $g_N$ is therefore also measurable.
4. $$ f - \sum_{k = 1}^n f_k = \sum_{k = 1}^n f_k \cdot (1_{N^c} - 1) + \sum_{k = n + 1}^\infty f_k \cdot 1_{N^c} $$
Thus, $$ \left\lVert f - \sum_{k = 1}^n f_k \right\rVert \le \left\lVert \sum_{k = 1}^n f_k \cdot (1_{N^c} - 1) \right\rVert + \left\lVert \sum_{k = n + 1}^\infty f_k \cdot 1_{N^c} \right\rVert $$
Notice that the first norm is zero because the sum inside is zero a.e. Now, we want to show that $$ \left\lVert \sum_{k = n + 1}^\infty f_k \cdot 1_{N^c} \right\rVert \le \sum_{k = n + 1}^\infty \lVert f_k \cdot 1_{N^c} \rVert $$
By definition, for each $N \in \mathbb{N}, N \ge n + 1$, and for every $x \in X$, we have $$ \left\vert \sum_{k = n + 1}^N f_k(x) \cdot 1_{N^c}(x) \right\vert \le \sum_{k = n + 1}^N \vert f_k(x) \cdot 1_{N^c}(x) \vert \le \sum_{k = n + 1}^N \lVert f_k \cdot 1_{N^c} \rVert \le \sum_{k = n + 1}^\infty \lVert f_k \cdot 1_{N^c}\rVert $$
Now, let $N \rightarrow \infty$, since $$ \sum_{k = n + 1}^N f_k(x) \cdot 1_{N^c}(x) \rightarrow \sum_{k = n + 1}^\infty f_k(x) \cdot 1_{N^c}(x) $$ we must have $$ \left\vert \sum_{k = n + 1}^\infty f_k(x) \cdot 1_{N^c}(x) \right\vert \le \sum_{k = n + 1}^\infty \lVert f_k \cdot 1_{N^c}\rVert \hspace{0.5cm} (\color{red}{1}) $$ By definition of the norm, we have $$ \left\lVert \sum_{k = n + 1}^\infty f_k \cdot 1_{N^c} \right\rVert = \inf\left\{M : \left\{\left\vert \sum_{k = n + 1}^\infty f_k(x) \cdot 1_{N^c}(x) \right\vert > M\right\} \text{ is locally } \mu-\text{null} \right\} $$ By $(\color{red}{1})$, the set $$ \left\{\left\vert \sum_{k = n + 1}^\infty f_k(x) \cdot 1_{N^c}(x) \right\vert > \sum_{k = n + 1}^\infty \lVert f_k \cdot 1_{N^c}\rVert\right\} $$ is an empty set, and thus is a (locally) $\mu$-null set. Therefore, $$ \left\lVert \sum_{k = n + 1}^\infty f_k \cdot 1_{N^c} \right\rVert \le \sum_{k = n + 1}^\infty \lVert f_k \cdot 1_{N^c}\rVert $$
Finally, notice that $$ f_k \cdot 1_{N^c} = f_k \text{ a.e} \Longrightarrow \lVert f_k \cdot 1_{N^c} \rVert = \lVert f_k \rVert $$ Thus, $$ \left\lVert f - \sum_{k = 1}^n f_k \right\rVert \le \sum_{k = n + 1}^\infty \lVert f_k \rVert $$
For the rest, since $$ 0 \le \left\lVert f - \sum_{k = 1}^n f_k \right\rVert \le \sum_{k = n + 1}^\infty \lVert f_k \rVert $$ and $$ \lim_{n \rightarrow \infty} \sum_{k = n + 1}^\infty \lVert f_k \rVert = 0 $$ the squeeze theorem implies the limit of $\left\lVert f - \sum_{k = 1}^n f_k \right\rVert$ exists and equals to $0$