Understanding Proof: Simple Functions In $\mathscr{L}^{\infty}(X,\mathscr{A},\mu)$ Form A Dense Subspace of $\mathscr{L}^{\infty}(X,\mathscr{A},\mu)$

Question

I have difficulties understanding the proof of the following proposition:

Proposition$\quad$ Let $(X,\mathscr{A},\mu)$ be a measure space, and let $p$ satisfy $1\leq p$. Then the simple functions in $\mathscr{L}^{\infty}(X,\mathscr{A},\mu)$ form a dense subspace of $\mathscr{L}^{\infty}(X,\mathscr{A},\mu)$

The book defines $\mathscr{L}^{\infty}$ as follows:

Definition$\quad$ Let $\mathscr{L}^{\infty}(X,\mathscr{A},\mu,\mathbb{R})$ be the set of all bounded real-valued $\mathscr{A}$-measurable functions on $X$. Define $\|\cdot\|_{\infty}$ by letting $\|f\|_{\infty}$ be the infimum of those nonnegative numbers $M$ such that $\{x\in X:|f(x)|>M\}$ is locally $\mu$-null.

We only consider real-valued functions. Here is the proof:

Proof$\quad$ Let $f$ belong to $\mathscr{L}^{\infty}(X,\mathscr{A},\mu,\mathbb{R})$, and let $\epsilon$ be a positive number. Choose real numbers $a_0,a_1,\dots,a_n$ such that \begin{align} a_0<a_1<\dots<a_n \end{align} and such that the intervals $(a_{i-1},a_i]$ cover the interval $[-\|f\|_{\infty},\|f\|_{\infty}]$ and have lengths at most $\epsilon$. Let $A_i=f^{-1}((a_{i-1},a_i])$ for $i=1,\dots,n$, and let $f_{\epsilon}=\sum_{i=1}^na_i\chi_{A_i}$. Then $f_{\epsilon}$ is a simple $\mathscr{A}$-measurable function that satisfies $\|f-f_{\epsilon}\|_{\infty}<\epsilon$. Since $f$ and $\epsilon$ are arbitrary, the proof is complete.

So my understanding of the idea of this proof is the following:

For each $f\in\mathscr{L}^{\infty}(X,\mathscr{A},\mu,\mathbb{R})$ and each $\epsilon>0$, we want to construct a simple function $f_{\epsilon}$ in $\mathscr{L}^{\infty}(X,\mathscr{A},\mu,\mathbb{R})$ so that $$ \|f-f_{\epsilon}\|_{\infty} = \inf\Big\{M\in\mathbb{R}_+:\{x: X:|f(x) - f_{\epsilon}(x)|>M\}\ \text{is locally $\mu$-null}\Big\}<\epsilon. $$ So, if we can construct $f_{\epsilon}$ such that $\{x\in X:|f(x)-f_{\epsilon}(x)|>\epsilon\}$ is locally $\mu$-null, then we are done.

I can see that in the proof, if $x\in A_i$ for some $i$, then $f(x)\in(a_{i-1},a_i]$ and $f_{\epsilon}(x)=a_i$, and so $|f(x)-f_{\epsilon}(x)|<\epsilon$. But I don't understand why we need to let the intervals $(a_{i-1},a_i]$ cover the interval $[-\|f\|_{\infty},\|f\|_{\infty}]$. I don't think I understand how $f_{\epsilon}$ is constructed either.

Could someone please help me out? Thanks a lot in advance!

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Reference:$\quad$ Proposition 3.4.2 from Measure Theory by Donald Cohn.

Answer 1

If the intervals don't cover, what are we really doing? The idea of the infinity norm is that $f$ should be essentially concentrated to the range $[-M,M]$, but if we dare to miss a bit we can't get an arbitrarily close estimate. Just think of a basic example like $f(x)=x$ on the measure space $[0,1]$. If we approximate $f$ by some $\sum_i a_i\chi_{A_i}$ where $\bigcup A_i\subseteq(\frac{1}{3},\frac{2}{3})$ (say), as an example where the intervals don't cover, then the difference between $f$ and this sum is going to be... well, the whole of $f$, for $x$ in the missed-out bits $(0,\frac{1}{3}),(\frac{2}{3},1)$ and $f$ has nonzero values there, nonzero norm.

Remark: this "locally $\mu$-null" thing is curious, I have not come across it but have looked at lots of stuff with measure theory. I'm very used to $L^\infty$ being defined via genuinely $\mu$-null sets, and the $\|\cdot\|_\infty$ norm also being defined in this way. But whatever - one basic idea to proving $\|f-f_\epsilon\|<\epsilon$ is to "get" $f$ on $[-\|f\|_\infty,\|f\|_\infty]$ and throw away what's left (again, this is the point of $\infty$-norm). So let $A:=\bigcup_i A_i=f^{-1}[-\|f_\infty\|,\|f\|_\infty]$; $f=f\chi_A+f\chi_{X\setminus A}=\sum_{i=0}^nf\chi_{A_i}+f\chi_{X\setminus A}$. Here I need the $A_i$ to cover $A$!

Then by the triangle inequality, $\|f-f_\epsilon\|_\infty\le\|f\chi_{X\setminus A}\|_\infty+\sum_{i=0}^n\|(f-a_i)\chi_{A_i}\|$ and hopefully the result is clear now. On each $A_i$, $|f-a_i|$ is literally less than $\epsilon$, our locally $\mu$-null set can just be the empty set... (with $M:=a_{i+1}-a_i<\epsilon$) and hopefully you can realise $M=0$ works, for $f\chi_{X\setminus A}$, so we see this has norm zero. You have to do a continuity of measure argument but it's not too bad; note $\|f\|_\infty$ is the limit of a decreasing sequence of feasible "$M$"s. I can elaborate if you like.

Answer 2

Without loss of generality, one can assume that $(X,\mathscr{A},\mu)$ is complete, that is all $\mu$-null sets are in $\mathscr{A}$ (otherwise use the completion $\sigma$-algebra $\overline{\mathscr{A}},$ and the completion measure $\overline{\mu}$. Doing so does not change the $\mathcal{L}_p(mu)$ spaces (modulo sets of measure $0$).

I gather the textbook you are using uses the weak--norm defined as $$\|f\|_\infty=\inf\{M>0: \mu(A\cap\{|f|>M\})=0,\,\forall A\in\mathscr{A},\,\text{wth}\,\mu(A)<\infty\}$$

Notice that $\mu(A\cap\{|f|>\|f\|_\infty\})=0$ for all measurable set $A$ with $\mu(A)<\infty$ for if $m_n>0$ is a strictly decreasing sequence converging to $\|f\|_\infty$ and such that $\{|f|>m_n\}$ is locally $\mu$-null, then for any measurable set $A$ of finite measure $\mu$ $$0=\mu(A\cap\{|f|>m_n\})\nearrow\mu(A\cap\{|f|>\|f\|_\infty\})$$

To make the argument simpler, suppose that $f\geq0$. The sequence of simple functions defined by $$s_n= \sum^{n2^n-1}_{k=0}2^{-n}k\mathbb{1}_{\{2^{-n}k\leq f<(k+1)2^{-n}\}} + n\mathbb{1}_{\{f>n\}} $$ converge point wise to $f$. Furthermore, on $\{0\leq f\leq \|f\|_\infty\}$, $s_n$ converges uniformly to $f$ (for the dyadic numbers $2^{-n}k$, $n\in\mathbb{N}$, $k\in\mathbb{N}$ are dense in $[0,\infty)$. Hence, for $\varepsilon>0$, there exists $N>0$ such that if $n\geq N$, and $x\in \{0\leq f\leq \|f\|_\infty\}$, $$|f(x)-s_n(x)|\leq \varepsilon$$ Let $A$ be a measurable set of finite measure. For $n\geq N$, $$A\cap\{|f-s_n|>\varepsilon\}\subset A\cap\{f>\|f\|_\infty\}$$ Consequently, for $n\geq N$ $$\mu(A\cap\{|f-s_n|>\varepsilon\})=0$$ That is, for $n\geq N$, $\|f-s_n\|_\infty\leq\varepsilon$. This shows that $s_n\xrightarrow{n\rightarrow\infty}f$ in $\mathscr{L}_\infty(\mu)$.

For general real-valued $f$, one can use the same argument for $f_+$ and $f_-$ separately.

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Comment: A similar argument works of $\mathcal{L}_\infty(\mu)$ is defined by the stronger norm

$$\|f\|^e_\infty=\inf\{M>0: \mu(|f|>M)=0\}$$

Of course, if the ambient space $(X,\mathscr{A},\mu)$ is $\sigma$-finite, the weak and strong $\mathscr{L}_\infty$ norms coincide.

Notice that although simple function are dense in $\mathscr{L}_\infty$, $L_\infty$ is not separable unless $dim(\mathscr{L}_\infty)<\infty$.