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Let's consider the space $\mathcal{L}^\infty(X,\Sigma, \mu)$ that consists of all essentially bounded functions $f$ on $X$, i.e. all such $f$ that $$\mathrm{ess}\sup f=\inf\{\alpha>0\ \colon\ \mu\{x\in X\ |\ |f(x)|>\alpha\}=0\} < \infty.$$ Here $X$ is a metric space, $\Sigma\subset 2^X$ — sigma-algebra on $X$ and $\mu$ — is the measure on $\Sigma$. As usual, functions that are equal almost everywhere considered to be equal, so technically we are dealing with classes $$[f]=\{g\in \mathcal{L}^\infty(X,\Sigma, \mu) \colon \mu \{x\in X\colon f(x)\neq g(x)\}=0\}.$$ This space is endowed with metric $\rho_\infty(f,g)=\mathrm{ess}\sup|f-g|$.

The question is what necessary and sufficient constraints should be laid on $\Sigma$ to make our space separable?

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    $\begingroup$ Whenever there is a sequence $A_n$ of pairwise disjoint measurable sets with $\mu(A_n)>0$ the mapping $\ell^\infty \to \mathcal L^\infty(X,\Sigma,\mu)$, $(x_n)_n \mapsto \sum_n x_n1_{A_n}$ is an isometry. Hence your space is not separable in that case. Therefore I would guess that $\mathcal L^\infty(X,\Sigma,\mu)$ is separable only if it is finite dimensional. $\endgroup$ – Jochen Sep 22 '15 at 10:34
  • $\begingroup$ @Jochen, what do mean by $1_{A_n}$? Indicator of $A_n$? $\endgroup$ – Glinka Sep 22 '15 at 11:36
  • $\begingroup$ Yes. The function associated to $(x_n)_n$ is constant and equal to $x_k$ on each $A_k$ and $0$ on the complement of $\bigcup_nA_n$. $\endgroup$ – Jochen Sep 22 '15 at 14:14
  • $\begingroup$ @Jochen, thanks, that gives a clue! $\endgroup$ – Glinka Sep 27 '15 at 13:55
  • $\begingroup$ The function $1_A$ is commonly denoted $\chi_A$ (Greek chi) and is called the characteristic function of $A$ $\endgroup$ – DanielWainfleet Oct 3 '15 at 7:03
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The space will be separable if and only if $\Sigma$ has only finitely many sets of positive measure (by ignoring null sets, from here on out I will just refer to $\Sigma$ as being "finite" or "infinite").

If $\Sigma$ is finite then there exist pairwise disjoint nonempty sets $A_1,\ldots,A_n\in\Sigma$ such that $\bigcup_iA_i=X$ and

$$\text{for all }A\in\Sigma\text{, if }A\subsetneq A_i\text{ then }\mu(A)=0\qquad(*)$$

Observe this implies that every $A\in\Sigma$ of positive measure can be written as a disjoint union of some $A_i$'s. The existence of such a decomposition can be shown as follows: choose $A_1\in\Sigma$ such that $(*)$ holds for $i=1$; certainly such a set exists since $\Sigma$ is finite. Now let $\Sigma_1=\{A\in\Sigma\,:\,A\subseteq A_1^c\}$. If $A_1^c$ has positive measure then this is a $\sigma$-algebra of subsets of $A_1^c$, so we can similarly choose a set $A_2\in\Sigma_1$ such that $(*)$ holds for $i=2$. Repeating this procedure we end up with a list $A_1,\ldots,A_n$ which must be finite since every $A_i$ is a distinct element of the finite set $\Sigma$. By construction $(*)$ holds for all $i$ and $\bigcup_iA_i=X$.

Now if $f\in L^\infty(\mu)$, choose $x_i\in A_i$ for each $i$ and let $a_i=f(x_i)$. Then $f=\sum_{i=1}^na_i\mathbf1_{A_i}$, where $\mathbf1_A$ is the indicator function of the set $A$. Indeed, $\{a_i\}$ is a Borel set and so $f^{-1}(\{a_i\})\in\Sigma$, and hence $f^{-1}(\{a_i\})=\bigcup_{j=1}^{k}A_{m_j}$ for some subcollection $A_{m_1},\ldots,A_{m_k}$ (again ignoring sets of measure zero). So for all $x\in\bigcup_jA_{m_j}$, $f(x)=a_i$ which implies $a_i=a_{m_j}$ for each $j$ and hence $\sum_{i=1}^na_i\mathbf1_{A_i}(x)=a_i=f(x)$. This implies that $L^\infty(\mu)$ is finite dimensional and hence separable. An explicit countable dense subset is $$\left\{\sum_{i=1}^nb_i\mathbf1_{A_i}\,:\,b_1,b_2,\ldots,b_n\in\mathbb Q\right\}$$ Conversely, suppose $\Sigma$ is infinite. Then we have a sequence $A_1,A_2,\ldots$ of pairwise disjoint elements of $\Sigma$ (this has been proved several times on MathSE, see for instance this question). For each subset $E\subseteq\mathbb N$, define $$f_j(x)=\sum_{j\in E}\mathbf1_{A_j}=\begin{cases} 1&\text{if }x\in A_j,\ j\in E\\ 0&\text{otherwise.} \end{cases}$$ Then $\{f_E\}_{E\subseteq\mathbb N}$ is an uncountable family of $L^\infty$ functions. However, if $E,F\subseteq N$ and $E\neq F$, there exists $x_0\in E\Delta F$ (where $\Delta$ denotes symmetric difference) and $|f_E(x_0)-f_F(x_0)|=1$; thus, $\|f_E-f_F\|_\infty=1$. Hence $L^\infty(\mu)$ cannot be separable.

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  • $\begingroup$ This is sort of morally correct: $\Sigma$ could contain infinitely many null sets. $\endgroup$ – Ian Sep 30 '15 at 23:37
  • $\begingroup$ Good point - I have added a slight addendum. $\endgroup$ – Jason Sep 30 '15 at 23:42

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