There is a theorem that for a given $p\in [1,\infty)$ a space $L^p(\mu)$ is finite dimensional iff the set of values of $\mu$ is finite.

Is a similar theorem for the space $L^\infty(\mu)$ for general positive measure $\mu$ ( finite or infinite)?

Edit. Idea of the proof for $L^p(\mu)$ with $p\in [1,\infty)$.

We show that if the set of values of $\mu$ is infinite then $dim L^p(\mu)=\infty$.
There exists a sequence of pairwise disjoint measurable subsets of $X$ such that $0<\mu(P_n)<\infty$ with the property that the following set is infinite $$ \{\mu(D): D \textrm{ is measurable, } D\subset X\setminus \bigcup_{k=1}^n P_k \} $$ For, we take measurable $D_1 $ be such that $0< \mu (D_1)<\mu(X)$ and $D_2=X\setminus D_1$. Let $$ W_1=\{\mu(D): D \textrm{ is measurable}, D \subset D_1 \}, $$ $$ W_2=\{\mu(D): D \textrm{ is measurable} , D \subset D_2 \}. $$ At least one among the sets $W_1, W_2$ is infinite, since for arbitrary measurable $D$ we have $D=(D\cap D_1) \cup (D\cap D_2)$. Let $P_1=D_1$ if $W_1$ is finite and $P_2=D_2$ if $W_2$ is finite. Let pairwise disjoint measurable $P_1,...,P_n $ with the property $0<\mu(P_i)<\infty$ and such that the set $\{\mu(D): D \textrm{ is measurable}, D\subset X\setminus \bigcup_{k=1}^n P_k \}$ is infinite be defined. Let $X_n=X\setminus \bigcup_{k=1}^n P_k$. Then there is a measurable $E_1\subset X_n$ such that $0<\mu(E_1)<\mu(X_n)$. Let $E_2:=X_n\setminus E_1$. We put $$ V_1=\{\mu(D): D \textrm{ is measurable}, D \subset E_1 \}, $$ $$ V_2=\{\mu(D): D \textrm{ is measurable }, D \subset E_2 \}. $$ We define $P_{n+1}=E_1$ or $E_2$ depending on $V_1$ or $V_2$ is finite. The characteristic functions of sets $P_1, P_2,...$ are linearly independent.

We show that if a set of values of $\mu$ is finite then $dim L^p(\mu)<\infty$. Let $x_1$ be a smallest positive finite value of measure and let $\mu(D_1)=x_1$. Then $D_1$ is an atom. Next we take the smallest positive finite value $x_2$ of the measure on $X\setminus D_1$ and a set $D_2 \subset X\setminus D_1$ with $\mu(D_2)=x_2$-it is an atom, and so on. The procedure have to finish after finite many steps, say $n$ steps. On the set $X \setminus (D_1\cup...\cup D_n)$ the measure takes at most two values: zero and infinity, hence each function from $L^p(\mu)$ is zero a.a on this set. On arbitrary atom measurable function is constant a.a. (because it is true for measurable simple functions). Hence arbitrary function from $L^p(\mu)$ is equal a.a to linear combination of characteristic function of atoms $D_1,...,D_n$.

  • $\begingroup$ Thank you very much for the proof sketch! $\endgroup$ – PhoemueX Nov 30 '15 at 6:53

If we allow $\mu$ to be an infinite measure, then at least precisely the same formulation will not work. Let $\mu = \infty \cdot \lambda$ be the Lebesgue measure "multiplied by $\infty$", i.e. $\mu(E) = 0$ if $\lambda(E) = 0$ and $\mu(E) = \infty$ otherwise. Then $L^\infty (\mu) = L^\infty (\lambda)$ is infinite dimensional, although $\mu$ only assumes two values.

Now, if $\mu$ is a finite measure and $\mu$ assumes only finitely many values, then $L^1$ is finite dimensional. Since $L^\infty \hookrightarrow L^1$ (with an injective(!) embedding), $L^\infty$ is finite dimensional as well.

Conversely, if $L^\infty$ is finite dimensional, then so is $L^1$, since $L^\infty \subset L^1$ is dense (approximate $f \in L^1$ by $f \cdot 1_{|f|\leq n}$). Since finite dimensional subsets of a normed vector space are closed, we get $L^1 = L^\infty$, so that $L^1$ is finite dimensional. By your result, this shows that $\mu$ only assumes finitely many values.

BTW: Do you have a convenient source for the result in case of $1 \leq p < \infty$?

  • $\begingroup$ Thanks. I do not know references on this topic in English. I edit the question and write the idea of the proof for $1 \leq p < \infty$. $\endgroup$ – Alex Nov 29 '15 at 22:37
  • $\begingroup$ A side-question: you say "injective embedding"; do you generally use a definition of "embedding" that does not assume injectivity? $\endgroup$ – Alex M. Nov 29 '15 at 22:40
  • $\begingroup$ @AlexM.: No, I do not. I just wanted to emphasize it here since it is important for the argument :) $\endgroup$ – PhoemueX Nov 30 '15 at 6:49

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