# $L^1(μ)$ is finite dimensional if it is reflexive

If $(X,\Omega,\mu)$ is a $\sigma -$ finite measure space, show that if $L^1(X,\Omega,\mu)$ is reflexive then it is finite dimensional.

My attempt: I want to show there is a copy of $\ell^1$ in $L^1(X,\Omega,\mu)$. For this suppose $L^1(X,\Omega,\mu)$ is infinite dimensional. there is a sequence $\{x_n\}$ of disjoint points of $X$. For every n, Put $\chi_n:=\chi(x_n)$ (characteristic function ). define $\phi:\ell^1\to L^1(X,\Omega,\mu)$ such that $\phi(\{a_n\})=\Sigma a_n \chi_n$. But I can not show that $\phi$ is an isometry.

I think I did not in a correct way. Please help me. Thanks in advance.

• You need to take sets of positive measure instead of points – Norbert Jul 13 '14 at 20:52
• Huh? Finite dimensional vector spaces are easy to describe, and you can use equivalence of norms there to get an inner product to dualize. – Adam Hughes Jul 13 '14 at 20:55
• Also you need to normalize $\chi_n$. Once you did it, you can show it is an isometry. Note: since $\chi_n$ have disjoint supports $|\sum a_n\chi_n|=\sum|a_n||\chi_n|$. – Norbert Jul 13 '14 at 20:56
• Adam Hughes :sorry,I made a mistake in the question. I correct it – niki Jul 13 '14 at 20:58
• Norbert: Thanks for your advice. I always like your answers. – niki Jul 13 '14 at 21:03

This is less smart solution than Christian's.

For the begining, every measure space $(\Omega,\mu)$ have continuous $(\Omega_c,\mu_c)$ and atomic part $(\Omega_a,\mu_a)$. Note, for $\sigma$-finite spaces, all atoms are of finite measure. See this question.

Case 1. If atomic part contains infinite amount of atoms we may choose a countalble collection of of disjoint atoms $(A_n)_{n\in\mathbb{N}}$. The desired embedding then would be $$i:\ell_1\to L_1(\Omega,\mu):(a_n)_{n\in\mathbb{N}}\mapsto \left(\omega\mapsto\sum_{n\in\mathbb{N}}a_n\mu(A_n)^{-1}\chi_{A_n}(\omega)\right)$$ Therefore $L_1(\Omega,\mu)$ can't be reflexive as it contains a copy of $\ell_1$ which is not reflexive.

Case 2. If $(\Omega_a,\mu_a)$ consist of finite amount of atoms and continuous part is non empty, then choose any set $A\subset\Omega_a$ of positive finite measure. By Sierpinski theorem, we can divide $A$ onto two parts of equal measure. Each of these aprts we can divide onto two parts of equal measure and etc. In the end we get a sequence of disjoint sets of positive measure. The desired embedding is $$i:\ell_1\to L_1(\Omega,\mu):(a_n)_{n\in\mathbb{N}}\mapsto \left(\omega\mapsto\sum_{n\in\mathbb{N}}a_n\mu(A_n)^{-1}\chi_{A_n}(\omega)\right)$$ Therefore $L_1(\Omega,\mu)$ can't be reflexive as it contains a copy of $\ell_1$ which is not reflexive.

Case 3. If continuous part of measure space is empty, and atomic part consist of finite amount of disjoint atoms $(A_n)_{n\in\mathbb{N}_m}$, then the $L_1(\Omega,\mu)=\operatorname{span}\{\chi_{A_n}:n\in\mathbb{N}_m\}$. So $\dim (L_1(\Omega,\mu))=m$, and $L_1(\Omega,\mu)$ is finite dimensional.

Call $A\in\Omega$ an atom if $\mu(A)>0$, but if $B\subset A$ is measurable, then $\mu(B)=0$ or $=\mu(A)$. Since $\mu$ is $\sigma$-finite, we can cover all atoms by at most countably many atoms $A_n$. Let $Y=X\setminus \bigcup A_n$. Then $\mu(Y)=0$ if $L^1(X,\mu)$ is a dual space because the unit ball of $L^1(Y,\mu)$ has no extreme points: if $\|f\|=1$, we can split $\{x\in Y: f(x)\not=0\}$ into two disjoint sets of positive measure (since we're away from the atoms) and write $f$ as a corresponding convex combination, which would contradict Krein-Milman + Banach-Alaoglu (we're applying KM to the relative topology induced by the weak $*$ topology on the subspace $L^1(Y)$).

However, if $\mu(Y)=0$, then $L^1(X)\cong \ell^1$ via the isomorphism $f\mapsto\int_{A_n}f\, d\mu$.