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Suppose that $(X,m,\mu)$ is a measure space.

Prove that $$L^{1}(\mu)=L^{\infty}(\mu) \iff \dim L^{1}(\mu)<\infty$$

It's obvious that if $\mu(X)<\infty$, then $L^{\infty}(\mu)\subseteq L^{1}(\mu)$ which does not hold in this case. Also don't know how to use the finiteness of $\dim L^{1}(\mu)$.

Any help would be appreciated. Thanks in advance!

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This is not quite true. Let $X = \{0\}$, and let $\mu(\{0\}) = \infty$. Then $L^{1}(\mu)$ contains only the zero function, hence has finite dimension, but $L^{\infty}(\mu)$ contains every function $f \colon \{0\} \rightarrow \mathbb{R}$.

However, like many results in measure theory, the statement becomes true if we restrict our attention to $\sigma$-finite measures. We can get away with less, if we want, but I'm going to stick with $\sigma$-finiteness.

A useful result here is that $L^{1} \nsubseteq L^{\infty}$ if and only if $X$ contains sets of arbitrarily small positive measure, and $L^{\infty} \nsubseteq L^{1}$ if and only if $\mu(X) = \infty$ (as you noted). This is essentially the content of Exercise 6.1.5 of Folland's Real Analysis.

With this in hand, proving the following three claims will give us our solution:

  1. If $X$ contains sets of arbitrarily small positive measure, then $\dim L^{1} = \infty$.
  2. If $\mu(X) = \infty$, then $\dim L^{1} = \infty$.
  3. If $X$ does not contain sets of arbitrarily small postive measure and $\mu(X) < \infty$, then $\dim L^{1} < \infty$.

Proof of 1:

Let $E_{1}$ be a set such that $0 < \mu(E_{1}) < 1$. Inductively, define $E_{n}$ such that $0 < \mu(E_{n}) < \frac{\mu(E_{n - 1})}{2}$. This is possible since $X$ contains sets of arbitrarily small positive measure. Set $F_{n} = E_{n} \backslash \cup_{k = n + 1}^{\infty} E_{k}$. Then the $F_{n}$'s are all disjoint and have positive finite measure, so the indicator functions $1_{F_{n}}$ are independent functions in $L^{1}$. Thus, $\dim L^{1} = \infty$.

Proof of 2:

Since $\mu$ is $\sigma$-finite, we can find $(E_{n})_{n \in \mathbb{N}}$ such that $\mu(E_{n}) < \infty$ for all $n$ and $\cup_{n = 1}^{\infty} E_{n} = X$. Let $F_{n} = E_{n} \backslash \cup_{k = n + 1}^{\infty} E_{k}$. Then the $F_{n}$'s are disjoint, have finite measure, and $\cup_{n = 1}^{\infty} F_{n} = X$. Since $\mu(X) = \infty$, there are infinitely many $F_{n}$'s with positive measure. Thus, the indicator functions $1_{F_{n}}$ contain an infinite family of independent functions in $L^{1}$.

Proof of 3:

Any measure which does not contain sets of arbitrarily small positive measure is purely atomic. Thus, there is a countable partition of $X$ into atoms (see wikipedia for slightly more information). Since $\mu(X) < \infty$ and there is a lower bound on the measure of any subset of $X$, that countable partition of $X$ into atoms is actually a finite partition. The indicator functions of these atoms is a finite family of functions that spans $L^{1}$, so $\dim L^{1} < \infty$.

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