# $L^p$ space: measure theory

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!

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 $$\dim L^{1} < \infty$$.