In this answer, the author shows that $L^p(X, \mathcal{M}, \mu)$ is not an inner product space when $p \neq 2$ under the assumption that there exist disjoint $A, B \in \mathcal{M}$ such that $0 < \mu(A), \mu(B) < \infty$.
My question is the following:
Suppose $(X, \mathcal{M}, \mu)$ is a measure space such that $\dim_{\mathbb{R}}L^p(X, \mathcal{M}, \mu) > 1$, then do there exist disjoint sets $A, B \in \mathcal{M}$ such that $0 < \mu(A), \mu(B) < \infty$?
The motivation for the problem is clear. The above being true would imply that $L^p(X, \mathcal{M}, \mu)$ is not an inner product space when $p \neq 2$ assuming $\dim_{\mathbb{R}}L^p(X, \mathcal{M}, \mu) > 1$. The assumption about the dimension is clearly necessary. This would also make it sufficient.
Do we have to make any assumption about $X$ being semi-finite? $\sigma$-finite?
My attempt (feel free to ignore this): $\DeclareMathOperator{\supp}{supp}$
Since $\dim L^p \ge 2$, there exists linearly independent $f, g \in L^p$. Note that $A = \supp f$ and $B = \supp g$ are measurable sets (since $f, g$ are measurable functions) and moreover, $\mu(A), \mu(B) \neq 0$ since $f$ and $g$ cannot be $0$ a.e. (since they are linearly independent).
We also have the sets $$C = \{x \in X: f(x) - g(x) \neq 0\} \in \mathcal{M}$$ and $$D = \{x \in X: f(x) + g(x) \neq 0\} \in \mathcal{M}.$$ Again, linear independence of $f$ and $g$ implies that $\mu(C), \mu(D) \neq 0$.
My hunch was that by considering these sets $A, B, C, D$ and considering their union/intersection/set-difference, we could construct the desired disjoint sets. However, I'm not sure how one could claim that any of these sets have finite measure to begin with.