Consider a probability space $(\Omega,\mathcal{F},P)$ and a measurable partition $(A_i)_{i=1,\ldots, n}$ of $\Omega$ with $P(A_i)>0$.
Moreover for any measurable subset $B\subset A_i$, it holds $$P(B)=P(A_i)\quad\text{or}\quad P(B)=0$$
I was able to show that $(\mathbb{1}_{A_i})_{i=1,\ldots, n}$ is a basis of $L^\infty(\Omega,\mathcal{F},P)$. Now I would like to show, that
$$\dim L^p(\Omega,\mathcal{F},P)=\sup\{n\in\mathbb{N}:\text{There is a partition }(A_i)_{i=1,\ldots, n}\text{ of }\Omega\text{, such that }P(A_i)>0\}$$
I know that $||\cdot||_p\rightarrow||\cdot||_{\infty}$ for $p\rightarrow\infty$ and I was thinking that I might have to use Young's inequality, but I do not find any good start to attack this problem. Also splitting the proof up into "$\le$" and "$\ge$", did not give me a proper idea.
I'd really appreciate some help on this one. Thanks for your attention!
If $\mathscr{F}$ finite, then not only is each $L_p$ finite dimensional, but also they all have the same dimension, and so by a well known fact about finite dimensional spaces, al $L_p$'s are isomorphic and homeomorphic.
The dimension is the same as the number of atoms (sets $A\in\mathscr{F}$ such that $\mathbb{P}[A]>0$ and whenever $B\subset A$ and $B\in\mathscr{F}$ wither $\mathbb{P}[B]=0$ or $\mathbb{P}[B]=\mathbb{P}[A]$.