Dimension of $L^p$ space Let $X = \{a,b,c\}$, $\mu$ is a measure on $2^X$ with $\mu(\{a\})=0, \mu(\{b\})=1, \mu(\{c\})=\infty$.
What is the dimension of $L^1(\mu)$, $L^2(\mu)$ and $L^\infty(\mu)$?

Consider a function $f$ on $2^X$, then we have
$||f||_{L^1} = |f(a)|\mu(\{a\})+|f(b)|\mu(\{b\})+|f(c)|\mu(\{c\}) = |f(b)|+|f(c)|\cdot\infty$.
Similarly,
$||f||_{L^2} = (|f(b)|^2+|f(c)|\cdot\infty)^{\frac{1}{2}}$, and $||f||_{L^\infty}=\text{esssup}\{|f(a)|,|f(b)|,|f(c)|\}=\sup\{|f(b)|,|f(c)|\}$.
For $f\in L^1(\mu)$ or $f\in L^2(\mu)$, $|f(b)|<\infty$, $f(c)=0$, $f(a)$ can be finite or infinite (two choices), so $L^1(\mu)=L^2(\mu)=3$.
For $f\in L^\infty(\mu)$, $|f(b)|<\infty$, $|f(c)|<\infty$, so $L^\infty(\mu)=4$.
Is this correct?

Note that any function space on $X$ is a subset (up to equivalence classes) of $\mathbb R^X$. The collection $\{\mathbb 1_{\{a\}},\mathbb 1_{\{b\}},\mathbb 1_{\{c\}}\}$ forms a basis for this space, so your dimension of any $L^p$ space is at most $3$. Also your functions shouldn't be allowed to have infinite values technically if you want to talk about the standard dimension of a vector space, because the extended real line is not a field.
Now let $f\in L^1(\mu)$. Then as you say $|f(a)|\mu(\{a\})+|f(b)|\mu(\{b\})+|f(c)|\mu(\{c\})<\infty$. This is only possible if $f(c)=0$. Thus we can write $f=f(a)\mathbb 1_{\{a\}}+f(b)\mathbb 1_{\{b\}}$. However remember that $L^1(\mu)$ is actually the quotient space $\mathcal L(\mu)/\ker(\|\cdot\|_1)=\mathcal L(\mu)/\operatorname{span}(\mathbb 1_{\{a\}})$, so the dimension of $L^1(\mu)$ must be $1$. I leave the rest to you.
