Is a set of jointly bounded functions over a compact domain compact under p-norm?

Question

Let $X$ be a metric space and a measurable space. Let $K$ be a compact set of nonzero measure and $r> 0$. Is a set $\{ f: K\rightarrow \mathbb R| |f|\leq r$ almost everywhere$\}$ compact with respect to $p$-norm for some kind of $X$?

Answer 1

If $K$ consists of finitely many atoms and a set of measure zero, then your set of functions is finite-dimensional, and compactness follows from the compactness of closed bounded subsets of $\mathbb R^n$. But then $X$ is not such an interesting space to look at.

If $K$ contains a set $A_0$ of positive measure with no atoms, then no. By a theorem of Sierpiński (op. cit.) there is a set $A_1\subset A_0$ with $\mu(A_1)=\mu(A_1^c)=\frac12 \mu(A_0)$ (here and below the complements are with respect to $A_0$). Subdividing both $A_1$ and $A_0\setminus A_1$ in equal parts, and joining them, we get a set $A_2\subset A_0$ with $$\mu(A_1\cap A_2) = \mu(A_1\cap A_2^c) = \mu(A_1^c\cap A_2)= \mu(A_1^c\cap A_2^c) = \frac14$$ and so forth. Consider the sequence $f_n=r \chi_{A_n}$. For distinct indices $n\ne m$ we have $$\int |f_n-f_m|^p = r^p \mu(A_n\triangle A_m) = \frac12r^p$$ Thus, the infinite family $(f_n)$ has no limit point; compactness fails.