Let $(X,\mu)$ be a probability measure space and $A$ be a measurable subset of $X$ such that $0 \le \mu(A) < p < 1$.


Question
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When is it true that there exists a measurable $B \subseteq X$ such that $A \subseteq B$ and $\mu(B)=p$ ?



If $0 \le \mu(A) < p < 1$, when is it true that there exists a measurable $B \supseteq A$ such that $\mu(B)=p$?
An equivalent condition is that the probability space $(X, \mathcal{M}, \mu)$ is non-atomic, meaning that for every $A\in \mathcal{M}$ with $\mu(A)>0$ there exists $B \in \mathcal{M}$ such that $B\subset A$ and $0<\mu(B)<\mu(A)$.


If $(X, \mathcal{M}, \mu)$ is non-atomic then $\mu: \mathcal{M}\to [0,1]$ has a monotone section: that is, a map $E:[0,1]\to \mathcal{M}$ such that for all $0\le t\le s\le1$ one has $\emptyset=E\_0\subset E\_t\subset E\_s\subset E\_1=X$ and $\mu(E\_t)=t$. 


In fact, any "partial monotone section of $\mu$" (meaning a family $E:S\to \mathcal{M}$ as above, but with possibly smaller domain $S\subset [0,1]$) can be extended to a monotone section of $\mu$ (which in particular solves your problem). 


I think this result is essentially due to Sierpiński; the proof uses the Zorn lemma on the set of graphs of the partial monotone sections of $\mu$, ordered by "extension" i.e. inclusion of graphs. 


By the way, the maximality argument via Zorn lemma generalizes to a proof of the Lyapunov convexity theorem: if a $\mathbb{R}^n$ valued bounded vector measure $\mu=(\mu\_1,\dots,\mu\_n)$ is non atomic (meaning, all $\mu\_i$ are non-atomic bounded measures), then the range of $\mu$ is a compact convex set. 


