2
$\begingroup$

Let $X$ be a borelian, finite measure space, WLOG, $\mu(X)=a$.

Are there any additional hypotheses that guarantee the following assertion is true?

" For any real number $0\leq b <a$, there exists a subset $X_1$ of $X$ such that $\mu(X_1)=b$"

Thanks in advance.

$\endgroup$
1
  • 2
    $\begingroup$ I think nonatomic measures guarantee this? $\endgroup$
    – Cameron Williams
    Commented Feb 14, 2021 at 20:32

2 Answers 2

Reset to default
4
$\begingroup$

As Cameron Williams mentioned, non-atomic measures satisfy this. The proof is due to Sierpinski and uses Zorn's Lemma; see this Mathoverflow discussion and this Math.SE answer.

$\endgroup$
1
  • 2
    $\begingroup$ And it's an if and only if: if a measure is atomic, then there is some subset $X_1$ such that $0<\mu(X_1)\le a$ and no measurable subset has measure $\frac{\mu(X_1)}2$. $\endgroup$
    – user239203
    Commented Feb 14, 2021 at 20:44
1
$\begingroup$

If the measure space is complete, then it contains all sets of measure zero. Since the empty set is a subset of every set, then you can always find a subset $X_1 = \emptyset \subset X$ such that your condition holds.

A less trivial example has to do with non-atomic measures. I refer you to the nice links given by angryavian.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .