4

A measure space $\left(X,\mathcal{M},\mu\right)$ is said to have Property A if for every $A\in\mathcal{M}$ with $\mu(A)>0$ there is a set $B\subset A$, $B\in\mathcal{M}$ such that $0<\mu(B)<\mu(A)$.

Let $\left(X,\mathcal{M},\mu\right)$ be a measure space with Property A. Show that for every $A\in\mathcal{M}$ with $0<\mu(A)<\infty$ and for every $\epsilon>0$ there is a set $B\subset A$, $B\in\mathcal{M}$ such that $0<\mu(B)<\varepsilon$.

I am not really sure how to go about doing this. I'd expect some argument relying on countable additivity to be made, e.g. $$\mu(B)<\sum_n \frac{\varepsilon}{2^n} = \varepsilon.$$ But I am just not sure overall how to do such problem.

vitamin d
  • 5,783
  • I wonder if the use of "Property A" is just an attempt to avoid somebody Googling this? These are frequently called non-atomic measure spaces, if that helps you (or anybody else who visits this question and is interested in Property A) find other resources about them, such as related questions on Math.SE. – leslie townes Feb 24 '21 at 23:54
  • For example https://math.stackexchange.com/questions/1937894/show-that-there-is-e-in-m-such-that-mue-alpha-for-any-0-lt-alpha-lt is a related result. – leslie townes Feb 24 '21 at 23:59
  • 1
    @leslietownes: It is just from some lecture notes...throughout the entirety of them "named" def., cor., lem., and even thms. are called "def. A" and such. E.g. Fatous Lemma was called $Lemma A2$ in these notes. But thanks for the actual name of this property. –  Feb 25 '21 at 00:10

1 Answers1

3

Let $\mu(A)=r$ for some $r\in\Bbb R$. By property $A$ we can find a $B\subseteq A$ with $0<\mu(B)<\mu(A)$.
Since $\mu(B)+\mu(A\setminus B)=r$ one of $B$ and $A\setminus B$ must have measure $\leq r/2$. Pick this one and apply property $A$ to it, do you see how to conclude inductively?

Alessandro Codenotti
  • 12,285
  • I see what is going on, but not sure where the $\epsilon$ comes into play. –  Feb 25 '21 at 00:07
  • If you keep halving the measure sooner or later you'll go below $\epsilon$ – Alessandro Codenotti Feb 25 '21 at 00:12
  • Yes. I suppose at the "$n^\text{th}$" step we will have a measurable subset of $A$, say $A_n$, such that $0<\mu(A_n)<r/2^n$? But also I'm not sure how I can go from saying, "let $\epsilon>0$" to "$\mu(B)<\epsilon$". –  Feb 25 '21 at 00:19
  • pick as $B$ any $A_n$ for $n$ big enough to have $\mu(A_n)<r/2^n<\epsilon$. – Alessandro Codenotti Feb 25 '21 at 00:21