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Let $(\Bbb{X}, \Sigma, \mu)$ be a measure space and Let $(A_n) \subset \Sigma$.

The following property is satisfied: $\mu(\liminf_{n\to \infty}A_n) \le \liminf_{n\to \infty}\mu(A_n)$ where $\liminf_{n\to \infty}A_n = \displaystyle \bigcup_{n=1}^\infty \bigcap_{k=n}^\infty A_k$.

I noticed that this property appears all the time (in Folland's book anyway) and I'm not sure I have grasped the notion correctly - or if to make it concrete and dumb - why aren't the two sides equal? Can anyone give an illustration (or a link to one), that will clarify this. Either a numerical illustration or a graphic one.

Thanks.

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  • $\begingroup$ Wonder what happens if e.g. the $A_n$ are disjoint. $\endgroup$
    – drhab
    Commented Jan 22, 2018 at 12:34

1 Answer 1

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Setting $B_{n}=\bigcap_{k=n}^{\infty}A_{k}$ it is evident that $B_{1}\subseteq B_{2}\subseteq\cdots$ with $\bigcup_{n=1}^{\infty}B_{n}=\liminf A_{n}$ so that $\lim_{n\to\infty}\mu\left(B_{n}\right)=\mu\left(\liminf A_{n}\right)$.

Further for every $k\geq n$ we have $B_{n}\subseteq A_{k}$ and consequently $\mu\left(B_{n}\right)\leq\inf_{k\geq n} \mu A_{k} $

So: $$\mu\left(\liminf A_{n}\right)=\lim_{n\to\infty}\mu\left(B_{n}\right)\leq\lim_{n\to\infty}\inf_{k\geq n} \mu A_{k} =\liminf\mu A_{n}$$

If $A_n=A$ if $n$ is odd and $A_n=B$ if $n$ is even, and $A\cap B=\varnothing$ then $B_n=\varnothing$ for every $n$ so that $\liminf A_n=\varnothing$ and consequently $\mu(\liminf A_n)=0$.

Further $\liminf\mu A_{n}=\min(\mu(A),\mu(B))$, so that - if at least one of $A$ and $B$ has non-zero measure: $$\liminf\mu A_{n}>0=\mu(\liminf A_n)$$

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  • $\begingroup$ Could you please explain what does the inequality $\mu\left(B_{n}\right)\leq\inf_{k\geq n} \mu A_{k}$ mean? Why is it smaller or equal to that inf? $\endgroup$
    – Friedman
    Commented Jan 22, 2018 at 14:19
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    $\begingroup$ Because $B_n\subseteq A_k$ for every $k\geq n$ we have $\mu(B_n)\leq\mu(A_k)$ for every $k\geq n$. That means that $\mu(B_n)$ is a lower bound of the set $\{\mu(A_k)\mid k\geq n\}$ and consequently does not exceed the greatest lower bound of this set. In symbols: $\mu(B_n)\leq\inf\{\mu(A_k)\mid k\geq n\}$. Here $\inf_{k\geq n}\mu(A_k)$ is another notation for $\inf\{\mu(A_k)\mid k\geq n\}$. In general for a set $S\subseteq\mathbb R$ the following statements are equivalen: $x\leq\inf S$ and $x\leq s$ for every $s\in S$. $\endgroup$
    – drhab
    Commented Jan 22, 2018 at 14:25
  • $\begingroup$ Got it, thanks @drhab. $\endgroup$
    – Friedman
    Commented Jan 22, 2018 at 14:31
  • $\begingroup$ You are very welcome. $\endgroup$
    – drhab
    Commented Jan 22, 2018 at 14:31

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