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.
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: $$\liminf\mu A_{n}>0=\mu(\liminf A_n)$$