Tuesday, 6 September 2016

real analysis - Can $mu(A) < liminf_{ntoinfty} mu (A_n)$?



Let $\left\{ A_{n}\right\} $
be a sequence of sets that Lebesgue measurable on $\mathbb{R}$
such that $\mu\left(A_{n}\right)<\infty$
for all $n$
(integer). Let



$$A={ \bigcup_{m=1}^{\infty}}{ \left(\bigcap_{k\geq m}^{\infty}A_{k}\right)}$$




Do we have the following inequality:



$$ \mu(A) \leq \liminf_{n\to\infty} \mu(A_n) ?$$



And can



$$\mu(A) < \liminf_{n\to\infty}\mu(A_n)?$$



My question is the second inequality (the first is well-known).




Thank you in advance.


Answer



$A$ is the set of points which are in infinitely many of the $A_k$. This gives us an idea: Make $A$ very small, but keep the $A_k$ at a fixed size.



In particular we can take



$\displaystyle\qquad
A_k = \begin{cases}
[0,1] & k \text{ odd} \cr
[1,2] & k \text{ even}

\end{cases}$



Now $A = \{1\}$ and $\mu(A_k) = 1$ for all $k$ giving us the desired sharp inequality.



You can make all kinds of variations on this theme. For example $A_k = [k,k+1]$.


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