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

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.

• Please don't use $$..$$ in the title of your questions. It messes up the front page. – kahen Jun 14 '14 at 2:12
• Okay I see. I'm sorry. – chuyenvien94 Jun 14 '14 at 2:13
• Further TeX tips: Use \liminf and do not overuse \left .. \right. Most of the time they're unnecessary and sometimes they make things look wrong. – kahen Jun 14 '14 at 2:18

## 1 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 $A_k = [k,k+1]$.