8
$\begingroup$

Let $(\Omega, \mathcal{A}, \mu)$ be a measure space, where $\mu(\Omega)< \infty$. Further $(A_n)_{n \in \mathbb{N}}$ is a a sequence of $\mathcal{A}$-measurable sets. I want to prove, that

$$ \mu ( \liminf_{n \rightarrow \infty} A_n) \leq \liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$ holds for any sequence $(A_n)_{n \in \mathbb{N}}$. I have no experience working with the limit superior/inferior. Clearly $$\mu ( \liminf_{n \rightarrow \infty} A_n) \leq \mu (\limsup_{n \rightarrow \infty} A_n)$$ holds, since it is easy to prove that the one is a superset of the other. Also $$\liminf_{n \rightarrow \infty} \mu (A_n) \leq \limsup_{n \rightarrow \infty} \mu (A_n)$$ holds, for any sqeuence. But I am stuck how to show the connection. I could use the Definitions, then I get $$ \mu (\bigcup_n^\infty \bigcap_{k=n}^\infty A_n) \leq \lim_{n \rightarrow \infty} \inf_{k \geq n} \mu(A_n) \leq \inf_{n \geq 0} \sup_{k \geq n} \mu(A_n) \leq \mu (\bigcap_n^\infty \bigcup_{k=n}^\infty A_n) $$ But I don't know if this helps. Anyone got a hint how to go on?

$\endgroup$
1
  • $\begingroup$ Since each measurable subset has a finite measure, what can you say about $\mu(\bigcap_{k\geq n}A_k)$? Anyway, +1 for showing your work. $\endgroup$ – Davide Giraudo Jul 16 '12 at 12:05
5
$\begingroup$

Consider sets $$ B_n=\bigcap\limits_{k=n}^\infty A_k $$ Obviously, $B_n\subset A_k$ for all $k\geq n$, so $\mu(B_n)\leq \mu(A_k)$ for all $k\geq n$ After taking infimum we get $\mu(B_n)\leq\inf_{k\geq n}\mu(A_k)$. Since $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$ then the sequence $\{\mu(B_n):n\in\mathbb{N}\}$ is non-decreasing, so there exist $\lim_{n\to\infty}\mu(B_n)$. Similarly the sequence $\{\inf_{k\geq n}\mu(A_k):n\in\mathbb{N}\}$ is non-decreasing hence there exist $\lim_{n\to\infty}\inf_{k\geq n}\mu(A_k)$. Since existence of limits is justified we can say $$ \lim\limits_{n\to\infty}\mu(B_n)\leq \lim\limits_{n\to\infty}\inf\limits_{k\geq n}\mu(A_k)=\liminf\limits_{n\to\infty}\mu(A_n)\tag{1} $$ Again recall that $B_n\subset B_{n+1}$ for all $n\in\mathbb{N}$, so $$ \mu\left(\bigcup\limits_{n=1}^\infty B_n\right)=\lim\limits_{n\to\infty}\mu(B_n)\tag{2} $$ It is remains to note that $$ \liminf\limits_{n\to\infty}A_n= \bigcup\limits_{n=1}^\infty \bigcap\limits_{k=n}^\infty A_k= \bigcup\limits_{n=1}^\infty B_n\tag{3} $$ From $(1)$, $(2)$ and $(3)$ we have $$ \mu\left(\liminf\limits_{n\to\infty}A_n\right)\leq \liminf\limits_{n\to\infty}\mu(A_n) $$ Now try to prove in the similar way the second inequality.

$\endgroup$
1
  • 1
    $\begingroup$ Thanks think I got it. The prove for the other inequality worked. $\endgroup$ – Haatschii Jul 16 '12 at 13:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.