1

If $(X,M,\mu)$ is a measure space and $\{E_j\}_{1}^{\infty}\subset M$, then $\mu(\liminf E_j) \leq \liminf \mu(E_j)$.

Attempted proof - Let $(X,M,\mu)$ be a measure space and $\{E_j\}_{1}^{\infty}\subset M$. Notice that $\bigcap_{j=k}^{\infty} E_j = \cap E_k\cap E_{k+1}\cap\ldots$ which is increasing so $\{\bigcap_{1}^{\infty}E_k\}$ is an increasing sequence of sets so $\mu(\bigcap_{j=k}^{\infty}E_j) \leq \mu(E_k)$. We know that $$\liminf E_j = \bigcup_{k=1}^{\infty}\bigcap_{j=k}^{\infty}E_j$$ So $$\mu(\liminf E_j) = \mu\left(\bigcup_{k=1}^{\infty}\left(\bigcap_{j=k}^{\infty}E_j\right)\right) = \lim_{k\rightarrow \infty}\mu\left(\bigcap_{j=k}^{\infty}E_j\right) = \liminf\mu\left(\bigcap_{j=k}^{\infty}E_j\right) \leq \liminf\mu(E_j)$$ Therefore we must have $$\mu(\liminf E_j) \leq \liminf \mu(E_j)$$

I am somewhat skeptical of this result if anyone could provide some reasoning I would greatly appreciate it.

Wolfy
  • 6,495

1 Answers1

5

Let $F_{k}=\bigcap\limits_{j=k}^{\infty} E_{j}$, then $\{F_{k}\}$ is an increasing sequence of sets. By definition $\lim\inf E_{j}=\bigcup\limits_{k=1}^{\infty} F_{k}$.

By Theorem 1.8.c (Continuity from below) in Folland's book, we get $$\mu\left(\lim\inf E_{j}\right)=\mu\left(\bigcup\limits_{k=1}^{\infty} F_{k}\right)=\lim\limits_{k\rightarrow \infty}\mu\left(F_{k}\right)=\lim\limits_{k\rightarrow \infty}\mu\left(\bigcap\limits_{j=k}^{\infty} E_{j}\right)$$ By definition $$\lim\inf\mu\left(E_{j}\right)=\lim\limits_{k\rightarrow \infty}\inf\{\mu\left(E_{j}\right) \mid j\geqslant k\}$$ Since $\mu\left(\bigcap\limits_{j=k}^{\infty} E_{j}\right)\leqslant \mu\left(E_{j}\right)$ for all $j \geqslant k$, we must have $\mu\left(\bigcap\limits_{j=k}^{\infty} E_{j}\right)\leqslant \inf\{\mu\left(E_{j}\right) \mid j\geqslant k\}$

Therefore, we have $$\mu\left(\lim\inf E_{j}\right)\leqslant\lim\inf\mu\left(E_{j}\right)$$

Yining Zhang
  • 717
  • It makes sense to that since $\mu\left(\bigcap_{j=k}^{\infty}E_j\right)\leq \mu(E_j)$ then we should have the result but I am somewhat skeptical – Wolfy May 14 '16 at 16:52
  • Which part of the argument confuse you? – Yining Zhang May 14 '16 at 18:00
  • Actually I think I get it now I made a edit to my proof can you see if it is correct? – Wolfy May 15 '16 at 12:10
  • $\lim\limits_{k\rightarrow \infty}\mu\left(\bigcap_{j=k}^{\infty}E_j\right)\leqslant\mu(E_j)$ is kind of unclear to me. It seems like you view $j$ as a fixed index. I think you should use $E_k$ instead of $E_j$. – Yining Zhang May 15 '16 at 15:46
  • $\lim\limits_{k\rightarrow \infty}\mu\left(\bigcap_{j=k}^{\infty}E_j\right)=\liminf\mu\left(\bigcap_{j=k}^{\infty}E_j\right)\leqslant \liminf \mu(E_{k})=\liminf \mu(E_{j})$ – Yining Zhang May 15 '16 at 15:49
  • $\lim_{k\rightarrow \infty}\mu\left(\bigcap_{j=k}^{\infty}E_j\right) = \liminf\mu\left(\bigcap_{j=k}^{\infty}E_j\right)$ how did you get that is that a definition I am not aware of? – Wolfy May 18 '16 at 13:07
  • If the limit exists, then we will have $\liminf\mu\left(\bigcap\limits_{j=k}^{\infty}E_{j}\right)=\lim_{k\rightarrow \infty}\mu\left(\bigcap\limits_{j=k}^{\infty}E_{j}\right)$. Theorem 1.8.c (Continuity from below) provide us that limit does exist. – Yining Zhang May 18 '16 at 15:51
  • Ok, I got it now thanks – Wolfy May 18 '16 at 17:56
  • edited my answer I think it is correct now based on your comment – Wolfy May 18 '16 at 17:59