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.

  • $\begingroup$ See here, continuity of measures $\endgroup$
    – Moya
    May 14 '16 at 15:25
  • $\begingroup$ I know that proposition but I do not understand how it equals $\liminf....$ $\endgroup$
    – Wolfy
    May 14 '16 at 15:28
  • $\begingroup$ $\mu(\bigcap_{j=k}^{\infty}E_j) \leq \mu(E_j)$ should be $\mu(\bigcap_{j=k}^{\infty}E_j) \leq \mu(E_k)$. The $j$ is a running index. $\endgroup$ May 14 '16 at 15:37

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 $$\mu\left(\lim\inf E_{j}\right)\leqslant\lim\inf\mu\left(E_{j}\right)$$

  • $\begingroup$ 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 $\endgroup$
    – Wolfy
    May 14 '16 at 16:52
  • $\begingroup$ Which part of the argument confuse you? $\endgroup$ May 14 '16 at 18:00
  • $\begingroup$ Actually I think I get it now I made a edit to my proof can you see if it is correct? $\endgroup$
    – Wolfy
    May 15 '16 at 12:10
  • $\begingroup$ $\lim\limits_{k\rightarrow \infty}\mu\left(\bigcap_{j=k}^{\infty}E_j\right)\leqslant\mu(E_j)$ 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 $E_j$. $\endgroup$ May 15 '16 at 15:46
  • $\begingroup$ $\lim\limits_{k\rightarrow \infty}\mu\left(\bigcap_{j=k}^{\infty}E_j\right)=\liminf\mu\left(\bigcap_{j=k}^{\infty}E_j\right)\leqslant \liminf \mu(E_{k})=\liminf \mu(E_{j})$ $\endgroup$ May 15 '16 at 15:49

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