If $(X,M,\mu)$ is a measure space and {$E_j$}, j=1 ,2....$\infty$.$\subset$ M, then $\mu$ (lim inf $E_j$) $\leq$ lim inf $\mu(E_j)$
Proof: $\mu(\liminf E_j)=\mu(\left(\bigcup_{k=1}^{\infty} \bigcap_{j=k}^{\infty} E_j) \right) = \lim_{k\to\infty}\mu(\bigcap_{j=k}^{\infty}( E_j))=\liminf_{k\to\infty}\mu(\bigcap_{j=k}^{\infty}( E_j))\leq\liminf_{k\to\infty}\mu( E_j)$.
Should I be taking the limit as k approaches infinity or should I be taking the limit as j approaches infinity? I'm not sure whether I'm correct. I'm using the property continuity from below. Please let me know whether I should taking limit as j approaches infinity or k approaches infinity. Thanks
For anybody who's curious about the answer to kemb's main question, it is fine to use "k" in the limit. Reason? Recall by the basic property of set intersection, we can deduce that $\bigcap_{j=k }^{\infty} (E_{j}) \subseteq E_{j}$ given any $j\in \mathbb{N}$. Thus, it necessarily must follow from subadditivity that, indeed, $\limsup_{k \to \infty} \mu (\bigcap_{j=k }^{\infty} (E_{j})) \leq \mu (\bigcap_{j=k }^{\infty} (E_{j})) $$\leq \mu (E_{j}) = \limsup_{k \to \infty} \mu (E_{j})$.