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Show measure of $\liminf$ is less than $\liminf$ of measure (Proof Verification)

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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


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