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I don't know how can I imply Fatou's lemma for any measurable sets $A_k$

that is..
$\lambda(\liminf A_k)\le\liminf\lambda(A_k)$

how can I prove it?

and is there any example in $R$ of sequence of measurable sets $A_k$ such that $A_k\subset[0,1]$, $lim\lambda(A_k)=1$, but $\liminf A_k=\varnothing$ ?

thx for your help!.

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What does $\liminf A_k$ even mean, since $A_k$ are sets? I thought $\liminf$ was only defined for real number sequences. – Ravi May 6 '12 at 11:09
$\liminf A_k = \bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_n$ – Stefan Hansen May 6 '12 at 11:14
@StefanHansen Ah, thank you. – Ravi May 6 '12 at 11:46

First, since by monotonicity we have $\lambda(\bigcap_{n=k}^{\infty}A_{n})\leq \lambda(A_{j})$ for all $j\geq k$, it follows that $\lambda(\bigcap_{n=k}^{\infty}A_{n})\leq \inf_{n\geq k}\lambda(A_{n})$.

Using convergence of measure to the nondecreasing sequence of sets $((\bigcap_{n=k}^{\infty}A_{n}))_{k=1}^{\infty}$, we obtain:

\begin{align*} \lambda(\liminf A_{n})=\lambda(\bigcup_{k=1}^{\infty}(\bigcap_{n=k}^{\infty}A_{n}))=\lim_{k\to\infty}\lambda(\bigcap_{n=k}^{\infty}A_{n})\leq \lim_{k\to\infty}\inf_{n\geq k}\lambda(A_{n})=\liminf \lambda(A_{n}). \end{align*}

Which is the result that we wanted.

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