Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!.

share|improve this question
    
What does $\liminf A_k$ even mean, since $A_k$ are sets? I thought $\liminf$ was only defined for real number sequences. –  Ravi Donepudi May 6 '12 at 11:09
2  
$\liminf A_k = \bigcup_{k=1}^\infty \bigcap_{n=k}^\infty A_n$ –  Stefan Hansen May 6 '12 at 11:14
1  
@StefanHansen Ah, thank you. –  Ravi Donepudi May 6 '12 at 11:46

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.