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.

Let $(X,\mathcal{M},\mu)$ be a measure space. Let $A_1, A_2, \ldots \in \mathcal{M}$.

Then, I want to show that: $$\mu\left(\bigcup_N \bigcap_{n=N}^{\infty} A_n \right) \leq \lim \inf \mu(A_n)$$

There is a solution in lecture notes:

Let $B_N = \bigcap_{n=N}^{\infty} A_n$. $B_N$ form an increasing sequence of elements, then by continuity from below:

$$\mu\left(\bigcup_N \bigcap_{n=N}^{\infty} A_n \right) = \mu\left(\bigcup_N B_N\right) = \lim_{N \to \infty} \mu(B_N) \leq \lim_{N \to \infty} \inf_{n \geq N} \mu(A_n) = \lim \inf \mu(A_n)$$

OK. I do not understand how $\mu\left(\bigcup_N B_N\right) = \lim_{N \to \infty} \mu(B_N)$. And then following inequality and equality. Can somebody give a detailed explanation? Thank you very much.

share|improve this question
add comment

3 Answers

up vote 5 down vote accepted

As for the equality $$ \mu\left(\bigcup\limits_{N=1}^\infty B_N\right)=\lim\limits_{N\to\infty}\mu(B_N) $$ the proof is the following. Since $\{B_N:N\in\mathbb{N}\}$ is the increasing sequence of sets we have $$ \bigcup\limits_{N=1}^\infty B_N=\coprod\limits_{N=1}^\infty C_N $$ where $C_1=B_1$ and $C_N=B_{n+1}\setminus B_N$ for $N>1$. Moreover for $N>1$ $$ \mu(C_N)=\mu(B_{N+1})-\mu(B_N) $$ hence from $\sigma$-additivity of measure we have $$ \begin{align} \mu\left(\bigcup\limits_{N=1}^\infty B_N\right) =\mu\left(\coprod\limits_{N=1}^\infty C_N\right) &=\mu(C_1)+\sum\limits_{N=2}^\infty\mu(C_N)\\ &=\mu(B_1)+\lim\limits_{M\to\infty}\sum\limits_{N=2}^M\mu(C_N)=\\ &=\mu(B_1)+\lim\limits_{M\to\infty}\sum\limits_{N=2}^M(\mu(B_{N+1})-\mu(B_N))\\ &=\mu(B_1)+\lim\limits_{M\to\infty}(\mu(B_{M+1})-\mu(B_1))\\ &=\lim\limits_{M\to\infty}\mu(B_{M+1})\\ &=\lim\limits_{M\to\infty}\mu(B_M) \end{align} $$ As for the inequality $\mu(B_N)\leq\inf\limits_{n\geq N}\mu(A_n)$ you need to note that $$ B_N=\bigcap\limits_{k=N}^\infty A_k\subset A_n $$ for all $n\geq N$, hence for the same $n$ we have $\mu(B_N)\leq \mu(A_n)$. Therefore $$ \mu(B_N)\leq\inf\limits_{n\geq N}\mu(A_n) $$

share|improve this answer
    
Thanks Norbert, but I have never seen the notation $\coprod$ before. :) –  Mark Nov 13 '12 at 8:07
3  
I believe it means union of disjoint sets. –  Stefan Hansen Nov 13 '12 at 8:10
    
@StefanHansen yes, exactly –  Norbert Nov 13 '12 at 8:10
    
thanks StefanHansen and Norbert for answers. –  Mark Nov 13 '12 at 8:36
add comment

In general: If $(A_n)_{n\geq 1}$ is an increasing sequence of sets, i.e. $A_1\subseteq A_2\subseteq \cdots$, then $$ \mu\left(\bigcup_{n\geq 1}A_n\right)=\lim_{n\to\infty}\mu(A_n). $$ This should be present in your lecture notes.

Since $B_N\subseteq A_n$ for every $n\geq N$, then $\mu(B_N)\leq \mu(A_n)$ for all $n\geq N$ and so $$ \mu(B_N)\leq \inf_{n\geq N}\mu(A_n). $$ Now take $\lim_{N\to\infty}$ on both sides. The last equality is the definition of $\liminf$.

share|improve this answer
add comment

For the first part, set $B_N'= B_N \setminus \bigcup_{i=1}^{N-1} B_i$,

Then $\bigcup B_N' = \bigcup B_N$ and so $\mu(\bigcup B_N) = \mu(\bigcup B_N')$, which by additivity of measure gives us $$\mu(\bigcup B_N) = \sum \mu(B_N') = \lim_{N\rightarrow\infty} \sum_1^N \mu(B_i') = \lim_{N\rightarrow\infty}\mu(B_N)$$ where the last equality follows from how we constructed the $B_N'$'s.

Then the inequality follows by monotonicity of measure because $B_N \subseteq A_n$ for all $n\geq N$.

share|improve this answer
add comment

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.