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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.

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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 $$ \mu(B_N)\leq\inf\limits_{n\geq N}\mu(A_n) $$

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  • $\begingroup$ Thanks Norbert, but I have never seen the notation $\coprod$ before. :) $\endgroup$ – user48547 Nov 13 '12 at 8:07
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    $\begingroup$ I believe it means union of disjoint sets. $\endgroup$ – Stefan Hansen Nov 13 '12 at 8:10
  • $\begingroup$ @StefanHansen yes, exactly $\endgroup$ – Norbert Nov 13 '12 at 8:10
  • $\begingroup$ thanks StefanHansen and Norbert for answers. $\endgroup$ – user48547 Nov 13 '12 at 8:36
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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 $\liminf$.

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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 $n\geq N$.

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