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The math.stackexchange community did a very good job at helping me in my previous question, especially Eric Wofsey. So I implore you to continue helping me spot flaws in my reasoning.

Let $(X, \mu)$ be a finite measure space, and let $\{E_i\}_{i = 1}^{\infty}$ be a sequence of measurable sets in $X$. Then the limit superior and limit inferior of $\{E_i\}_{i = 1}^{\infty}$ is defined such that

$\limsup\limits_{n\to\infty} E_n = \bigcup\limits_{n=1}^{\infty}\bigcap\limits_{k\geq n} E_k, \;\;\; \liminf\limits_{n\to\infty} E_n = \bigcap\limits_{n=1}^{\infty}\bigcup\limits_{k\geq n} E_k.$

Prove that

$\mu(\liminf\limits_{n\to\infty} E_n) \leq \liminf\limits_{n\to\infty} \mu(E_n) \leq \limsup\limits_{n\to\infty} \mu(E_n) \leq \mu(\limsup\limits_{n\to\infty} E_n)$

Attempted proof. Let $\{L_n\}_{n=1}^\infty = \bigcap\limits_{k\geq n} E_k$ and let $\{M_n\}_{n=1}^\infty = \bigcup\limits_{k\geq n} E_k$. Then $\{L_n\}_{n=1}^\infty$ is an increasing sequence and $\{M_n\}_{n=1}^\infty$ is a decreasing sequence.

Using the continuity of measures, we know that $\mu\left(\limsup\limits_{n\to\infty} E_n\right) = \mu\left(\bigcup\limits_{n=1}^\infty L_n\right) = \lim\limits_{n\to\infty} \mu(L_n) = \mu(L)$ where $L$ is the maximal element of the chain and that $\mu\left(\liminf\limits_{n\to\infty} E_n\right) = \mu\left(\bigcap\limits_{n=1}^\infty M_n\right) = \lim\limits_{n\to\infty} \mu(M_n) = \mu(M)$ where $\mu(M)$ is the minimal element of the chain.

$\mu(M) \leq \liminf\limits_{n\to\infty} \mu(E_n)$ because the minimal element of the chain is less than the minimal element of any subsequence of the chain. $\limsup\limits_{n\to\infty} \mu(E_n) \leq \mu(L)$ because the maximal element is greater than the maximal element of any subsequence of the chain. And $\liminf\limits_{n\to\infty} \mu(E_n) \leq \limsup\limits_{n\to\infty} \mu(E_n)$ is obvious. Taken together this proves it voila.

Alon Amit
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ArtIntoNihonjin.
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1 Answers1

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The definition of limit superior and limit inferior may be false? It seems different from the ordinary textbook of measure theory. \begin{equation} \varliminf_{n\to\infty}E_n=\bigcup_{n=1}^{\infty}\bigcap_{k\geq n}E_k,\varlimsup_{n\to\infty}E_n=\bigcap_{n=1}^{\infty}\bigcup_{k\geq n}E_k \end{equation}

HQR
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  • This problem is from "A Crash Course in the Lebesgue Integral and Measure Theory" and definitions can't be false only unconventional. – ArtIntoNihonjin. Aug 08 '18 at 04:11
  • This should be a comment, not an answer. And it's not even a helpful comment, since it doesn't say how the definitions given in the question differ from the usual definitions. But @TomislavOstojich are you sure you didn't get liminf and limsup backward? Usually liminf is a union of intersections, and limsup is an intersection of unions. – Alex Kruckman Aug 08 '18 at 04:16
  • $\forall x \in \bigcup\limits_{n=1}^{\infty}\bigcap\limits_{k\geq n}E_n\rightarrow \exists n_0, x\in \bigcap\limits_{k\geq n_0}E_k\rightarrow \forall k\geq n_0 ,x\in E_k \rightarrow x\in \bigcap\limits_{n=n_0}\bigcup\limits_{k\geq n_0}E_k =\bigcap\limits_{n=1}\bigcup\limits_{k\geq n_0}E_k.$ which means $\lim supE_n\subset \liminf E_n$ in your definition. – HQR Aug 08 '18 at 04:20
  • @AlexKruckman you guys are probably right. – ArtIntoNihonjin. Aug 08 '18 at 06:06