The question is

Is $\left| {\mathop {\lim \sup }\limits_{n \to \infty } {A_n}\Delta \mathop {\lim \sup }\limits_{n \to \infty } {B_n}} \right| = 0$ if given $\left| {{A_n}\Delta {B_n}} \right| = 0$ and $\left| {\bigcup\nolimits_{n = 1}^\infty {{A_n}} \Delta \cup \bigcup\nolimits_{n = 1}^\infty {{B_n}} } \right| = 0$?

Here $|*|$ means Lebesgue measure and $\Delta $ means symmetric difference, i.e. $A\Delta B = (A\backslash B) \cup (B\backslash A)$.

Let $E_1={\bigcup\nolimits_{n = 1}^\infty {{A_n}} \Delta \cup \bigcup\nolimits_{n = 1}^\infty {{B_n}} }$, $E_2={\mathop {\lim \sup }\limits_{n \to \infty } {A_n}\Delta \mathop {\lim \sup }\limits_{n \to \infty } {B_n}} $.

I found neither $E_1\subseteq E_2$ nor $E_2\subseteq E_1$. If $x\in E_1$, then $x$ belongs to some $A_n$ but none of $B_n$ or vice versa. If $x\in E_2$, then $x$ belongs to infinite many $A_n$ but at most finite many $B_n$ or vise versa.

Then $x\in E_1$ cannot imply $x\in E_2$ if $x$ only appears in finite many $A_n$ or $B_n$; $x\in E_2$ cannot imply $x\in E_1$ if $x$ appears if $x$ appears in both some $A_n$ and some $B_n$.

But $E_1$ and $E_2$ has an intersection in which $x$ belongs to infinite many $A_n$ but none of $B_n$ or vise versa, and the measure of this intersection is zero since it is a subset of $E_1$.

How do I go on from here to infer if $E_2$ has measure zero or not? Or there is other way to solve it? Thank you!

  • $\begingroup$ Set $C_n = A_n \cap B_n$. Then $M_n = A_n\setminus C_n$, $N_n = B_n \setminus C_n$. Find some relation between $\limsup A_n$ and $\limsup C_n$ etc. $\endgroup$ – Daniel Fischer Jul 21 '15 at 13:33
  • $\begingroup$ Yes that's what I have difficulty in. Can you provide some more hint? $\endgroup$ – Tony Jul 21 '15 at 14:40

The crucial property is $\lvert A_n \triangle B_n\rvert = 0$ for all $n$. That means that $A_n$ and $B_n$ are "essentially the same set", and in any countable sequence of operations, replacing $A_n$ with $B_n$ can only lead to changes of measure $0$.

To make things precise, define $C_n = A_n \cap B_n$, and $M_n = A_n \setminus C_n$, $N_n = B_n \setminus C_n$. By definition, $A_n \triangle B_n = M_n \cup N_n$, so $M_n$ and $N_n$ are null sets. Hence $M = \bigcup M_n$ and $N = \bigcup N_n$ are null sets.

For the task at hand, recall the definition of the limes superior of a sequence of sets,

$$\limsup_{n\to\infty} A_n = \bigcap_{n = 1}^\infty \Biggl(\bigcup_{k = n}^\infty A_k\Biggr).$$

Since we have $C_k \subset A_k$ for all $k$, we have

$$\bigcup_{k = n}^\infty C_k \subset \bigcup_{k = n}^\infty A_k$$

for all $n$, whence

$$\limsup_{n\to\infty} C_n \subset \limsup_{n\to\infty} A_n.$$

But we also have $A_k = C_k \cup M_k \subset C_k \cup M$ for all $k$, so

$$\bigcup_{k = n}^\infty A_k \subset \bigcup_{k = n}^\infty (C_k\cup M) = M \cup \bigcup_{k = n}^\infty C_k,$$

and therefore

$$\limsup_{n\to\infty} A_n = \bigcap_{n = 1}^\infty \Biggr(\bigcup_{k = n}^\infty A_k\Biggr) \subset \bigcap_{n = 1}^\infty \Biggr( M \cup \bigcup_{k = n}^\infty C_k\Biggr) = M \cup \bigcap_{n = 1}^\infty \Biggr(\bigcup_{k = n}^\infty C_k\Biggr) = M \cup \limsup_{n\to\infty} C_n.$$

So $\limsup\limits_{n\to\infty} A_n$ and $\limsup\limits_{n\to\infty} C_n$ differ only by a null set (a subset of $M$).

Doing the same for $\limsup\limits_{n\to \infty} B_n$ shows that indeed

$$\Bigl\lvert \limsup_{n\to\infty} A_n \triangle \limsup_{n\to\infty} B_n \Bigr\rvert = 0.$$


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