Intuitive interpretation of limsup and liminf of sequences of sets?

Question

What is an intuitive interpretation of the 'events' $$\limsup A_n:=\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}A_k$$ and $$\liminf A_n:=\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}A_k$$ when $A_n$ are subsets of a measured space $(\Omega, F,\mu)$. Of the first it should be that 'an infinite number of those events is verified', but I don't see how to explain (or interpret this). Thanks for any help!

Answer 1

Try reading it piece by piece. Recall that $A\cup B$ means that at least one of $A$, $B$ happens and $A\cap B$ means that both $A$ and $B$ happen. Infinite unions and intersections are interpreted similarly. In your case, $\bigcup_{k=n}^{\infty}A_k$ means that at least one of the events $A_k$ for $k\geq n$ happens. In other words "there exists $k\geq n$ such that $A_k$ happens".

Now, let $B_n=\bigcup_{k=n}^{\infty}A_k$ to simplify notation a bit. This gives us $\bigcap_{n=0}^{\infty}\bigcup_{k=n}^{\infty}A_k = \bigcap_{n=0}^{\infty}B_n$. This is interpreted as "all of the events $B_n$ for $n\geq 0$ happen" which is the same as "for each $n\geq 0$ the event $B_n$ happens". Combined with the above interpretation, this tells us that that $\limsup A_n$ means "for each $n\geq 0$ it happens that there is a $k\geq n$ such that $A_k$ happens". This is precisely the same as saying that infinitely many of the events $A_k$ happen.

The other one is interpreted similarly: $\bigcap_{k=n}^{\infty}A_k$ means that for all $k\geq n$ the event $A_k$ happens. So, $\bigcup_{n=0}^{\infty}\bigcap_{k=n}^{\infty}A_k$ says that for at least one $n\geq0$ the event $\bigcap_{k=n}^{\infty}A_k$ will happen, i.e.: there is a $n\geq 0$ such that for all $k\geq n$ the event $A_k$ happens. In other words: $\liminf A_n$ is the event that from some point on, every event happens.

Edit: As requested by Diego, I'm adding a further explanation. Sets are naturally ordered by inclusion $\subseteq$. This is a partial order, even a lattice. (Putting aside the fact that the universe of sets is not a set.) In fact, every family of sets has an $\inf$ and $\sup$ with respect to $\subseteq$, which can be defined by: $$\inf_{\lambda\in\Lambda}A_\lambda =\bigcap_{\lambda\in\Lambda}A_\lambda$$ and $$\sup_{\lambda\in\Lambda}A_\lambda =\bigcup_{\lambda\in\Lambda}A_\lambda.$$

Now, the usual definition of $\limsup$ and $\liminf$ (of sequences of real numbers) can be rephrased in terms of infima and suprema as follows: $$\liminf_{n\to\infty}a_n=\sup_{n\geq 0}\inf_{k\geq n} a_n$$ and $$\limsup_{n\to\infty}a_n=\inf_{n\geq 0}\sup_{k\geq n} a_n.$$

We can now use the same definition for sets: $$\liminf_{n\to\infty}A_n=\sup_{n\geq 0}\inf_{k\geq n} A_n$$ and $$\limsup_{n\to\infty}A_n=\inf_{n\geq 0}\sup_{k\geq n} A_n.$$ Rewriting this in terms of $\bigcup$ and $\bigcap$, we get precisely the definitions from the question.

Answer 2

The $\limsup$ is the collection of all elements which appear in every tail of the sequence, namely results which occur infinitely often in the sequence of events.

The $\liminf$ is the union of all elements appearing in all events from a certain point in time, namely results which occur in all but finitely many events of the sequence.

Answer 3

For the begining note that $$ s\in\bigcap\limits_{\lambda\in\Lambda} S_\lambda\Longleftrightarrow \forall\lambda\in\Lambda\quad s\in S_\lambda $$ $$ s\in\bigcup\limits_{\lambda\in\Lambda} S_\lambda\Longleftrightarrow \exists \lambda\in\Lambda\quad s\in S_\lambda $$ Now we can show more formal explanation of what $\limsup$ is. $$ x\in\limsup\limits_{n\to\infty}A_n \Longleftrightarrow x\in\bigcap\limits_{n=0}^\infty\bigcup\limits_{k=n}^\infty A_k \Longleftrightarrow \forall n\in\mathbb{N}\quad x\in\bigcup\limits_{k=n}^\infty A_k \Longleftrightarrow $$ $$ \forall n\in\mathbb{N}\quad\exists k\geq n\quad x\in A_k $$ This means exactly that $x$ occurs in the sequence of sets $\{A_n:n\in\mathbb{N}\}$ infinitely many times.

Here is a formal explanation for $\liminf$ $$ x\in\liminf\limits_{n\to\infty}A_n \Longleftrightarrow x\in\bigcup\limits_{n=0}^\infty\bigcap\limits_{k=n}^\infty A_k \Longleftrightarrow \exists n\in\mathbb{N}\quad x\in\bigcap\limits_{k=n}^\infty A_k \Longleftrightarrow $$ $$ \exists n\in\mathbb{N}\quad\forall k\geq n\quad x\in A_k $$ This means exactly that $x$ occurs in all the sets of sequence $\{A_n:n\in\mathbb{N}\}$ except maybe in the first $n$ sets.

Answer 4

Read 'forall' with intersection, 'exists' with union. I prefer to think of events as follows:

$x \in\limsup A_n$ iff $\forall n$ $\exists k \geq n$ such that $x \in A_k$.

$x \in\liminf A_n$ iff $\exists n$ $\forall k \geq n$ we have $x \in A_k$.

Answer 5

Its the same intuition you have for sequence of real numbers $\{x_n \in \mathbb{R}\}$.

Recall for $\{x_n \in \mathbb{R}\}$,

$y_n = \sup\{x_{n\ge 1}\}, \sup\{x_{n\ge 2}\}, ....$ we sup'ing less & less pts, so $y_n \downarrow$

$z_n = \inf\{x_{n\ge 1}\}, \inf\{x_{n\ge 2}\}, ....$ we inf'ing less & less pts, so $z_n \uparrow$

So lim sup := limit of $y_n$, must be an $y_n$ that is smaller than all $y_n$, so lim sup $= \inf\{y_n\}$

So lim inf := limit of $z_n$, must be an $z_n$ that greater than all $z_n$, so lim inf $= \sup\{z_n\}$

-------End of recall.

Now for sequence of sets,$\{X_n \subseteq \mathbb{R}\}$

$Y_n = \bigcup_{n\ge 1}X_{n}, \bigcup_{n\ge 2}X_n,...$ we union'ing less & less subsets, so $Y_n \downarrow$

$Z_n = \bigcap_{n\ge 1}X_n, \bigcap_{n\ge 2}X_n,...$ we intersecting less & less subsets, so $Z_n \uparrow$

So lim sup := limit of $Y_n$, must be an $Y_n$ that is smaller than (so subset to) all $Y_n$,

so lim sup $= \bigcap_{n=1} Y_n$, and, $\bigcap_{n=1} Y_n = \bigcap_{n=1} (\bigcup_{n \ge 1} X_n)$

So lim inf := limit of $Z_n$, must be an $Z_n$ that greater than (so include) all $Z_n$,

so lim inf $= \bigcup Z_n$, and, $\bigcup_{n=1} Z_n = \bigcup_{n=1} (\bigcap_{n \ge 1} X_n)$

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In conclusion, $\le $ becomes $\subseteq$, $\inf$ becomes $\bigcap$, $\sup$ becomes $\bigcup$.