Proving that $$1_{\limsup A_n}=\limsup1_{A_n}$$

$$1_{\limsup A_n}= \begin{cases}1, x \in \bigcap_{n=1}^{+ \infty}\bigcup_{k=n}^{+\infty}A_n \\ 0, x \notin \ldots\end{cases}$$

or that $x$ is in infinitely many $A_n$. And also:

$$\limsup1_{A_n}=\inf_{n\geq1}\{\sup_{k \geq n}\{1_{A_n}\}\}$$ this function seems to be $1$ when $x\in A_n$ (just one set). I clearly am not understanding this $\inf\{\sup\{\ldots\}\}$ definition correctly. Can someone enlighten this for me. I have a test very soon.

  • $\begingroup$ If $x\in A_n$ and no other $A_i$ (or even "no later $A_i$"), then $\sup_{k\geq n+1}[1_{A_k}(x)]=0$. $\endgroup$
    – Arthur
    Sep 5, 2016 at 10:02
  • $\begingroup$ That's right, I have been so blind. $\endgroup$ Sep 5, 2016 at 10:03
  • $\begingroup$ Ill leave the question open still, for a formal complete answer. I understand what you right, but not exactly the answer fully. Now it seems to me that the function is $1$ when $x\in$ every $A_n$ $\endgroup$ Sep 5, 2016 at 10:04

2 Answers 2


For an arbitrary $x$ the following statements are equivalent:

  • $1_{\limsup A_{n}}\left(x\right)=1$

  • $x\in\limsup A_{n}$

  • $\forall n\in\mathbb{N}\exists k\geq n\left[x\in A_{k}\right]$

  • $\forall n\in\mathbb{N}\exists k\geq n\left[1_{A_{k}}\left(x\right)=1\right]$

  • $\forall n\in\mathbb{N}\sup_{k\geq n}1_{A_{k}}\left(x\right)=1$

  • $\lim_{n\to\infty}\sup_{k\geq n}1_{A_{k}}\left(x\right)=1$

The last expression can be written as $\limsup1_{A_{n}}\left(x\right)=1$.

So we have: $$\forall x\left[1_{\limsup A_{n}}\left(x\right)=1\iff\limsup1_{A_{n}}\left(x\right)=1\right]$$ or equivalently: $$1_{\limsup A_{n}}=\limsup1_{A_{n}}$$

If $s_n:=\sup_{k\geq n}1_{A_{k}}\left(x\right)$ then $(s_n)_n$ by definition is a decreasing sequence in $\{0,1\}$ so that it will converge and: $$\lim s_n=\inf\{s_n\mid n\in\mathbb N\}$$

  • $\begingroup$ formality at it's finest. $\endgroup$ Sep 5, 2016 at 10:33
  • 1
    $\begingroup$ Glad to help you. $\endgroup$
    – drhab
    Sep 5, 2016 at 10:33

Let $f_n(x)=\sup_{k\geq n}\boldsymbol 1_{A_k}(x)$. First, $$f_n(x)\in\{0,1\},$$ for all $x$ and all $n$. For $x$ fixed, the sequence $(f_n(x))_n$ is decreasing, therefore $$\lim_{n\to \infty }f_n(x)=\inf_{n\in\mathbb N} f_n(x)\in \{0,1\}$$ since an infimum is an adherent point. Let fix $x$. Now, $$\inf_{n\in\mathbb N}f_n(x)=0\iff \exists k:f_k(x)=0.$$ In other word, if only if there is a $k$ s.t. for all $m\geq k$, $\boldsymbol 1_{A_m}(x)=0$. Therefore, $$\lim_{n\to \infty }f_n(x)=0\iff \exists k\in\mathbb N: \forall m\geq k, \boldsymbol 1_{A_m}(x)=0\iff x\in \bigcup_{k}\bigcap_{m\geq k}A_m^c\iff x\notin \bigcap_{k}\bigcup_{m\geq k}A_m\iff x\notin \limsup A_m^c\iff \boldsymbol 1_{\limsup A_n}(x)=0.$$

I let you do the case where $\lim_{n\to \infty }f_n(x)=1$.


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