Prove that there exists a limit of the sequence of sets $E_n$ if and only if there exists a limit of the sequence of the characteristic functions of $E_n$.


$$\exists \lim_{n \to \infty}E_n \iff \exists \lim_{n \to \infty}\lambda E_n$$

This seems very obvious at first glance but to prove it, not so much . I have been taught that a limit of a function exists if: (and it is equal to)$$\overline{\lim_{n \to \infty}}f_n(x)= \inf_{n\geq1}\{\sup_{k\geq n}f_k(x)\}=\underline{\lim_{n \to \infty}}f_n(x)= \sup_{n\geq1}\{\inf_{k\geq n}f_k(x)\}=\lim_{n\to \infty}f_n(x)$$ And for sets $E_n$ in a way, analogously: $$\overline{\lim_{n \to \infty}}E_n= \bigcap_{n=1}^{+ \infty}\bigcup_{k=n}^{+\infty} E_k=\underline{\lim_{n \to \infty}}f_n(x)= \bigcup_{n=1}^{+ \infty}\bigcap_{k=n}^{+\infty} E_k=\lim_{n\to \infty}E_n$$

How would I use these two definitions to put this together?

  • $\begingroup$ Can you say explicitly what you mean by the "limit of the sequence of sets"? Perhaps it is said implicitly by some equations you wrote, but if so, those are not standard definitions, and it would help greatly if you were explicit about this point. $\endgroup$ – Lee Mosher Aug 24 '16 at 13:13
  • $\begingroup$ @LeeMosher A sequence of sets $(A_n)_n$ is convergent if $\limsup A_n=\liminf A_n$. If so then the limit equals both sides of the equality. $\endgroup$ – Vera Aug 24 '16 at 14:51


It is enough to prove that:

$$1_{\limsup A_n}=\limsup1_{A_n}$$ and: $$1_{\liminf A_n}=\liminf1_{A_n}$$

There is convergence of characteristic functions if the right-hand-sides are equal:$$\limsup1_{A_n}=\liminf1_{A_n}$$

There is convergence of sets if the left-hand-sides are equal:$$1_{\limsup A_n}=1_{\liminf A_n}$$


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