I have been wrestling with this question and I am not sure how to solve it.


Let $(X, s)$ be a measure space and $\{f_n\}$ a sequence of measurable functions such that $f_n:X\to R$ with the normal Borel sigma algebra on $R$.

Let $$A = \{x\in X | \lim_{n\to\infty} f_n(x)\text{ exists}\}$$

Show $A$ is measurable.


I know we need to show $A$ is in s but I don't know how, we don't have integrals yet so this needs to be proved only using facts about measurable functions.

Help would be greatly appreciated.( I saw another post on this but it did not help)

  • 4
    $\begingroup$ Hint: $$A(n,m,k) = \left\{ x : \lvert f_n(x) - f_m(x)\rvert \leqslant \frac{1}{k}\right\}$$ is measurable for all $n,m,k$. $\endgroup$ – Daniel Fischer Nov 15 '14 at 23:13
  • $\begingroup$ I don't think integrals would help here to be honest. $\endgroup$ – Patrick Da Silva Nov 16 '14 at 12:49
  • $\begingroup$ $\mathbb{R}$ is complete so it suffices to consider Cauchy sequences, and the difference of measurable functions is measurable. $\endgroup$ – user682705 Oct 21 at 9:57

Hint : Show that $\liminf f_n$ is measurable. The proof for $\limsup f_n$ is similar, and $$ \{ x \in X \, | \, \lim_{n \to \infty} f_n \text{ exists} \} = \{ x \in X \, | \, \liminf_{n\to \infty} f_n = \limsup_{n\to \infty} f_n\}. $$ Hope that helps,

  • $\begingroup$ But lim sup f_n is not measurable as it may not be finite.... $\endgroup$ – Mathstudent Nov 16 '14 at 2:38
  • $\begingroup$ @Mathstudent : You can either work with $\pm \infty$ in your $\sigma$-algebra for $\mathbb R$ or take the subset of points where it is finite. (It is less of a problem as in the limit case, because at least the $\limsup$ is always defined, just sometimes taking the $\infty$/$-\infty$ value.) $\endgroup$ – Patrick Da Silva Nov 16 '14 at 12:49

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