Let $(f_n)_n$ a sequence of measurable function. Prove that {$\omega \in \Omega:(f_n(\omega))_n$ converges} is measurable.

My idea is write problem as follows:

For all $\ n \in \mathbb{N}$, exist $\ k \in \mathbb{N}$ such that for all $\ i,j>k \ $ then $\ |f_i (\omega)- f_j (\omega)|< 1/n$, last part because $\mathbb{R}$ is complete.


Let $A_{n,i,j}:=$ {$\omega \in \Omega:|f_i (\omega)- f_j (\omega)|< 1/n$},

then $\cap_{n\in \mathbb{N}} \ \cup_{k\in \mathbb{N}} \cap_{i,j>k}$ $ \ A_{n,i,j}$


Thanks for any help!

  • $\begingroup$ The last part does not follow because $\mathbb{R}$ is complete. It is true in any metric space that any convergent sequence is Cauchy. $\endgroup$ – mathworker21 Jun 26 '17 at 6:56
  • $\begingroup$ I believe you want $\cap_n\cup_k\cap_{i,j,>k}A_{n,i,j}$. As a rule of thumb - "for every" is translated to intersections, "there exists" - to unions. $\endgroup$ – uniquesolution Jun 26 '17 at 6:57

The good observation you partially made is the following:

$(f_n(\omega))_n$ is convergent if and only if for all $n \ge 1$ there is some $k$ such that for all $i,j > k$, $|f_i(\omega)-f_j(\omega)| < \frac{1}{n}$. The only if direction holds in any metric space, while the if direction uses completeness of $\mathbb{R}$.

Therefore, $\{\omega \in \Omega : (f_n(\omega))_n \hspace{1mm} \text{converges}\} = \cap_{n \ge 1}\cup_{k \ge 1} \cap_{i,j > k} \{\omega : |f_i(\omega)-f_j(\omega)| < \frac{1}{n}\}$. I think you had some unions and intersections mixed up. But the point is $\{\omega : |f_i(\omega)-f_j(\omega)| < \frac{1}{n}\}$ is measurable since $f_i-f_j$ is measurable, so if we do intersections and unions, the result is measurable.


Maybe you already know that for a sequence of measurable functions also $\limsup f_n$ and $\liminf f_n$ are measurable functions then it holds

$$\{ f_n \text{ converges} \} = \{\limsup f_n = \liminf f_n\} $$ and because $\limsup f_n$ and $\liminf f_n$ are measurable functions $\{\limsup f_n = \liminf f_n\} $ is measurable… so is $\{ f_n \text{ converges} \}$


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