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I have a question about an answer to this, it is to show $$\{ x \in X: \lim_{n\rightarrow \infty}f_{n}(x)\text{ exists and is finite}\}$$

is measurable, given $f_n$ is measurable and $X$ is a measure space.The first answer appeals to Cauchy sequences to avoid dealing with the limit function, but given that the limit function can be shown to be measurable can this be avoided so the answer becomes as follows: Let $f_n \to f$ ($f$ might be different for each $x$ in that set) then

$$ \begin{align*} \{x \in X\,&:\,\lim_{n\to\infty} f_n(x) \text{ exists and is in }\mathbb{R}\} \\ & = \{x \in X\,:\,\forall k \in \mathbb{N},\,\exists N \in \mathbb{N},\,\forall n \geq N \; |f_n(x) - f(x)| \lt 1/k\} \\ & = \bigcap_{k = 1}^\infty \bigcup_{N = 1}^\infty \bigcap_{n \geq N} \{x \in X\,:\,\lvert f_n(x)-f(x)\rvert \lt 1/k\} \end{align*} $$ The last set is measurable since $f_n-f$ is a difference of measurable finite functions and taking the absolute value does not remove measurability.

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I think once you prove $\{x \in X : \lim \limits_{n \to \infty} f_{n}(x) \text{ exists and is in } \mathbb{R} \} = \bigcap \limits_{k = 1}^{\infty} \bigcup \limits_{N = 1}^{\infty} \bigcap \limits_{n \geq N} \{ x \in X : |f_{n}(x) - f(x) | < \frac{1}{k} \}$, then since the set on the right hand side is measurable (if $f$ is measurable), we do get the desired result.

I had to prove that the two sets are equal because I wasn't immediately sure. Here is my argument:

Let $x \in \{x \in X : \lim \limits_{n \to \infty} f_{n}(x) \text{ exists and is in } \mathbb{R} \}$. Then $\forall k \in \mathbb{N}$, $\exists N(k)$ such that for all $n \geq N$, $|f_{n} - f| < \frac{1}{k}$. This means given $k$, finding the $N(k)$ that exists implies $x \in \bigcap \limits_{n \geq N(k)} \{ x \in X : |f_{n} - f| < \frac{1}{k} \}$.

But $x \in \bigcap \limits_{n \geq N(k)} \{ x \in X : |f_{n} - f| < \frac{1}{k} \} \implies x \in \bigcup \limits_{N = 1}^{\infty} \bigcap \limits_{n \geq N} \{ x \in X : |f_{n} - f| < \frac{1}{k} \}$, and this is true for all $k$, which means $x \in \bigcap \limits_{k = 1}^{\infty} \bigcup \limits_{N = 1}^{\infty} \bigcap \limits_{n \geq N} \{ x \in X : |f_{n}(x) - f(x) | < \frac{1}{k} \}$.

Now for the reverse direction:

Let $x \in \bigcap \limits_{k = 1}^{\infty} \bigcup \limits_{N = 1}^{\infty} \bigcap \limits_{n \geq N} \{ x \in X : |f_{n}(x) - f(x) | < \frac{1}{k} \}$. Then for each $k$, $x \in \bigcup \limits_{N = 1}^{\infty} \bigcap \limits_{n \geq N} \{ x \in X : |f_{n}(x) - f(x) | < \frac{1}{k} \}$. But this implies for some $N(k)$, $x \in \bigcap \limits_{n \geq N(k)} \{ x \in X : |f_{n}(x) - f(x)|< \frac{1}{k} \}$. But this means for each $k$, we found $N(k)$ such that $n \geq N(k) \implies |f_{n} - f| < \frac{1}{k}$. Thus, the limit of $f_{n}$ exists and is equal to $f$ for that $x$, which means $x \in \{x \in X : \lim \limits_{n \to \infty} f_{n}(x) \text{ exists and is in } \mathbb{R} \} $.

Now that we have proved the two sets are equal, if you prove $f$ is a measurable function, then as you said, the set on the right hand side is a countable intersection of countable unions of measurable sets, so it is measurable. Thus, the set on the left hand side is measurable. So it looks like the argument you suggested does work, provided you know $f$ is measurable.

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  • Thank you for your answer, I'm sorry you had to type a lot. My question is about not using the fact that $f_n$ is Cauchy, not about showing that the absolute value is measurable . – user63697 Sep 28 '14 at 12:38
  • @user63697 Ah, I see. I think what you are asking is: if you can show that the limit of a sequence of measurable functions is measurable, would the argument you posted above be enough to prove that set is measurable? – layman Sep 28 '14 at 13:36
  • Yes that is it, sorry for the misunderstanding – user63697 Sep 28 '14 at 13:41
  • @user63697 I think the whole point of the answer in which Cauchy sequences were used is to avoid showing that the limit $f$ is measurable. Once you have that the limit is measurable (by proving that $\liminf f_{n}$ and $\limsup f_{n}$ are measurable), then we know the function $\liminf f_{n} - \limsup f_{n}$ is meausurable, and so ${ x \mid \liminf f_{n} - \limsup f_{n} = 0 }$ is a measurable set. Finally, this set could include $\pm \infty$ since the lim inf and lim sup can output infinite values. So we intersect ${ x \mid \liminf f_{n} - \limsup f_{n} = 0 }$ with... – layman Sep 28 '14 at 13:48
  • @user63697 $(\liminf f_{n})^{-1}(\mathbb{R})$ and $(\limsup f_{n})^{-1}(\mathbb{R})$, both of which are measurable sets, to get the set where the limit function exists. And this would show that the set is measurable. – layman Sep 28 '14 at 13:49
  • @user63697 At least, that is what I think is going on. I hope someone else decides to answer this question so we know for sure. – layman Sep 28 '14 at 13:52
  • No need to take the difference, just consider the set on which they are equal. I'm just trying to see if this way works. – user63697 Sep 28 '14 at 13:57
  • @user63697 I updated my answer. Please review it and let me know if my logic is sound. It looks like your suggestion does work, assuming the function $f$ is measurable. – layman Sep 28 '14 at 14:15