# $\lbrace \lim f_n(x) \rbrace$ is a Borel set if each $f_n$ is borel

Suppose for all $n$ that $f_n:\mathbb{R}\to \mathbb{R}$ is Borel measurable. What follows is an attempt of the proof that $\lbrace x: \lim_{n\to \infty} f_n\rbrace$ is Borel measurable, but I am a bit hesitant since I don't see how this proof would be any different if the word "Borel" was not included in the problem (not ever measurable set is Borel).

I want to show $A=\lbrace x | \lim f_n \,\,\text{exists}\rbrace$ is a Borel set. Let $g(x)=\limsup f_n(x),h(x)=\liminf f_n(x).$ Then $A=\lbrace x:g(x)=h(x)\rbrace$. Since $h\le g$, $A=\lbrace x:g(x)=h(x)\rbrace \cup \lbrace h(x)<g(x) \rbrace$ and the sets in the union are disjoint so it is enough to show that one of them is Borel. $\lbrace x: h(x) < g(x)\rbrace=\cup_{q\in \mathbb{Q}}\lbrace x: h(x)<q\rbrace\cap \lbrace x: q<g(x)\rbrace$. The last two sets are Borel since they are just $h^{-1}((-\infty,q]), g^{-1}([q,\infty)$. So $A$ is Borel.

Does my "proof" have any chance?

• You have to show that $g,h$ are Borel measurable, in order to justify your last step. However, this is an obvious consequence of the fact that all $f_n$ are Borel. – Crostul May 29 '15 at 8:48

You have to show that $g,h$ are Borel measurable, in order to justify your last step (otherwise you would have that $h^{-1}((- \infty , q]), g^{-1}([q, + \infty))$ are only measurable). However, this is an obvious consequence of the fact that all $f_n$ are Borel.