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Let {$f_n$} be a sequence of measurable real-valued functions. Prove that the set of points $\omega$ for which the sequence of values $f_n(\omega)$ changes sign infinitely often is measurable.

My thoughts: I know this basically means that $f_n(\omega)>0$ & $f_n(\omega)<0$ for infinitely many $\omega$ values. My idea is to write this statement as,

{$\omega$:$f_n(\omega)$ changes the sign infinitely often}.

Then, express this set in terms of unions & intersection to come to conclusion.

But, the problem is 'infinitely often' part.Ideally, I would wanna consider the negation of the statement and then prove that the complement of the set is measurable. How would we express the negation of this statement?. Is it

$f_n(\omega)\leq 0$ countably many times OR $f_n(\omega)\geq 0$ countably many times?.

Any help would be appreciated.

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The set of points at which $f_n$ is positive for infinitely many values of $n$ can be written as $\bigcap _n \bigcup _{m \geq n} \{x:f_m(x) >0\}$ and this set is measurable. Similarly, the set of points at which $f_n$ is negative for infinitely many values of $n$ can be written as $\bigcap _n \bigcup _{m \geq n} \{x:f_m(x) <0\}$ and this set is also measurable. The given set is the intersection of these two sets.

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  • $\begingroup$ So, no need to consider the negation of the statement?. Is it possible to conclude the set, {$\omega$: $f_n(ω) \leq 0$ countably many times OR $f_n(ω)\geq 0$ countably many times} is measurable? $\endgroup$ – SL_MathGuy Dec 19 '19 at 7:58
  • $\begingroup$ @SL_MathGuy When you say 'countable many' do you really mean infinitely many? $\endgroup$ – Kavi Rama Murthy Dec 19 '19 at 8:07
  • $\begingroup$ I meant, it should be countably infinite OR finite right?. $\endgroup$ – SL_MathGuy Dec 19 '19 at 8:13
  • $\begingroup$ @SL_MathGuy Any subset of $\mathbb N$ is countably infinite or finite, so your set is the entire space $\Omega$. $\endgroup$ – Kavi Rama Murthy Dec 19 '19 at 8:15
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    $\begingroup$ @SL_MathGuy The statement '$f_n(x)>0$ for infinitely many values of $n$' is equivalent to the statement 'for every $n$ there exists $m \geq n$ such that $f_m(x) >0$'. I have just written this in set theoretic notation. $\endgroup$ – Kavi Rama Murthy Dec 19 '19 at 8:20
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First of all notice that "$f_n(\omega)>0$ & $f_n(\omega)<0$ for infinitely many $\omega$ values" this statement is not wahat you want.

since for a given $n$ you cannot have $f_n(\omega)>0$ & $f_n(\omega)<0$.

The statement should be, $f_n(\omega)>0$ & $f_m(\omega)<0$ for infinitely many $n,m$ values.

So if $A= \{\omega:$ there are infinitely many $n$ s.t. $ f_n(w)>0 \}$ and $B= \{\omega:$ there are infinitely many $n$ s.t. $ f_n(w)<0 \}$

then your required set is $A \cap B$.

For the negation, your word "countably many" is misleading. It should be "finite".

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  • $\begingroup$ I don't think your statement 'for a given $n$ you cannot have $f_n(ω)>0$ & $f_n(ω)<0$' is quite true. This is a sequence of functions. For ex: Suppose $f_n(x)=x/n$ on $[-1,1]$ and let $n=1$. Then, $f_1(x)=x$.There are $x$ values s.t $f_1(x)<0$ & $f_1(x)>0$. But, I agree with the rest of the stuff. I should've used 'finite'. $\endgroup$ – SL_MathGuy Dec 19 '19 at 8:29
  • $\begingroup$ @SL_MathGuy Yes. That's why I did not use the word "wrong". It's simply not what you want. First, you have to fix a $\omega$, then check are there infinitely many $n,m$ such that $f_n(\omega)>0$ & $f_m(\omega)<0$ for that $\omega$. If yes $\omega$ is in your set. if not then $\omega$ is not in your set. $\endgroup$ – Gune Dec 19 '19 at 8:39

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