# Proving the set is measurable

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.

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.

• 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? – SL_MathGuy Dec 19 '19 at 7:58
• @SL_MathGuy When you say 'countable many' do you really mean infinitely many? – Kavi Rama Murthy Dec 19 '19 at 8:07
• I meant, it should be countably infinite OR finite right?. – SL_MathGuy Dec 19 '19 at 8:13
• @SL_MathGuy Any subset of $\mathbb N$ is countably infinite or finite, so your set is the entire space $\Omega$. – Kavi Rama Murthy Dec 19 '19 at 8:15
• @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. – Kavi Rama Murthy Dec 19 '19 at 8:20

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".

• 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'. – SL_MathGuy Dec 19 '19 at 8:29
• @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. – Gune Dec 19 '19 at 8:39