Prove that $E$ is sigma algebra on $mathbb{R}$.












1












$begingroup$


Let
$E={usubset mathbb{R} : text{$u$ is finite or $complement u$ is finite}}$
Prove that $E$ is a sigma algebra on $mathbb{R}$.



I had checked the three conditions of a sigma algebra, and actually I got the proof.



But I'm confused because for example
${A_i={i}}$ belong to $E$ for all $i$, but the union doesn't!
And that doesn't match the condition of closed under countable union.
What is my mistake?
I think my example is false, but why?






measure-theory







share|cite|improve this question











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  • $begingroup$
    You made no mistake. In fact, you proved correctly that $E$ is not a $sigma$-algebra.
    $endgroup$
    – José Carlos Santos
    Oct 4 '18 at 17:28


















1












$begingroup$


Let
$E={usubset mathbb{R} : text{$u$ is finite or $complement u$ is finite}}$
Prove that $E$ is a sigma algebra on $mathbb{R}$.



I had checked the three conditions of a sigma algebra, and actually I got the proof.



But I'm confused because for example
${A_i={i}}$ belong to $E$ for all $i$, but the union doesn't!
And that doesn't match the condition of closed under countable union.
What is my mistake?
I think my example is false, but why?






measure-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    You made no mistake. In fact, you proved correctly that $E$ is not a $sigma$-algebra.
    $endgroup$
    – José Carlos Santos
    Oct 4 '18 at 17:28
















1












1








1





$begingroup$


Let
$E={usubset mathbb{R} : text{$u$ is finite or $complement u$ is finite}}$
Prove that $E$ is a sigma algebra on $mathbb{R}$.



I had checked the three conditions of a sigma algebra, and actually I got the proof.



But I'm confused because for example
${A_i={i}}$ belong to $E$ for all $i$, but the union doesn't!
And that doesn't match the condition of closed under countable union.
What is my mistake?
I think my example is false, but why?






measure-theory







share|cite|improve this question











$endgroup$




Let
$E={usubset mathbb{R} : text{$u$ is finite or $complement u$ is finite}}$
Prove that $E$ is a sigma algebra on $mathbb{R}$.



I had checked the three conditions of a sigma algebra, and actually I got the proof.



But I'm confused because for example
${A_i={i}}$ belong to $E$ for all $i$, but the union doesn't!
And that doesn't match the condition of closed under countable union.
What is my mistake?
I think my example is false, but why?






measure-theory




measure-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 4 '18 at 17:56









Henrik

5,99792030




5,99792030










asked Oct 4 '18 at 17:27









Duaa HamzehDuaa Hamzeh

664




664












  • $begingroup$
    You made no mistake. In fact, you proved correctly that $E$ is not a $sigma$-algebra.
    $endgroup$
    – José Carlos Santos
    Oct 4 '18 at 17:28




















  • $begingroup$
    You made no mistake. In fact, you proved correctly that $E$ is not a $sigma$-algebra.
    $endgroup$
    – José Carlos Santos
    Oct 4 '18 at 17:28


















$begingroup$
You made no mistake. In fact, you proved correctly that $E$ is not a $sigma$-algebra.
$endgroup$
– José Carlos Santos
Oct 4 '18 at 17:28






$begingroup$
You made no mistake. In fact, you proved correctly that $E$ is not a $sigma$-algebra.
$endgroup$
– José Carlos Santos
Oct 4 '18 at 17:28












1 Answer
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$begingroup$

As pointed out by José Carlos Santos, you proved that $E$ is not a $sigma$-algebra.



A small modification of the definition of $E$ would make is a $sigma$-algebra, namely, replacing the word "finite" by "at most countable".






share|cite|improve this answer









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    0












    $begingroup$

    As pointed out by José Carlos Santos, you proved that $E$ is not a $sigma$-algebra.



    A small modification of the definition of $E$ would make is a $sigma$-algebra, namely, replacing the word "finite" by "at most countable".






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      As pointed out by José Carlos Santos, you proved that $E$ is not a $sigma$-algebra.



      A small modification of the definition of $E$ would make is a $sigma$-algebra, namely, replacing the word "finite" by "at most countable".






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        As pointed out by José Carlos Santos, you proved that $E$ is not a $sigma$-algebra.



        A small modification of the definition of $E$ would make is a $sigma$-algebra, namely, replacing the word "finite" by "at most countable".






        share|cite|improve this answer









        $endgroup$



        As pointed out by José Carlos Santos, you proved that $E$ is not a $sigma$-algebra.



        A small modification of the definition of $E$ would make is a $sigma$-algebra, namely, replacing the word "finite" by "at most countable".







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 15:07









        Davide GiraudoDavide Giraudo

        126k16150261




        126k16150261






























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