Mixing
Prove that $E$ is sigma algebra on $mathbb{R}$.
$begingroup$
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
edited Oct 4 '18 at 17:56
Henrik
5,99792030
asked Oct 4 '18 at 17:27
Duaa HamzehDuaa Hamzeh
664
$endgroup$
add a comment |
$begingroup$
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
edited Oct 4 '18 at 17:56
Henrik
5,99792030
asked Oct 4 '18 at 17:27
Duaa HamzehDuaa Hamzeh
664
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Oct 4 '18 at 17:56
Henrik
5,99792030
asked Oct 4 '18 at 17:27
Duaa HamzehDuaa Hamzeh
664
$endgroup$
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
edited Oct 4 '18 at 17:56
Henrik
5,99792030
asked Oct 4 '18 at 17:27
Duaa HamzehDuaa Hamzeh
664
edited Oct 4 '18 at 17:56
Henrik
5,99792030
asked Oct 4 '18 at 17:27
Duaa HamzehDuaa Hamzeh
664
edited Oct 4 '18 at 17:56
Henrik
5,99792030
edited Oct 4 '18 at 17:56
Henrik
5,99792030
edited Oct 4 '18 at 17:56
Henrik
5,99792030
5,99792030
asked Oct 4 '18 at 17:27
Duaa HamzehDuaa Hamzeh
664
asked Oct 4 '18 at 17:27
Duaa HamzehDuaa Hamzeh
664
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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".
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
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1 Answer
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1 Answer
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oldest
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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".
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
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".
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
add a comment |
$begingroup$
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".
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
$endgroup$
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".
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
answered Jan 8 at 15:07
Davide GiraudoDavide Giraudo
126k16150261
126k16150261
add a comment |
add a comment |
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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