Show that the family $A^*$ of $mu^*$-measurable sets is $sigma$-algebra (proof verification)
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This is from Bartle text book, and this page proves that the family $A^*$ of $mu^*$-measurable sets is $sigma$-algebra. I don't understand the red line. Could you elaborate on this?
measure-theory elementary-set-theory
asked Aug 12 at 5:09
Sihyun Kim
716211
add a comment |Â
up vote
-1
down vote
favorite
This is from Bartle text book, and this page proves that the family $A^*$ of $mu^*$-measurable sets is $sigma$-algebra. I don't understand the red line. Could you elaborate on this?
measure-theory elementary-set-theory
asked Aug 12 at 5:09
Sihyun Kim
716211
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
This is from Bartle text book, and this page proves that the family $A^*$ of $mu^*$-measurable sets is $sigma$-algebra. I don't understand the red line. Could you elaborate on this?
measure-theory elementary-set-theory
asked Aug 12 at 5:09
Sihyun Kim
716211
This is from Bartle text book, and this page proves that the family $A^*$ of $mu^*$-measurable sets is $sigma$-algebra. I don't understand the red line. Could you elaborate on this?
measure-theory elementary-set-theory
asked Aug 12 at 5:09
Sihyun Kim
716211
asked Aug 12 at 5:09
Sihyun Kim
716211
asked Aug 12 at 5:09
Sihyun Kim
716211
asked Aug 12 at 5:09
Sihyun Kim
716211
716211
add a comment |Â
add a comment |Â
1 Answer
1
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up vote
0
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accepted
The simplest is just to draw a Venn diagram of $E,A,F$ and shade the areas involved.
$B$ is defined as $A setminus (E cap F)$. Show that $B cap F = (A cap F) setminus E$ by two inclusions:
Suppose $x in B cap F$, then $x in F$ and $x in B$, which means $x in A$ and $x notin E cap F$. We already know $x in F$ so $x notin E cap F$ implies that $x notin E$. So $x in A cap F$ and $x notin E$ so $x in (A cap F)setminus E$.
Suppose that $x in (A cap F) setminus F$ then $x in A$, $x in F$ and $x notin E$. So $x notin E cap F$ and as we also have $x in A$ we know that $x in B$ and we already knew $x in F$ so that $x in B cap F$.
Now $B setminus F = A setminus F$ can be seen in a similar way:
$x in B setminus F$, then $x in B$ and $x notin F$. As $B subseteq A$ we know that $x in A setminus F$.
If $x in A setminus F$, then $x notin E cap F$ so $x in A setminus (E cap F) = B$. Hence $x in B setminus F$.
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
Hi. Sorry for late comment, but how can we justify the last line "$xnot in Ecap F$, so $x in A cap (E cap F)^c$" ??
– Sihyun Kim
Aug 19 at 6:06
@SihyunKim $x in A setminus F$, so $x in A$ and $x notin F$. From $x notin F$ we know $x notin (E cap F)$. We still have $x in A$, so $x in A setminus (E cap F)$ by definition. The last set is the definition of $B$.
– Henno Brandsma
Aug 19 at 6:09
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
The simplest is just to draw a Venn diagram of $E,A,F$ and shade the areas involved.
$B$ is defined as $A setminus (E cap F)$. Show that $B cap F = (A cap F) setminus E$ by two inclusions:
Suppose $x in B cap F$, then $x in F$ and $x in B$, which means $x in A$ and $x notin E cap F$. We already know $x in F$ so $x notin E cap F$ implies that $x notin E$. So $x in A cap F$ and $x notin E$ so $x in (A cap F)setminus E$.
Suppose that $x in (A cap F) setminus F$ then $x in A$, $x in F$ and $x notin E$. So $x notin E cap F$ and as we also have $x in A$ we know that $x in B$ and we already knew $x in F$ so that $x in B cap F$.
Now $B setminus F = A setminus F$ can be seen in a similar way:
$x in B setminus F$, then $x in B$ and $x notin F$. As $B subseteq A$ we know that $x in A setminus F$.
If $x in A setminus F$, then $x notin E cap F$ so $x in A setminus (E cap F) = B$. Hence $x in B setminus F$.
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
Hi. Sorry for late comment, but how can we justify the last line "$xnot in Ecap F$, so $x in A cap (E cap F)^c$" ??
– Sihyun Kim
Aug 19 at 6:06
@SihyunKim $x in A setminus F$, so $x in A$ and $x notin F$. From $x notin F$ we know $x notin (E cap F)$. We still have $x in A$, so $x in A setminus (E cap F)$ by definition. The last set is the definition of $B$.
– Henno Brandsma
Aug 19 at 6:09
add a comment |Â
up vote
0
down vote
accepted
The simplest is just to draw a Venn diagram of $E,A,F$ and shade the areas involved.
$B$ is defined as $A setminus (E cap F)$. Show that $B cap F = (A cap F) setminus E$ by two inclusions:
Suppose $x in B cap F$, then $x in F$ and $x in B$, which means $x in A$ and $x notin E cap F$. We already know $x in F$ so $x notin E cap F$ implies that $x notin E$. So $x in A cap F$ and $x notin E$ so $x in (A cap F)setminus E$.
Suppose that $x in (A cap F) setminus F$ then $x in A$, $x in F$ and $x notin E$. So $x notin E cap F$ and as we also have $x in A$ we know that $x in B$ and we already knew $x in F$ so that $x in B cap F$.
Now $B setminus F = A setminus F$ can be seen in a similar way:
$x in B setminus F$, then $x in B$ and $x notin F$. As $B subseteq A$ we know that $x in A setminus F$.
If $x in A setminus F$, then $x notin E cap F$ so $x in A setminus (E cap F) = B$. Hence $x in B setminus F$.
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
Hi. Sorry for late comment, but how can we justify the last line "$xnot in Ecap F$, so $x in A cap (E cap F)^c$" ??
– Sihyun Kim
Aug 19 at 6:06
@SihyunKim $x in A setminus F$, so $x in A$ and $x notin F$. From $x notin F$ we know $x notin (E cap F)$. We still have $x in A$, so $x in A setminus (E cap F)$ by definition. The last set is the definition of $B$.
– Henno Brandsma
Aug 19 at 6:09
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
The simplest is just to draw a Venn diagram of $E,A,F$ and shade the areas involved.
$B$ is defined as $A setminus (E cap F)$. Show that $B cap F = (A cap F) setminus E$ by two inclusions:
Suppose $x in B cap F$, then $x in F$ and $x in B$, which means $x in A$ and $x notin E cap F$. We already know $x in F$ so $x notin E cap F$ implies that $x notin E$. So $x in A cap F$ and $x notin E$ so $x in (A cap F)setminus E$.
Suppose that $x in (A cap F) setminus F$ then $x in A$, $x in F$ and $x notin E$. So $x notin E cap F$ and as we also have $x in A$ we know that $x in B$ and we already knew $x in F$ so that $x in B cap F$.
Now $B setminus F = A setminus F$ can be seen in a similar way:
$x in B setminus F$, then $x in B$ and $x notin F$. As $B subseteq A$ we know that $x in A setminus F$.
If $x in A setminus F$, then $x notin E cap F$ so $x in A setminus (E cap F) = B$. Hence $x in B setminus F$.
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
The simplest is just to draw a Venn diagram of $E,A,F$ and shade the areas involved.
$B$ is defined as $A setminus (E cap F)$. Show that $B cap F = (A cap F) setminus E$ by two inclusions:
Suppose $x in B cap F$, then $x in F$ and $x in B$, which means $x in A$ and $x notin E cap F$. We already know $x in F$ so $x notin E cap F$ implies that $x notin E$. So $x in A cap F$ and $x notin E$ so $x in (A cap F)setminus E$.
Suppose that $x in (A cap F) setminus F$ then $x in A$, $x in F$ and $x notin E$. So $x notin E cap F$ and as we also have $x in A$ we know that $x in B$ and we already knew $x in F$ so that $x in B cap F$.
Now $B setminus F = A setminus F$ can be seen in a similar way:
$x in B setminus F$, then $x in B$ and $x notin F$. As $B subseteq A$ we know that $x in A setminus F$.
If $x in A setminus F$, then $x notin E cap F$ so $x in A setminus (E cap F) = B$. Hence $x in B setminus F$.
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
answered Aug 12 at 8:10
Henno Brandsma
91.8k342100
91.8k342100
Hi. Sorry for late comment, but how can we justify the last line "$xnot in Ecap F$, so $x in A cap (E cap F)^c$" ??
– Sihyun Kim
Aug 19 at 6:06
@SihyunKim $x in A setminus F$, so $x in A$ and $x notin F$. From $x notin F$ we know $x notin (E cap F)$. We still have $x in A$, so $x in A setminus (E cap F)$ by definition. The last set is the definition of $B$.
– Henno Brandsma
Aug 19 at 6:09
add a comment |Â
Hi. Sorry for late comment, but how can we justify the last line "$xnot in Ecap F$, so $x in A cap (E cap F)^c$" ??
– Sihyun Kim
Aug 19 at 6:06
@SihyunKim $x in A setminus F$, so $x in A$ and $x notin F$. From $x notin F$ we know $x notin (E cap F)$. We still have $x in A$, so $x in A setminus (E cap F)$ by definition. The last set is the definition of $B$.
– Henno Brandsma
Aug 19 at 6:09
Hi. Sorry for late comment, but how can we justify the last line "$xnot in Ecap F$, so $x in A cap (E cap F)^c$" ??
– Sihyun Kim
Aug 19 at 6:06
Hi. Sorry for late comment, but how can we justify the last line "$xnot in Ecap F$, so $x in A cap (E cap F)^c$" ??
– Sihyun Kim
Aug 19 at 6:06
@SihyunKim $x in A setminus F$, so $x in A$ and $x notin F$. From $x notin F$ we know $x notin (E cap F)$. We still have $x in A$, so $x in A setminus (E cap F)$ by definition. The last set is the definition of $B$.
– Henno Brandsma
Aug 19 at 6:09
@SihyunKim $x in A setminus F$, so $x in A$ and $x notin F$. From $x notin F$ we know $x notin (E cap F)$. We still have $x in A$, so $x in A setminus (E cap F)$ by definition. The last set is the definition of $B$.
– Henno Brandsma
Aug 19 at 6:09
add a comment |Â
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