Prove a composite function involving cantor set is measurable
Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by
$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$
and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.
I know that $f$ is a measurable function. Than I proceed by
$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.
I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?
real-analysis measure-theory lebesgue-measure
asked Nov 27 '18 at 3:31
Ricky Ng
438
add a comment |
Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by
$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$
and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.
I know that $f$ is a measurable function. Than I proceed by
$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.
I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?
real-analysis measure-theory lebesgue-measure
asked Nov 27 '18 at 3:31
Ricky Ng
438
Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36
yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41
What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55
I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58
add a comment |
Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by
$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$
and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.
I know that $f$ is a measurable function. Than I proceed by
$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.
I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?
real-analysis measure-theory lebesgue-measure
asked Nov 27 '18 at 3:31
Ricky Ng
438
Let $cal K$ be the Cantor set. Let $f : mathbb R to mathbb R$ be defined by
$$f(x)=
begin{cases}
1, &mbox{ if }x in mathcal K\
0, &mbox{ otherwise }\
end{cases}
$$
and let $g : mathbb R to mathbb R$ be continuous.
Prove that $f circ g = h$ is measurable.
I know that $f$ is a measurable function. Than I proceed by
$$h^{-1}([a, infty])=
begin{cases}
mathbb R, &text{ if $a le 0$} \
B, &text{ if $0<ale 1$} \
varnothing, &text{ if $a>1$.}
end{cases}
$$
where $B = {x mid g(x) ∈ K}$.
I feel it very hard to show $B$ is measurable. I know $m(g(B)) = 0$ is. As it is subset of Cantor set which is measure zero. But than I cannot proceed. What should I do?
real-analysis measure-theory lebesgue-measure
real-analysis measure-theory lebesgue-measure
asked Nov 27 '18 at 3:31
Ricky Ng
438
asked Nov 27 '18 at 3:31
Ricky Ng
438
edited Nov 27 '18 at 4:01
asked Nov 27 '18 at 3:31
Ricky Ng
438
asked Nov 27 '18 at 3:31
Ricky Ng
438
asked Nov 27 '18 at 3:31
Ricky Ng
438
438
Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36
yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41
What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55
I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58
add a comment |
Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36
yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41
What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55
I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58
Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36
Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36
yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41
yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41
What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55
What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55
I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58
I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58
add a comment |
2 Answers
2
active
oldest
votes
For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
– Ricky Ng
Nov 27 '18 at 6:13
@RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
– Guacho Perez
Nov 27 '18 at 6:28
add a comment |
If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$
$g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set
answered Nov 27 '18 at 3:59
ZAF
4307
add a comment |
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2 Answers
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2 Answers
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oldest
votes
For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
– Ricky Ng
Nov 27 '18 at 6:13
@RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
– Guacho Perez
Nov 27 '18 at 6:28
add a comment |
For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
– Ricky Ng
Nov 27 '18 at 6:13
@RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
– Guacho Perez
Nov 27 '18 at 6:28
add a comment |
For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
For a Lebesgue set $L$:$$h^{-1}(L)=begin{cases}g^{-1}(mathcal K) & text{if $1in L$ and $0notin L$} \ g^{-1}(Bbb R)& text{if $1in L$ and $0in L$} \g^{-1}(mathcal K^c)& text{if $1notin L$ and $0in L$} \ g^{-1}(varnothing)& text{if $1notin L$ and $0notin L$} \end{cases}$$
Since $mathcal K$ is closed, $mathcal K^c$ is open. The continuity of $g$ shows $g^{-1}(mathcal K)$ is closed and $g^{-1}(mathcal K^c)$ is open. The other two cases are trivial.
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
answered Nov 27 '18 at 4:04
Guacho Perez
3,88911131
3,88911131
How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
– Ricky Ng
Nov 27 '18 at 6:13
@RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
– Guacho Perez
Nov 27 '18 at 6:28
add a comment |
How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
– Ricky Ng
Nov 27 '18 at 6:13
@RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
– Guacho Perez
Nov 27 '18 at 6:28
How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
– Ricky Ng
Nov 27 '18 at 6:13
How can we deduce from $K^{c}$ is open to it is measurable. I have only learned that close sets are measurable and there are no other material mentioning that on the net
– Ricky Ng
Nov 27 '18 at 6:13
@RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
– Guacho Perez
Nov 27 '18 at 6:28
@RickyNg If $U$ is open, then $U^c$ is closed, hence $U^c in mathcal M$ (the sigma algebra of measurable sets). But $mathcal M$ is closed under complements so $U=(U^c)^c in mathcal M$.
– Guacho Perez
Nov 27 '18 at 6:28
add a comment |
If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$
$g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set
answered Nov 27 '18 at 3:59
ZAF
4307
add a comment |
If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$
$g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set
answered Nov 27 '18 at 3:59
ZAF
4307
add a comment |
If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$
$g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set
answered Nov 27 '18 at 3:59
ZAF
4307
If $B = {x : g(x) in K}$, then $B = g^{-1}(K)$
$g$ is a continuous function, $K$ is a closed set, then $g^{-1}(K)$ is a closed set so is a measurable set
answered Nov 27 '18 at 3:59
ZAF
4307
answered Nov 27 '18 at 3:59
ZAF
4307
answered Nov 27 '18 at 3:59
ZAF
4307
answered Nov 27 '18 at 3:59
ZAF
4307
4307
add a comment |
add a comment |
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Is $f=chi_K$ ?
– Chinnapparaj R
Nov 27 '18 at 3:36
yes $f$ is characterstic function of $K$
– Ricky Ng
Nov 27 '18 at 3:41
What is your definition of measurable? Borel measurable? Lebesgue measurable? something else? What do you mean by $h^{-1}(a)$? Presumably, you mean the preimage of some set, but it is not clear which set you are considering.
– Xander Henderson
Nov 27 '18 at 3:55
I consider Lebesgue measurable. $h$ is measurable if $h^{-1}([a,b])$ is measurable, so i am taking advantage of this
– Ricky Ng
Nov 27 '18 at 3:58