Mixing
The Lebesgue integral of certain function on the set of irrational numbers of the unit interval
Clash Royale CLAN TAG#URR8PPP
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We define a function on $J$, the set of irrational points of the interval $(0,1)$ with $$f(0/a_1a_2a_3a_4ldots)=0/a_2a_1a_4a_3ldots$$ What is the value of Lebesgue integral $$int_J f $$ Furthermore is there an algebraic formulation for $f$?
calculus real-analysis lebesgue-integral
asked Jul 21 at 20:18
Ali Taghavi
148329
 |Â
show 2 more comments
up vote
1
down vote
favorite
We define a function on $J$, the set of irrational points of the interval $(0,1)$ with $$f(0/a_1a_2a_3a_4ldots)=0/a_2a_1a_4a_3ldots$$ What is the value of Lebesgue integral $$int_J f $$ Furthermore is there an algebraic formulation for $f$?
calculus real-analysis lebesgue-integral
asked Jul 21 at 20:18
Ali Taghavi
148329
Is $0/a_1a_2a_3a_4$ supposed to be a decimal expansion?
– Eric Wofsey
Jul 21 at 20:21
@EricWofsey Yes it is the unique decimal expansion.
– Ali Taghavi
Jul 21 at 20:22
Are you swapping the digits in each successive pair past the decimal point?
– MPW
Jul 21 at 20:31
@MPW I am sorry I do not understand your question. We swap with the formula $0/a_1a_2a_3a_4a_5a_6ldots mapsto 0/a_2a_1a_4a_3a_6a_5ldots $.
– Ali Taghavi
Jul 21 at 20:36
1
Cf. the Rademacher functions.
– copper.hat
Jul 21 at 21:07
 |Â
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
We define a function on $J$, the set of irrational points of the interval $(0,1)$ with $$f(0/a_1a_2a_3a_4ldots)=0/a_2a_1a_4a_3ldots$$ What is the value of Lebesgue integral $$int_J f $$ Furthermore is there an algebraic formulation for $f$?
calculus real-analysis lebesgue-integral
asked Jul 21 at 20:18
Ali Taghavi
148329
We define a function on $J$, the set of irrational points of the interval $(0,1)$ with $$f(0/a_1a_2a_3a_4ldots)=0/a_2a_1a_4a_3ldots$$ What is the value of Lebesgue integral $$int_J f $$ Furthermore is there an algebraic formulation for $f$?
calculus real-analysis lebesgue-integral
asked Jul 21 at 20:18
Ali Taghavi
148329
edited Jul 21 at 20:21
asked Jul 21 at 20:18
Ali Taghavi
148329
asked Jul 21 at 20:18
Ali Taghavi
148329
asked Jul 21 at 20:18
Ali Taghavi
148329
148329
Is $0/a_1a_2a_3a_4$ supposed to be a decimal expansion?
– Eric Wofsey
Jul 21 at 20:21
@EricWofsey Yes it is the unique decimal expansion.
– Ali Taghavi
Jul 21 at 20:22
Are you swapping the digits in each successive pair past the decimal point?
– MPW
Jul 21 at 20:31
@MPW I am sorry I do not understand your question. We swap with the formula $0/a_1a_2a_3a_4a_5a_6ldots mapsto 0/a_2a_1a_4a_3a_6a_5ldots $.
– Ali Taghavi
Jul 21 at 20:36
1
Cf. the Rademacher functions.
– copper.hat
Jul 21 at 21:07
 |Â
show 2 more comments
Is $0/a_1a_2a_3a_4$ supposed to be a decimal expansion?
– Eric Wofsey
Jul 21 at 20:21
@EricWofsey Yes it is the unique decimal expansion.
– Ali Taghavi
Jul 21 at 20:22
Are you swapping the digits in each successive pair past the decimal point?
– MPW
Jul 21 at 20:31
@MPW I am sorry I do not understand your question. We swap with the formula $0/a_1a_2a_3a_4a_5a_6ldots mapsto 0/a_2a_1a_4a_3a_6a_5ldots $.
– Ali Taghavi
Jul 21 at 20:36
1
Cf. the Rademacher functions.
– copper.hat
Jul 21 at 21:07
Is $0/a_1a_2a_3a_4$ supposed to be a decimal expansion?
– Eric Wofsey
Jul 21 at 20:21
Is $0/a_1a_2a_3a_4$ supposed to be a decimal expansion?
– Eric Wofsey
Jul 21 at 20:21
@EricWofsey Yes it is the unique decimal expansion.
– Ali Taghavi
Jul 21 at 20:22
@EricWofsey Yes it is the unique decimal expansion.
– Ali Taghavi
Jul 21 at 20:22
Are you swapping the digits in each successive pair past the decimal point?
– MPW
Jul 21 at 20:31
Are you swapping the digits in each successive pair past the decimal point?
– MPW
Jul 21 at 20:31
@MPW I am sorry I do not understand your question. We swap with the formula $0/a_1a_2a_3a_4a_5a_6ldots mapsto 0/a_2a_1a_4a_3a_6a_5ldots $.
– Ali Taghavi
Jul 21 at 20:36
@MPW I am sorry I do not understand your question. We swap with the formula $0/a_1a_2a_3a_4a_5a_6ldots mapsto 0/a_2a_1a_4a_3a_6a_5ldots $.
– Ali Taghavi
Jul 21 at 20:36
1
1
Cf. the Rademacher functions.
– copper.hat
Jul 21 at 21:07
Cf. the Rademacher functions.
– copper.hat
Jul 21 at 21:07
 |Â
show 2 more comments
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
Hints: (Thanks to @EricWofsey for catching an egregious error.)
Let $d_k(x)$ be the $k$th
element of decimal expansion of $x$. Then
$int d_k = 1 over 10 (0+1+cdots + 9)= 9 over 2$.
We have $x = sum_k=1^infty d_k(x) 1 over 10^k$.
Note that for any nice functions $g_k$ that satisfy $int g_k = 9 over 2$ we have
$int ( sum_k=1^infty g_k(x) 1 over 10^k ) dx =sum_k=1^infty int g_k(x) 1 over 10^k dx = 1 over 2$.
answered Jul 21 at 20:32
copper.hat
122k557156
If we integrate your formula $x = sum_k=1^infty d_k(x) 1 over 10^k$ we get contradiction assuming $int d_k=1/10$
– Ali Taghavi
Jul 21 at 20:49
1
That's 9/2, not 9.
– Eric Wofsey
Jul 21 at 21:07
1
@EricWofsey: Something is wrong with my computation
– copper.hat
Jul 21 at 21:08
1
@EricWofsey: Thanks for catching that, I was computing the integral as $int 1 dx$ instead of $int x dx$ to compute the constant. I didn't actually compute the sum :-(.
– copper.hat
Jul 21 at 21:14
1
@copper.hat Thank you very much for this very interesting answer.Sorry I did not read the first version of your answer carefully so I did not got your idea.
– Ali Taghavi
Jul 22 at 5:50
 |Â
show 2 more comments
up vote
3
down vote
Here is a probabilistic take on essentially the same method as copper.hat's answer. Lebesgue measure on $[0,1]$ is the same as the probability measure where you independently choose each digit uniformly from $0,1,dots,9$. Now, $f(x)$ (thought of as a random variable) has this same property: each digit of $f(x)$ is chosen independently and uniformly from $0,1,dots,9$, since the digits of $f(x)$ are just the digits of $x$ in a different order. This means that random variable $f(x)$ has the same probability distribution as the identity random variable $g(x)=x$, and so they have the same expectation. That is, $$int_J f(x),dx=int_J x,dx=frac12.$$
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
1
There is a wonderful monograph by Marc Kac (Statistical Independence in Probability, Analysis and Number Theory, I think) that starts with a similar argument (base 2). Worth a glance I believe...
– copper.hat
Jul 21 at 21:38
@Eric Thank you very much for your very interesting answer. i am sorry that i can not accept two answers simultaneously
– Ali Taghavi
Jul 22 at 5:51
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Hints: (Thanks to @EricWofsey for catching an egregious error.)
Let $d_k(x)$ be the $k$th
element of decimal expansion of $x$. Then
$int d_k = 1 over 10 (0+1+cdots + 9)= 9 over 2$.
We have $x = sum_k=1^infty d_k(x) 1 over 10^k$.
Note that for any nice functions $g_k$ that satisfy $int g_k = 9 over 2$ we have
$int ( sum_k=1^infty g_k(x) 1 over 10^k ) dx =sum_k=1^infty int g_k(x) 1 over 10^k dx = 1 over 2$.
answered Jul 21 at 20:32
copper.hat
122k557156
If we integrate your formula $x = sum_k=1^infty d_k(x) 1 over 10^k$ we get contradiction assuming $int d_k=1/10$
– Ali Taghavi
Jul 21 at 20:49
1
That's 9/2, not 9.
– Eric Wofsey
Jul 21 at 21:07
1
@EricWofsey: Something is wrong with my computation
– copper.hat
Jul 21 at 21:08
1
@EricWofsey: Thanks for catching that, I was computing the integral as $int 1 dx$ instead of $int x dx$ to compute the constant. I didn't actually compute the sum :-(.
– copper.hat
Jul 21 at 21:14
1
@copper.hat Thank you very much for this very interesting answer.Sorry I did not read the first version of your answer carefully so I did not got your idea.
– Ali Taghavi
Jul 22 at 5:50
 |Â
show 2 more comments
up vote
4
down vote
accepted
Hints: (Thanks to @EricWofsey for catching an egregious error.)
Let $d_k(x)$ be the $k$th
element of decimal expansion of $x$. Then
$int d_k = 1 over 10 (0+1+cdots + 9)= 9 over 2$.
We have $x = sum_k=1^infty d_k(x) 1 over 10^k$.
Note that for any nice functions $g_k$ that satisfy $int g_k = 9 over 2$ we have
$int ( sum_k=1^infty g_k(x) 1 over 10^k ) dx =sum_k=1^infty int g_k(x) 1 over 10^k dx = 1 over 2$.
answered Jul 21 at 20:32
copper.hat
122k557156
If we integrate your formula $x = sum_k=1^infty d_k(x) 1 over 10^k$ we get contradiction assuming $int d_k=1/10$
– Ali Taghavi
Jul 21 at 20:49
1
That's 9/2, not 9.
– Eric Wofsey
Jul 21 at 21:07
1
@EricWofsey: Something is wrong with my computation
– copper.hat
Jul 21 at 21:08
1
@EricWofsey: Thanks for catching that, I was computing the integral as $int 1 dx$ instead of $int x dx$ to compute the constant. I didn't actually compute the sum :-(.
– copper.hat
Jul 21 at 21:14
1
@copper.hat Thank you very much for this very interesting answer.Sorry I did not read the first version of your answer carefully so I did not got your idea.
– Ali Taghavi
Jul 22 at 5:50
 |Â
show 2 more comments
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Hints: (Thanks to @EricWofsey for catching an egregious error.)
Let $d_k(x)$ be the $k$th
element of decimal expansion of $x$. Then
$int d_k = 1 over 10 (0+1+cdots + 9)= 9 over 2$.
We have $x = sum_k=1^infty d_k(x) 1 over 10^k$.
Note that for any nice functions $g_k$ that satisfy $int g_k = 9 over 2$ we have
$int ( sum_k=1^infty g_k(x) 1 over 10^k ) dx =sum_k=1^infty int g_k(x) 1 over 10^k dx = 1 over 2$.
answered Jul 21 at 20:32
copper.hat
122k557156
Hints: (Thanks to @EricWofsey for catching an egregious error.)
Let $d_k(x)$ be the $k$th
element of decimal expansion of $x$. Then
$int d_k = 1 over 10 (0+1+cdots + 9)= 9 over 2$.
We have $x = sum_k=1^infty d_k(x) 1 over 10^k$.
Note that for any nice functions $g_k$ that satisfy $int g_k = 9 over 2$ we have
$int ( sum_k=1^infty g_k(x) 1 over 10^k ) dx =sum_k=1^infty int g_k(x) 1 over 10^k dx = 1 over 2$.
answered Jul 21 at 20:32
copper.hat
122k557156
edited Jul 21 at 21:50
answered Jul 21 at 20:32
copper.hat
122k557156
answered Jul 21 at 20:32
copper.hat
122k557156
answered Jul 21 at 20:32
copper.hat
122k557156
122k557156
If we integrate your formula $x = sum_k=1^infty d_k(x) 1 over 10^k$ we get contradiction assuming $int d_k=1/10$
– Ali Taghavi
Jul 21 at 20:49
1
That's 9/2, not 9.
– Eric Wofsey
Jul 21 at 21:07
1
@EricWofsey: Something is wrong with my computation
– copper.hat
Jul 21 at 21:08
1
@EricWofsey: Thanks for catching that, I was computing the integral as $int 1 dx$ instead of $int x dx$ to compute the constant. I didn't actually compute the sum :-(.
– copper.hat
Jul 21 at 21:14
1
@copper.hat Thank you very much for this very interesting answer.Sorry I did not read the first version of your answer carefully so I did not got your idea.
– Ali Taghavi
Jul 22 at 5:50
 |Â
show 2 more comments
If we integrate your formula $x = sum_k=1^infty d_k(x) 1 over 10^k$ we get contradiction assuming $int d_k=1/10$
– Ali Taghavi
Jul 21 at 20:49
1
That's 9/2, not 9.
– Eric Wofsey
Jul 21 at 21:07
1
@EricWofsey: Something is wrong with my computation
– copper.hat
Jul 21 at 21:08
1
@EricWofsey: Thanks for catching that, I was computing the integral as $int 1 dx$ instead of $int x dx$ to compute the constant. I didn't actually compute the sum :-(.
– copper.hat
Jul 21 at 21:14
1
@copper.hat Thank you very much for this very interesting answer.Sorry I did not read the first version of your answer carefully so I did not got your idea.
– Ali Taghavi
Jul 22 at 5:50
If we integrate your formula $x = sum_k=1^infty d_k(x) 1 over 10^k$ we get contradiction assuming $int d_k=1/10$
– Ali Taghavi
Jul 21 at 20:49
If we integrate your formula $x = sum_k=1^infty d_k(x) 1 over 10^k$ we get contradiction assuming $int d_k=1/10$
– Ali Taghavi
Jul 21 at 20:49
1
1
That's 9/2, not 9.
– Eric Wofsey
Jul 21 at 21:07
That's 9/2, not 9.
– Eric Wofsey
Jul 21 at 21:07
1
1
@EricWofsey: Something is wrong with my computation
– copper.hat
Jul 21 at 21:08
@EricWofsey: Something is wrong with my computation
– copper.hat
Jul 21 at 21:08
1
1
@EricWofsey: Thanks for catching that, I was computing the integral as $int 1 dx$ instead of $int x dx$ to compute the constant. I didn't actually compute the sum :-(.
– copper.hat
Jul 21 at 21:14
@EricWofsey: Thanks for catching that, I was computing the integral as $int 1 dx$ instead of $int x dx$ to compute the constant. I didn't actually compute the sum :-(.
– copper.hat
Jul 21 at 21:14
1
1
@copper.hat Thank you very much for this very interesting answer.Sorry I did not read the first version of your answer carefully so I did not got your idea.
– Ali Taghavi
Jul 22 at 5:50
@copper.hat Thank you very much for this very interesting answer.Sorry I did not read the first version of your answer carefully so I did not got your idea.
– Ali Taghavi
Jul 22 at 5:50
 |Â
show 2 more comments
up vote
3
down vote
Here is a probabilistic take on essentially the same method as copper.hat's answer. Lebesgue measure on $[0,1]$ is the same as the probability measure where you independently choose each digit uniformly from $0,1,dots,9$. Now, $f(x)$ (thought of as a random variable) has this same property: each digit of $f(x)$ is chosen independently and uniformly from $0,1,dots,9$, since the digits of $f(x)$ are just the digits of $x$ in a different order. This means that random variable $f(x)$ has the same probability distribution as the identity random variable $g(x)=x$, and so they have the same expectation. That is, $$int_J f(x),dx=int_J x,dx=frac12.$$
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
1
There is a wonderful monograph by Marc Kac (Statistical Independence in Probability, Analysis and Number Theory, I think) that starts with a similar argument (base 2). Worth a glance I believe...
– copper.hat
Jul 21 at 21:38
@Eric Thank you very much for your very interesting answer. i am sorry that i can not accept two answers simultaneously
– Ali Taghavi
Jul 22 at 5:51
add a comment |Â
up vote
3
down vote
Here is a probabilistic take on essentially the same method as copper.hat's answer. Lebesgue measure on $[0,1]$ is the same as the probability measure where you independently choose each digit uniformly from $0,1,dots,9$. Now, $f(x)$ (thought of as a random variable) has this same property: each digit of $f(x)$ is chosen independently and uniformly from $0,1,dots,9$, since the digits of $f(x)$ are just the digits of $x$ in a different order. This means that random variable $f(x)$ has the same probability distribution as the identity random variable $g(x)=x$, and so they have the same expectation. That is, $$int_J f(x),dx=int_J x,dx=frac12.$$
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
1
There is a wonderful monograph by Marc Kac (Statistical Independence in Probability, Analysis and Number Theory, I think) that starts with a similar argument (base 2). Worth a glance I believe...
– copper.hat
Jul 21 at 21:38
@Eric Thank you very much for your very interesting answer. i am sorry that i can not accept two answers simultaneously
– Ali Taghavi
Jul 22 at 5:51
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Here is a probabilistic take on essentially the same method as copper.hat's answer. Lebesgue measure on $[0,1]$ is the same as the probability measure where you independently choose each digit uniformly from $0,1,dots,9$. Now, $f(x)$ (thought of as a random variable) has this same property: each digit of $f(x)$ is chosen independently and uniformly from $0,1,dots,9$, since the digits of $f(x)$ are just the digits of $x$ in a different order. This means that random variable $f(x)$ has the same probability distribution as the identity random variable $g(x)=x$, and so they have the same expectation. That is, $$int_J f(x),dx=int_J x,dx=frac12.$$
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
Here is a probabilistic take on essentially the same method as copper.hat's answer. Lebesgue measure on $[0,1]$ is the same as the probability measure where you independently choose each digit uniformly from $0,1,dots,9$. Now, $f(x)$ (thought of as a random variable) has this same property: each digit of $f(x)$ is chosen independently and uniformly from $0,1,dots,9$, since the digits of $f(x)$ are just the digits of $x$ in a different order. This means that random variable $f(x)$ has the same probability distribution as the identity random variable $g(x)=x$, and so they have the same expectation. That is, $$int_J f(x),dx=int_J x,dx=frac12.$$
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
answered Jul 21 at 21:26
Eric Wofsey
162k12189300
162k12189300
1
There is a wonderful monograph by Marc Kac (Statistical Independence in Probability, Analysis and Number Theory, I think) that starts with a similar argument (base 2). Worth a glance I believe...
– copper.hat
Jul 21 at 21:38
@Eric Thank you very much for your very interesting answer. i am sorry that i can not accept two answers simultaneously
– Ali Taghavi
Jul 22 at 5:51
add a comment |Â
1
There is a wonderful monograph by Marc Kac (Statistical Independence in Probability, Analysis and Number Theory, I think) that starts with a similar argument (base 2). Worth a glance I believe...
– copper.hat
Jul 21 at 21:38
@Eric Thank you very much for your very interesting answer. i am sorry that i can not accept two answers simultaneously
– Ali Taghavi
Jul 22 at 5:51
1
1
There is a wonderful monograph by Marc Kac (Statistical Independence in Probability, Analysis and Number Theory, I think) that starts with a similar argument (base 2). Worth a glance I believe...
– copper.hat
Jul 21 at 21:38
There is a wonderful monograph by Marc Kac (Statistical Independence in Probability, Analysis and Number Theory, I think) that starts with a similar argument (base 2). Worth a glance I believe...
– copper.hat
Jul 21 at 21:38
@Eric Thank you very much for your very interesting answer. i am sorry that i can not accept two answers simultaneously
– Ali Taghavi
Jul 22 at 5:51
@Eric Thank you very much for your very interesting answer. i am sorry that i can not accept two answers simultaneously
– Ali Taghavi
Jul 22 at 5:51
add a comment |Â
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Is $0/a_1a_2a_3a_4$ supposed to be a decimal expansion?
– Eric Wofsey
Jul 21 at 20:21
@EricWofsey Yes it is the unique decimal expansion.
– Ali Taghavi
Jul 21 at 20:22
Are you swapping the digits in each successive pair past the decimal point?
– MPW
Jul 21 at 20:31
@MPW I am sorry I do not understand your question. We swap with the formula $0/a_1a_2a_3a_4a_5a_6ldots mapsto 0/a_2a_1a_4a_3a_6a_5ldots $.
– Ali Taghavi
Jul 21 at 20:36
1
Cf. the Rademacher functions.
– copper.hat
Jul 21 at 21:07