Let f and g be real functions defined on [0, 1], and denote by λ the Lebesgue measure on [0, 1]. In probability theory, f and g are called independent (on [0, 1]) if λ(f∈I, g∈J) = λ(f∈I) · λ(g∈J) for any open intervals I,J⊆ℝ.
1. It is easy to see that, if f or g is constant, then f and g are independent.
2. I know that, if either f or g are discontinuous, then f and g may be independent.
3. I know how to prove that f and g cannot be independent if they are non-constant, continuous and with bounded variation.
4. I don't know, and would like to: if one requires f and g to be non-constant and continuous, does it follow that f and g cannot be independent?
Dear Mr. Niknamfar,
Your first statement is (in the everyday speach) the same as claim 2. of George; thus:: why the triple acclamation sign?
Your second paragraph is completely unclear since George has given no arguments. Perhaps, your are asking for the proof of the proposition presented at 3. But this is not incorrect to hide own proof!
The question you are asking for is given at 4.
Regards.
I am so sorry to claim from some "people" instead of how to understand or to answer questions to learn how to talk, it will help them a lot in their daily life.
Dear George,
I like the question! It seems to me that the Cantor staircase function f
https://en.wikipedia.org/wiki/Cantor_function
helps a lot. This function is continuous, but has only countable set of "essential" values {tn} in the sense that the measure of f-1(A) is zero for every A that do not intersect the set {tn}. Now denote An = f-1(tn). Each An is an interval of the form [an, bn]. Let us define the requested g as follows: outside of the union of all [an, bn] function g equals zero, g(an) = g(bn) = 0, g((an+bn)/2) = 1 and on each of the intervals [an, (an+bn)/2] and [(an+bn)/2, bn] our function g is linear. It seems to me that with this construction we get the desired pair (f, g) of independent continuous functions.
All the best, Vladimir
What a series of slight but substantially missleading statements!
A. Function g in Vladimir's example is not continuous
B. Despite this, Vladimir's construction shows (after some generalization) that for the Cantor step function no continuous function independent of f exists; more generally, under additional restriction that one of the two continuous functions assumes countably many values with probability 1 the aswer to no. 4 is positive, i.e. the second function cannot be independent. Treat this as a conjecture, however I do not try to prove it in details.
C. Still open is the problem for the remaining cases, i.e. when neither f nor g assumes contably many values with probability 1.
D. An obvious counter-example can be constructed, based on a continuous and measure preserving bijection \phi from [0,1]^2 onto [0,1] {means: \lambda^2(\phi^{-1} (A)) = \lambda^1(A) for any open interval A \subset [0,1] }. Does such a \phi exist? And if not - would this solve Georege's problem no. 4?
Regards
Dear George,
Many thanks for your very, very nice job! I have felt that something with the Sierpiński curves and/or carpets can be in common with your problem, but I (bad guy) didn't study this in detail when there was time for this:( Your info gives me a possibility to read the papers with respect to the existence of the strange function \phi: I^2 \to I.
Treat this remark, please, as a cordial congratulation in being so decided in the investigaions successfully finished with localization of the deep theorem by Hugo Steinhaus.
Joachim
Dear Joachim,
Thank you! You have right, the function g in my example is not continuous, in fact every point of Cantor set is a point of discontinuity. Shame on me for such a childish mistake!
Best, Vladimir
Dear contributors, please do not upvote this my contribution. If you think the same, try to find a way to appreciate George.
Dear George, You have my great appreciation and recognition. Not only Your questions are interesting and make us think, speculate and learn, but You are an awesome director! The discussions started by Your questions possess all the qualities of a successful dramatic composition. With a happy end in most cases :)
Your method of keeping the debate is a celebration of mathematics and humanity at the same time. I hope this will last for many years.
Thank you dear Viera for these kind words. I am so happy to follow George and you, really, I benefit a lot. I want to say: Who want to sit to the table of George, should respect him as a good scientist and as a good Human, humble and generous. who know to answer his questions it is good, who do not know, it is also good to learn. We still learn until the last of our life. Arrogance with knowledge is unacceptable, however with ignorance? !
Take two independent Brownian motion, B_1 and B_2. For any omega, B_1(t) and B_2(t) will be independent continuous functions, with infinite variation.
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