Where is the Cantor function differentiable?
The Cantor Function warns us that there exists a function f : [0,1] → [0,1] that is continuous, non-decreasing, differentiable almost everywhere, with f(0) = 0, f(1) = 1, and wherever it is differentiable, the derivative is 0.
What makes the Cantor set a function?
In mathematics, the Cantor function is an example of a function that is continuous, but not absolutely continuous. It is a notorious counterexample in analysis, because it challenges naive intuitions about continuity, derivative, and measure.
What is the integral of the Cantor function?
EDIT: More generally, if F is any continuous distribution function (the Cantor function f is a particular example), then ∫RF(x)dF(x)=1/2. As before, this can be proved using integration by parts, which is allowed since F is continuous and non-decreasing.
Is the Cantor function measurable?
The Cantor-Lebesgue function ϕ is an increasing continuous function that maps [0,1] onto [0,1]. Its derivative exists on the open set O = [0,1] \ C and ϕ (x) = 0 for x ∈ O. Proposition 2.21. There is a measurable set, a subset of the Cantor set, that is not a Borel set.
Why is the Cantor set uncountable?
A simple way to see that the cantor set is uncountable is to observe that all numbers between 0 and 1 with ternary expansion consisting of only 0 and 2 are part of cantor set. Since there are uncountably many such sequences, so cantor set is uncountable.
Is the characteristic function of the Cantor set Riemann integrable?
Is it true that the characteristic function of the Cantor set is Lebesgue integrable in [0, 1] but not Riemann integrable? False. The characteristic function of the Cantor set is continuous on the com- plement to the Cantor set that complement consists of open intervals on which the function is identically zero).
What makes the Cantor set special?
For a number to be in the Cantor set, it must not be excluded at any step, it must admit a numeral representation consisting entirely of 0s and 2s. if it again its ternary expansion contains no 1’s and “ends” in infinitely many recurring 2s.
What is Cantor set in measure theory?
A general Cantor set is a closed set consisting entirely of boundary points. Such sets are uncountable and may have 0 or positive Lebesgue measure. The Cantor set is the only totally disconnected, perfect, compact metric space up to a homeomorphism (Willard 1970).
Why is Cantor set measurable?
Since the Cantor set is the complement of a union of open sets, it itself is a closed subset of the reals, and therefore a complete metric space.
Is Cantor function bounded variation?
and the cantor function is well-known example of function of bounded variation which is not absolutely continuous.
Is Cantor set countable or uncountable?
The Cantor set is uncountable.
Is the Cantor set integrable?
Since the Cantor set is of measure zero, the Lebesgue integral of its characteristic function is 0.
What is the significance of the Cantor function?
The Cantor function helps us understand what “nice enough” means. At every point not in the Cantor set, the Cantor function is flat, so it’s easy to draw a tangent line to the graph at that point. (The derivative there is 0.)
How do you find the Cantor distribution?
The Cantor function can also be seen as the cumulative probability distribution function of the 1/2-1/2 Bernoulli measure μ supported on the Cantor set: c ( x ) = μ ( [ 0 , x ] ) {\extstyle c(x)=\\mu ([0,x])} . This probability distribution, called the Cantor distribution, has no discrete part.
How do you find the Cantor function for z = 1/3?
be the dyadic (binary) expansion of the real number 0 ≤ y ≤ 1 in terms of binary digits bk ∈ {0,1}. Then consider the function For z = 1/3, the inverse of the function x = 2 C1/3 ( y) is the Cantor function. That is, y = y ( x) is the Cantor function. In general, for any z < 1/2,…
What is the Cantor-Lebesgue function?
Cantor-Lebesgue Function In this section, we define the Cantor set which gives us an example of anuncountable set of measure zero. We use the Cantor-Lebesgue Function to showthere are measurable sets which are not Borel; soB(M.