Suppose the real valued f is monotone on an interval I. Let S be the set of points of I where f is not differentiable. Show S is at most countable.
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Thank you.
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I think this is actually false. There's a popular counterexample--- a badly behaved function that nevertheless has a number of good properties--- known as the "Cantor function", which is monotone increasing on [0,1], continuous on [0,1], and yet fails to be differentiable on an uncountable set. The proofs of all of these assertions are somewhat complicated because the function itself has a complicated definition, but an example certainly exists. See the link below. (If by "monotone" you mean "strictly increasing" or "strictly decreasing", the Cantor function isn't that, but if you let f(x) be the Cantor function, the function g(x) = f(x) + x is strictly increasing, and still fails to be differentiable on an uncountable set.)
What you can say about a monotone function on an interval is that its set of discontinuities is at most countable. I can sketch the proof of why. Say f is monotone increasing. You can use this to show that for any c in the domain of f, the one-sided limits lim_{x to c-} f(x) and lim_{x to c+} f(x) always exist--- that they are sup {f(x): x < c} and inf {f(x): x > c} respectively--- and the increasingness of f implies that the first limit is always less than or equal to the second. If f fails to be continuous at such a c, the left-hand limit of f has to be strictly less than the right-hand limit of f, and so there is an interval of numbers between them: so you can choose a rational number, call it r_c, satisfying sup {f(x): x < c} < r_c < inf {f(x): x > c}. Doing this for every discontinuity defines a function r from the set of discontinuities of c to the rational numbers. You can check that r_c is itself strictly increasing: if c < c' are discontinuities of f, then since c' > sup {f(x): x < c'} by definition, the monotonicity of f implies that sup {f(x): x < c'} = sup {f(x): c < x < c'} >= inf {f(x): c < x < c'} = inf {f(x): x > c}, and you know r_c is less than this by definition, so r_c' > r_c. A strictly increasing function from one subset of R to another must be one-to-one. So, since there is a one-to-one function from the set of discontinuities of f to a subset of the rational numbers it follows that f has at most countably many discontinuities. [If f is monotone decreasing, apply the argument just given to -f, which is monotone increasing, and note that f and -f have the same set of discontinuities.]
What you can say about a monotone function on an interval is that its set of discontinuities is at most countable. I can sketch the proof of why. Say f is monotone increasing. You can use this to show that for any c in the domain of f, the one-sided limits lim_{x to c-} f(x) and lim_{x to c+} f(x) always exist--- that they are sup {f(x): x < c} and inf {f(x): x > c} respectively--- and the increasingness of f implies that the first limit is always less than or equal to the second. If f fails to be continuous at such a c, the left-hand limit of f has to be strictly less than the right-hand limit of f, and so there is an interval of numbers between them: so you can choose a rational number, call it r_c, satisfying sup {f(x): x < c} < r_c < inf {f(x): x > c}. Doing this for every discontinuity defines a function r from the set of discontinuities of c to the rational numbers. You can check that r_c is itself strictly increasing: if c < c' are discontinuities of f, then since c' > sup {f(x): x < c'} by definition, the monotonicity of f implies that sup {f(x): x < c'} = sup {f(x): c < x < c'} >= inf {f(x): c < x < c'} = inf {f(x): x > c}, and you know r_c is less than this by definition, so r_c' > r_c. A strictly increasing function from one subset of R to another must be one-to-one. So, since there is a one-to-one function from the set of discontinuities of f to a subset of the rational numbers it follows that f has at most countably many discontinuities. [If f is monotone decreasing, apply the argument just given to -f, which is monotone increasing, and note that f and -f have the same set of discontinuities.]
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Mcbengt is right. The claim is false. The mentioned set can be uncountable. Like Mcbengt showed, the at most countable set is the set of discontinuities
What is true about the derivative is: if the real valued f is monotone on an interval I, then f is differentiable almost everywhere on I. That is, there is a set D with Lebesgue measure 0 such that f is differentiable on I - D.
The proof of this non trivial fact can be found on the books on measure theory.
What is true about the derivative is: if the real valued f is monotone on an interval I, then f is differentiable almost everywhere on I. That is, there is a set D with Lebesgue measure 0 such that f is differentiable on I - D.
The proof of this non trivial fact can be found on the books on measure theory.