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Why is a function defined on a closed interval and monotone, then it must be integrable? What are the necessary conditions for monotonicity?
The condition of integrability is very broad, basically as long as there is no dense "dot hole"
If and only if a function is monotone
For x1 ≠ X2, f (x1) - f (x2) is not always zero
A function defined everywhere must be a bounded function on the interval
Give me a counterexample
f(x)=1/x
There are definitions everywhere on the interval (0,1), but they are unbounded
On closed interval, the same is a false proposition
For example, f (x) is defined as follows
F (x) = 1 / X if 0
"Function is bounded on an interval", please give an example
Answer: sunnykirby1111 you are too irresponsible. Don't give the wrong answer casually. It has nothing to do with the edge. For example, if the value range of a function is (1,2) (note the value range), its maximum value does not exist, and its minimum value does not exist (less than 1 and 2), but it is bounded
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