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How to prove that the N-dimensional nondegenerate closed interval is not a zero measure set with the definition of (Leberg) zero measure set?
With the definition of (Leberg) zero measure set, it is proved that the N-dimensional nondegenerate closed interval is not a zero measure set. Firstly, the definition of measurable function is clear, and let the function be f(x), then f measurable means that if for any real number T, e (f >; T)(E makes f >;; T) measurable, then f is a measurable function. Adopt this definition.

A continuous function is set to F. A continuous function has one property: for any λ∈R, the set {x | f (x) >; λ} are all open sets. This is a theorem, only defined by continuous functions in mathematical analysis. Then for any real number t, e (f >; T) is an open set, of course it is measurable, so f is measurable.

Lebesgue measure

It is a standard method to give a subset of Euclidean space a length, an area or a volume. It is widely used in real analysis, especially in defining Lebesgue integral. The set that can give a volume is called Lebegmeasurable Lebegger measurable set A, and the volume or measure is denoted as λ(A).

The Lebesgue measure with a value of ∞ is possible, but even so, under the assumption that axiom of choice holds, all subsets of R are not Lebesgue measurable. No, measurable set's "strange" behavior led to the Barna-Taskey paradox, which was a result of axiom of choice.