![]() In this case we have that Because as we have shown in step three, is Lipschitz with constant. Let’s look at for the following cases to see that it indeed bounded by a constant, and so is Lipschitz: So overall if we have that the derivative in is bounded (in absolute value, and it is because it’s continuous and our domain is a closed set) by some number then for all we have that, and so is Lipschitz.Ĭonsider the function defined by if, if, and if. This is true for any subinterval as well. (Proof is in the corollary, bottom of the page) Our function is continuously differentiable, and as such it has the property that such that. So, but, but since are constants and is arbitrarily small, is also arbitrarily small, hence is zero measure. This is the definition of being zero measure.īut because is Lipschitz, it maps to a different cube with side length. Since is zero measure, there are open cubes with side lengths such that and. We wish to prove that is also zero measure. Let be Lipschitz with constant, and be zero measure. Notice that is a subset of the box but the volume of this box is, so is zero measure subset of.
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