What does the term regularity mean? - Mathematics Stack Exchange The regularity a solution can inherit depends on the properties of the problem, i e , the smoothness of a domain boundary, the smoothness of boundary conditions and initial data Notice here that the smoothness of the initial data falls under the category of "regularity of initial data"
set theory - Help understanding the axiom of regularity - Mathematics . . . Usual mathematical results (like propositions on analysis and geometry) would not need the axiom of regularity, but some deeper set theory heavily relies on the axiom of regularity As a side note, a regularity-free proof of "the existence of a basis on any vector spaces implies AC" is not known $\endgroup$ –
On the regularity of the boundary of an open set I'm considering the following regularity conditions for the boundary of $\Omega$: $\overline{\Omega}$ is a topological manifold with boundary (in other words, around each boundary point, $\Omega$ looks like a Euclidean half-space --up to homeomorphism)
Need good reference or a proof on regularity of solution to Neumann problem The only reference I am aware of which gives a proof of this result, is the textbook by Mikhailov [1] Precisely, he deals with the regularity problem for the first boundary problem (Dirichlet problem) and the second boundary problem (Neumann problem) for the Poisson equation in §2 3, pp 216-226
regularity of a measure - Mathematics Stack Exchange Help save this proof about the regularity of the lebesgue measure on $\mathbb{R}^d$ 0 An example of a measure and of a set s t its measure cannot be approximated by measures of its open subsets
real analysis - Help save this proof about the regularity of the . . . In a previous homework our task was to prove the regularity of the Lebegue-measure on $\mathbb{R}^d$ More precisely: More precisely: Let $(\mathbb{R}^d, \mathcal{M}^*, \lambda)$ be a measure space and $\mathcal{M}^* := \mathcal{M}^*(\mathbb{R}^d)$ be the $\sigma$ -algebra of the $\lambda^*$ -measurable sets, with $\lambda$ being the by the