TheoremBase

Ulam's Theorem: a Finite Borel Measure on a Complete Separable Metric Space is Tight

Every finite Borel measure on a complete separable metric space is tight (Ulam).

Statement

In the setting of The Real Numbers: Standing Notation and Background, let (X,d)(X,d) be a complete and separable metric space, let Td\mathcal{T}_{d} be the collection of subsets of XX that are open in (X,d)(X,d), a topology on XX by Metric Open Sets Form a Topology, and let μ\mu be a Borel measure on (X,d)(X,d) with μ(X)<∞\mu(X)<\infty.

Then the set {μ}\{\mu\} is tight in (X,d)(X,d): for every real ε>0\varepsilon>0 there is a set K⊆XK\subseteq X, compact in (X,Td)(X,\mathcal{T}_{d}), with μ(X∖K)≤ε\mu(X\setminus K)\le\varepsilon.

Proofs

Log in to submit a proof.

Loading...

Citations

Loading…

Dependencies

Loading…

Related

0 relations

Curated associations between results. These are editable and subjective — they do not replace the dependency graph, which is derived from the references in the text.

No relations recorded yet.

Comments

Log in to comment.

Loading…