The Metric Subspace: Continuity, the Borel Sigma-Algebra, and the Restriction of a Borel Measure
lemmaAnalysisTopologylem:borel-subspace-metric-2026aThe Borel sets of a metric subspace are the traces of the ambient Borel sets; when the subspace is itself Borel they are the ambient Borel sets it contains, the restriction of a Borel measure is a Borel measure on it, and an integral of a function vanishing off the subspace is computed there.
In the setting of Measure Spaces and the Lebesgue Integral: Standing Notation, whose measure space is instantiated at each use below by the measure space named there, so that the letter is free for the use fixed here: let be a metric space, let be the collection of subsets of that are open in , which is a topology on by Metric Open Sets Form a Topology, and let be the Borel -algebra of .
Let , let be the restriction of to , which is a metric on by claim 1 of that lemma, and let be the Borel -algebra of the metric space . Let denote the real numbers, regarded as a metric space through the absolute-value metric, and let be the Borel -algebra of the real line. Measurability of a real-valued or -valued function on a measurable space is that of Measure Spaces and the Lebesgue Integral: Standing Notation §measurable.
Then the following hold.
1. (Continuity and a subspace)¶ Let be a metric space.
(a)¶ Let . Then is continuous on as a map from to if and only if is continuous on as a map from the metric space to .
(b)¶ Let , let be the restriction of to , which is a metric on by claim 1 of The Restriction of a Metric to a Subset Induces the Subspace Topology read for the metric space and the subset , let satisfy for every , and let be the map with for every . Then is continuous on as a map into if and only if is continuous on as a map into .
2. (Trace of the Borel sets)¶ .
3. (A Borel subspace, and the restriction of a measure)¶ Assume that . Then
which is the -algebra of claim 1 of Assembly of Measure Spaces: Restriction, Transport, One-Point Spaces, and Countable Disjoint Unions, read for the measurable space and the set . Consequently, for every Borel measure on , the restriction of that claim is a Borel measure on , with for every ; in particular .
4. (Integrating a function that vanishes off )¶ Assume that , let be a Borel measure on , and let be measurable with respect to and , integrable with respect to , and such that for every . Then the restriction is measurable with respect to and , is integrable with respect to , and
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