Alphabeta MathHow statement and proof provenance work
The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.
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Subsets and countable unions of null subsets of are null
Statement
Every subset of a null subset of is null. Assuming countable choice, every countable union of null subsets of is null.
Facts & Assumptions
Given: Null sets , .
Nullity is the cube-cover condition of Measure zero and content zero in by countable and finite cube covers.
Countable choice selects one cover for each (The Axiom of Countable Choice ()), and is countable ().
A nonnegative series converges with sum at most whenever all of its finite partial sums are at most (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series).
Proof
Subset closure follows because any cover of a set covers every subset.
Given , choose for each a cube cover of with total volume at most . This simultaneous selection uses [L2].
Enumerate the doubly indexed cubes through a bijection . Every finite partial sum is contained in a finite rectangle of indices and is at most .
By [L4], the enumerated nonnegative volume series converges with sum at most . The cubes cover , proving nullity.
Depends on
- Measure zero and content zero in $\mathbb{R}^m$ by countable and finite cube covers
- The Axiom of Countable Choice ($\mathrm{AC}_\omega$)
- $\mathbb{N} \times \mathbb{N} \approx \mathbb{N}$
- For $|r| < 1$, $\sum_{k \ge 0} r^k = 1/(1-r)$, and for $|r| \ge 1$ the series diverges
- Series, partial sums, convergence and the sum, divergence, and the tail series
- A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum
- Finite sums and finite products, by recursion
- Laws of finite sums and finite products
- Integer powers $a^m$
Used by
- The rational points of [0,1]² form a bounded null set that is not Jordan measurable Counterexample
- A C¹ map sends a compact set of content zero to a set of content zero Lemma
- Every smooth manifold embeds in some finite-dimensional Euclidean space Theorem
- Lebesgue's criterion in ℝᵐ: a bounded function on a closed nondegenerate rectangle is Riemann integrable iff its discontinuity set is null Theorem
Dependency tree · two levels
72 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- J. Lebl, Basic Analysis, Riemann Integral in Several Variables (standard reference, not scraped)
- J. Lebl, Basic Analysis, The Riemann-Lebesgue Criterion (standard reference, not scraped)
- J. Lebl, Basic Analysis, Outer Measure and Null Sets (standard reference, not scraped)