equivalent intervals of measurability for measurable functions

[[concept]]

[!themes] Topics

Evaluation Error: SyntaxError: Unexpected token '>'


at DataviewInlineApi.eval (plugin:dataview:19027:21)

at evalInContext (plugin:dataview:19028:7)

at asyncEvalInContext (plugin:dataview:19038:32)

at DataviewJSRenderer.render (plugin:dataview:19064:19)

at DataviewJSRenderer.onload (plugin:dataview:18606:14)

at DataviewJSRenderer.load (app://obsidian.md/app.js:1:1182416)

at DataviewApi.executeJs (plugin:dataview:19607:18)

at DataviewCompiler.eval (plugin:digitalgarden:10763:23)

at Generator.next (<anonymous>)

at fulfilled (plugin:digitalgarden:77:24)
Theorem

Let E⊂R be measurable and let f:E→[−∞,+∞]. Then the following are equivalent for all α∈R,

  1. f−1((α,∞])∈M
  2. f−1([α,∞])∈M
  3. f−1([−∞,α))∈M
  4. f−1([−∞,α])∈M
Proof

Not hard!

(1)⟹(2)

Suppose (1) holds. Then for all α∈R

[α,∞]=⋂n(α−1n,∞]⟹f−1([α,∞])=⋂nf−1((α−1n,∞])

RHS is a countable intersection of measurable sets and is thus measurable.

(2)⟹(1)

Suppose (2) holds. Then for all α∈R,

(α,∞]=⋃n[α+1n,∞]⟹f−1(α,∞]=⋃nf−1([α+1n,∞])

And this is a countable union of measurable sets and is thus Lebesgue measurable

ie (1)⟺(2)

(3)⟺(4) uses the same argument.

(2)⟺(3)

Note that [−∞,α)=([α,∞])c. Thus by taking the preimages and noting that complements of measurable sets are measurable yields the desired result.

◼

References

References

See Also

Mentions

Mentions

File Last Modified
Functional Analysis Lecture 9 2025-07-14
inverse image of infinity of measurable functions is measurable 2025-07-14
sums and products of measurable functions are measurable 2025-07-14
sups and infs of measurable functions are measurable 2025-07-14
the limit of a convergent sequence of measurable functions is measurable 2025-07-14

{ .block-language-dataview}
Created 2025-07-08 ֍ Last Modified 2025-09-11