TheoremBase

A Measurable Real Function Has Null Level Sets at Levels Arbitrarily Close to Any Point, for a Finite Measure

For a finite measure and a measurable real function, every open interval of levels contains a level whose level set has measure zero; in particular there are such levels converging to any given real number from above.

Statement

In the setting of Measurable Functions and the Lebesgue Integral: Standing Notation and Routine Facts, Continued: Typed Arithmetic, Maps and Predicativity, let (X,F,μ)(X,\mathcal{F},\mu) be a measure space with μ\mu finite, and let g:X→Rg:X\to\mathbb{R} be measurable. For t∈Rt\in\mathbb{R}, {g=t}={x∈X:g(x)=t}\{g=t\}=\{x\in X:g(x)=t\} belongs to F\mathcal{F} (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (v)).

Null levels in every interval. For all a,b∈Ra,b\in\mathbb{R} with a<ba<b there is t∈Rt\in\mathbb{R} with a<t<ba<t<b and

μ({g=t})=0.\mu(\{g=t\})=0 .

Null levels converging from above. For every a∈Ra\in\mathbb{R} there is a sequence (tk)(t_{k}) in R\mathbb{R} with a<tk<a+1/ka<t_{k}<a+1/k and μ({g=tk})=0\mu(\{g=t_{k}\})=0 for every k∈Nk\in\mathbb{N}.

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