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An open dense set of measure less than is the monotone -limit of Riemann integrable indicators, but its indicator is not Riemann integrable
Example
Assume the Axiom of Countable Choice. There exist open sets such that, with ,
- each is Riemann integrable on ;
- pointwise and
- is open and dense with ;
- is not Riemann integrable on .
Facts & Assumptions
Given: The Axiom of Countable Choice.
The rationals are countably infinite, and both the rationals and the irrationals are dense in . ( is countably infinite, Both and are dense in , and every nonempty open subset of is uncountable)
Countable subadditivity bounds the measure of a countable union by the sum of the individual measures. (Finite and countable subadditivity of measures)
The geometric series satisfies (For , , and for the series diverges)
For an increasing sequence of measurable sets , (Continuity from below for measures)
A bounded function on that is continuous except at finitely many points is Riemann integrable. (A bounded function on that is continuous except at finitely many points is Riemann integrable)
A bounded function on is Riemann integrable exactly when its discontinuity set has Lebesgue measure . (A bounded function on a closed bounded interval, or on a closed nondegenerate rectangle, is Riemann integrable exactly when its discontinuity set has Lebesgue measure zero)
If are measurable with , then (Measure of a set difference when the smaller set has finite measure)
The interval has Lebesgue measure , and every open interval has its usual length. (A box in with parameters is Lebesgue measurable of measure , whichever of its faces are included)
Every Borel subset of is Lebesgue measurable. (Assuming countable choice, every Borel subset of is Lebesgue measurable)
Verification
By [L1], fix an enumeration of . For each , let put for , and put . Each and is open, and every rational point of lies in , so is dense in and hence in . Each interval has length at most , so [L2], [L3], [L8], and [L9] give
Each is a finite union of open intervals, so is continuous away from the finitely many endpoints of those intervals. Thus [L5] makes every Riemann integrable on . Also , so pointwise and By [L4] and [L7],
Let . Step 1.1 and [L8] give Because is open, every point of is a continuity point of . Because is dense, every point of is a boundary point of , hence every neighbourhood of such a point meets both and ; so is discontinuous at every point of . Therefore the discontinuity set of contains the positive-measure set , and [L6] shows that is not Riemann integrable on .
Depends on
- $\mathbb{Q}$ is countably infinite
- Both $\mathbb{Q}$ and $\mathbb{R} \setminus \mathbb{Q}$ are dense in $\mathbb{R}$, and every nonempty open subset of $\mathbb{R}$ is uncountable
- Finite and countable subadditivity of measures
- For $|r| < 1$, $\sum_{k \ge 0} r^k = 1/(1-r)$, and for $|r| \ge 1$ the series diverges
- Continuity from below for measures
- A bounded function on $[a,b]$ that is continuous except at finitely many points is Riemann integrable
- A bounded function on a closed bounded interval, or on a closed nondegenerate rectangle, is Riemann integrable exactly when its discontinuity set has Lebesgue measure zero
- Measure of a set difference when the smaller set has finite measure
- A box in $\mathbb{R}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure $\prod_{i<n}(b_i-a_i)$, whichever of its faces are included
- Assuming countable choice, every Borel subset of $\mathbb{R}^n$ is Lebesgue measurable
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
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Sources
- Richard F. Bass, Real Analysis for Graduate Students, Version 5.0, Section 9.1 (standard reference, not scraped)