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An open dense set of measure less than $1$ is the monotone $L^{1}$ -limit of Riemann integrable indicators, but its indicator is not Riemann integrable

Example

Assume the Axiom of Countable Choice. There exist open sets $U_{1} \subseteq U_{2} \subseteq \dots \subseteq (0, 1)$ such that, with $U : = ⋃_{n \geq 1} U_{n}$ ,

each $1_{U_{n}}$ is Riemann integrable on $[0, 1]$ ;
$1_{U_{n}} ↑ 1_{U}$ pointwise and $\int_{[0, 1]} ∣ 1_{U} - 1_{U_{n}} ∣ d λ_{1} \to 0;$
$U$ is open and dense with $λ_{1} (U) < 1$ ;
$1_{U}$ is not Riemann integrable on $[0, 1]$ .

Facts & Assumptions

Given: The Axiom of Countable Choice.

[L1]

The rationals are countably infinite, and both the rationals and the irrationals are dense in $R$ . ( $Q$ is countably infinite, Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable)

[L2]

Countable subadditivity bounds the measure of a countable union by the sum of the individual measures. (Finite and countable subadditivity of measures)

[L3]

The geometric series satisfies $\sum_{k = 0}^{\infty} 2^{- k} = 2.$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges)

[L4]

For an increasing sequence of measurable sets $(E_{n})$ , $λ_{1} ⁣ (⋃_{n} E_{n}) = \sup_{n} λ_{1} (E_{n}) .$ (Continuity from below for measures)

[L5]

A bounded function on $[0, 1]$ that is continuous except at finitely many points is Riemann integrable. (A bounded function on $[a, b]$ that is continuous except at finitely many points is Riemann integrable)

[L6]

A bounded function on $[0, 1]$ is Riemann integrable exactly when its discontinuity set has Lebesgue measure $0$ . (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)

[L7]

If $A \subseteq B$ are measurable with $λ_{1} (A) < + \infty$ , then $λ_{1} (B ∖ A) = λ_{1} (B) - λ_{1} (A) .$ (Measure of a set difference when the smaller set has finite measure)

[L8]

The interval $[0, 1]$ has Lebesgue measure $1$ , and every open interval has its usual length. (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included)

[L9]

Every Borel subset of $R$ is Lebesgue measurable. (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable)

Verification

technique · direct

1.1L1L2L3L8L9construct

By [L1], fix an enumeration $(q_{k})_{k \geq 0}$ of $Q \cap (0, 1)$ . For each $k \geq 0$ , let $I_{k} : = (q_{k} - 2^{- k - 4}, q_{k} + 2^{- k - 4}) \cap (0, 1),$ put $U_{n} : = ⋃_{k < n} I_{k}$ for $n \geq 1$ , and put $U : = ⋃_{k \geq 0} I_{k}$ . Each $U_{n}$ and $U$ is open, and every rational point of $(0, 1)$ lies in $U$ , so $U$ is dense in $(0, 1)$ and hence in $[0, 1]$ . Each interval $I_{k}$ has length at most $2^{- k - 3}$ , so [L2], [L3], [L8], and [L9] give $λ_{1} (U) \leq \sum_{k = 0}^{\infty} λ_{1} (I_{k}) \leq \sum_{k = 0}^{\infty} 2^{- k - 3} = 2^{- 3} \sum_{k = 0}^{\infty} 2^{- k} = 1 / 4 < 1.$

2.1step 1.1L4L5L7

Each $U_{n}$ is a finite union of open intervals, so $1_{U_{n}}$ is continuous away from the finitely many endpoints of those intervals. Thus [L5] makes every $1_{U_{n}}$ Riemann integrable on $[0, 1]$ . Also $U_{n} ↑ U$ , so $1_{U_{n}} ↑ 1_{U}$ pointwise and $∣ 1_{U} - 1_{U_{n}} ∣ = 1_{U ∖ U_{n}} .$ By [L4] and [L7], $\int_{[0, 1]} ∣ 1_{U} - 1_{U_{n}} ∣ d λ_{1} = λ_{1} (U ∖ U_{n}) = λ_{1} (U) - λ_{1} (U_{n}) \to 0.$

3.1step 1.1step 2.1L6L7L8∎

Let $C : = [0, 1] ∖ U$ . Step 1.1 and [L8] give $λ_{1} (C) = λ_{1} ([0, 1]) - λ_{1} (U) = 1 - λ_{1} (U) > 0.$ Because $U$ is open, every point of $U$ is a continuity point of $1_{U}$ . Because $U$ is dense, every point of $C$ is a boundary point of $U$ , hence every neighbourhood of such a point meets both $U$ and $C$ ; so $1_{U}$ is discontinuous at every point of $C$ . Therefore the discontinuity set of $1_{U}$ contains the positive-measure set $C$ , and [L6] shows that $1_{U}$ is not Riemann integrable on $[0, 1]$ .

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Sources

Richard F. Bass, Real Analysis for Graduate Students, Version 5.0, Section 9.1 (standard reference, not scraped)

An open dense set of measure less than 11 is the monotone L1L1-limit of Riemann integrable indicators, but its indicator is not Riemann integrable