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FALSE: a pointwise limit of a sequence of Riemann integrable functions on [a,b][a,b] is Riemann integrable

Statement

False claim: if (fn)nN(f_n)_{n \in \mathbb{N}} is a sequence of Riemann integrable functions on [a,b][a,b] (The lower and upper Darboux integrals of a bounded ff on [a,b][a,b] as supPL(f,P)\sup_P L(f,P) and infPU(f,P)\inf_P U(f,P), Darboux integrability as their equality, and the notation abf\int_a^b f, Sequences of reals: bounded, eventually, frequently, tails, subsequences) and f:[a,b]Rf : [a,b] \to \mathbb{R} satisfies

fn(x)f(x)for every x[a,b]f_n(x) \longrightarrow f(x) \qquad \text{for every } x \in [a,b]

(Limits and Cauchy sequences of reals), then ff is Riemann integrable on [a,b][a,b].

The witness below is the standard one: an increasing sequence of indicators of finite sets of rationals, each integrable because it has only finitely many discontinuities, whose pointwise limit is the Dirichlet function, which is not integrable at all. Every fnf_n takes values in {0,1}\{0,1\}, so no unboundedness is involved, and the convergence is even monotone.

Facts & Assumptions

Given: The set E:=Q[0,1]E := \mathbb{Q} \cap [0,1], a surjection s:NEs : \mathbb{N} \to E, the finite sets Fn:={s(k):k<n}F_n := \{\, s(k) : k < n \,\} for nNn \in \mathbb{N}, and the indicators fn:[0,1]Rf_n : [0,1] \to \mathbb{R} with fn(x)=1f_n(x) = 1 for xFnx \in F_n and fn(x)=0f_n(x) = 0 otherwise.

[A1]

The false claim: a pointwise limit of Riemann integrable functions on a closed bounded interval with distinct endpoints is Riemann integrable.

[L1]

Q\mathbb{Q} is countably infinite and every subset of an at most countable set is at most countable, so EE is at most countable; EE is nonempty, since 0E0 \in E; and a nonempty at most countable set admits a surjection from N\mathbb{N} (Q\mathbb{Q} is countably infinite, Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable, A nonempty set is at most countable iff it is a surjective image of N\mathbb{N}).

[L2]

A bounded function on [a,b][a,b] that is continuous at every point other than rr listed points is Riemann integrable (A bounded function on [a,b][a,b] that is continuous except at finitely many points is Riemann integrable, Lower bound, bounded below, bounded set).

[L3]

The Dirichlet function restricted to [0,1][0,1], that is g:[0,1]Rg : [0,1] \to \mathbb{R} with g(x)=1g(x) = 1 for rational xx and g(x)=0g(x) = 0 for irrational xx, is bounded and not Riemann integrable: its lower Darboux integral is 00 and its upper Darboux integral is 11 (FALSE: every bounded function on [a,b][a,b] is Riemann integrable, The Dirichlet function 1Q1_{\mathbb{Q}}, and Thomae's function tt with t(x)=1/qt(x) = 1/q at a rational x=p/qx = p/q in lowest terms with q1q \ge 1 and t(x)=0t(x) = 0 at every irrational xx).

[L4]

A sequence of reals converges to xx when for every rational ε>0\varepsilon > 0 there is KK with xkx<ε|x_k - x| < \varepsilon for all kKk \ge K; an eventually constant sequence converges to that constant (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences, Basic properties of the absolute value).

[L6]

Continuity at a point: for every real ε>0\varepsilon > 0 there must be a real δ>0\delta > 0 with h(y)h(x)<ε|h(y) - h(x)| < \varepsilon for every yy in the domain with yx<δ|y - x| < \delta (Continuity of f:ARf : A \to \mathbb{R} at a point of AA and on AA: the ε\varepsilon-δ\delta condition, its agreement with limxcf(x)=f(c)\lim_{x \to c} f(x) = f(c) at a limit point, and continuity at an isolated point, The ε\varepsilon-neighbourhood and the punctured ε\varepsilon-neighbourhood of a point of R\mathbb{R}).

[L7]

Ordered-field arithmetic: the order is total and transitive, uv>0|u - v| > 0 for uvu \ne v, and 0<10 < 1 (Basic properties of the absolute value, Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property), Intervals of R\mathbb{R}: the nine order-convex forms, nondegeneracy, and length). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Refutation

technique · direct
1.1

By [L1] fix a surjection s:NEs : \mathbb{N} \to E and define FnF_n and fnf_n as in the Given. Each fnf_n takes only the values 00 and 11, so it is bounded.

givenL1chooseconstruct
2.1

Each fnf_n is continuous at every point of [0,1][0,1] outside the nn listed points s(0),,s(n1)s(0), \dots, s(n-1). Let x[0,1]x \in [0,1] with xs(k)x \ne s(k) for all k<nk < n, so fn(x)=0f_n(x) = 0. If n=0n = 0 put δ:=1\delta := 1; otherwise put δ:=min{xs(k):k<n}\delta := \min\{\, |x - s(k)| : k < n \,\}, which exists by [L5] and is positive by [L7]. Every y[0,1]y \in [0,1] with yx<δ|y - x| < \delta then differs from each s(k)s(k) with k<nk < n, so fn(y)=0f_n(y) = 0 and fn(y)fn(x)=0<ε|f_n(y) - f_n(x)| = 0 < \varepsilon for every ε>0\varepsilon > 0.

step 1.1L5L6L7
2.2

(fn)(f_n) converges pointwise to gg on [0,1][0,1]. Let x[0,1]x \in [0,1]. If xx is rational then xEx \in E, so x=s(k)x = s(k) for some kNk \in \mathbb{N} by surjectivity, and then xFnx \in F_n and fn(x)=1f_n(x) = 1 for every n>kn > k; the sequence is eventually constant with value 1=g(x)1 = g(x), so it converges to g(x)g(x) by [L4]. If xx is irrational then xEx \notin E and hence xFnx \notin F_n for any nn, so fn(x)=0=g(x)f_n(x) = 0 = g(x) for every nn and again the sequence converges to g(x)g(x).

step 1.1L1L3L4
3.1

By [L2], applied with the nn listed points s(0),,s(n1)s(0),\dots,s(n-1), each fnf_n is Riemann integrable on [0,1][0,1].

step 1.1step 2.1L2
4.1

So (fn)(f_n) is a sequence of Riemann integrable functions on [0,1][0,1], an interval with 0<10 < 1, converging pointwise to gg, and gg is not Riemann integrable by [L3]. Hence [A1] fails at this sequence and the claim is false.

step 3.1step 2.2A1L3

Remarks

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