CounterexampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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Shrinking rectangles converge pointwise to zero while every integral equals one

Statement refuted

Refuted claim: if Riemann-integrable functions on $[0,1]$ converge pointwise to $0$ , then their integrals converge to $0$ .

For $k\in\mathbb{N}$ put $a_k:=\iota(k+1)$ , the positive canonical natural in $\mathbb{R}$ , and define

$r_k(x):=\begin{cases}a_k,&0<x\le1/a_k,\\0,&x=0\text{ or }1/a_k<x\le1.\end{cases}$

Then $r_k\to0$ pointwise while $\int_0^1r_k=1$ for every $k$ .

Facts & Assumptions

Given: The functions $r_k$ in the Statement, with $a_k=\iota(k+1)>0$ .

[L1]

For every real $\varepsilon>0$ there is $N\ge1$ with $1/\iota(N)<\varepsilon$ ; canonical naturals increase and their positive reciprocals decrease (For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ , The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order).

[L2]

A bounded function on a closed interval with only finitely many possible discontinuities is Riemann integrable (A bounded function on $[a,b]$ that is continuous except at finitely many points is Riemann integrable, The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$ , Darboux integrability as their equality, and the notation $\int_a^b f$ ).

[L3]

Changing an integrable function at finitely many points preserves its integrability and integral (Changing an integrable function at finitely many points changes neither its integrability nor its integral).

[L4]

A constant $c$ has integral $c(v-u)$ on $[u,v]$ , and integrals add over adjacent subintervals, including the oriented convention at coincident endpoints (If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$ , For $a<c<b$ : $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$ , and then $\int_a^b f = \int_a^c f + \int_c^b f$ ; with the oriented form for arbitrary $a,b,c$ , The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

[L5]

Pointwise convergence of $(f_k)$ to $f$ means that for every $x$ and every $\varepsilon>0$ there is an $N$ such that $k\ge N$ implies $|f_k(x)-f(x)|<\varepsilon$ ; uniform convergence requires one such $N$ for every $x$ simultaneously (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Counterexample

technique · direct

1.1

Each $r_k$ is bounded and is continuous except possibly at $0$ and $1/a_k$ , so it is integrable by [L2].

1.2

Let $q_k$ equal $a_k$ on $[0,1/a_k]$ and $0$ on $(1/a_k,1]$ . The functions $q_k$ and $r_k$ differ only at $0$ , so they have the same integral by [L3].

L3construct

1.3

At $x=0$ one has $r_k(0)=0$ for all $k$ . If $x>0$ , choose $N$ with $1/\iota(N)<x$ ; for $k\ge N$ , monotonicity of the canonical naturals gives $1/a_k<x$ , hence $r_k(x)=0$ . Thus $r_k\to0$ pointwise.

L1L5choose

1.4

To see explicitly that the convergence is not uniform, take $\varepsilon:=1/2$ . For every proposed $N\in\mathbb N$ , choose $k:=N$ and $x_N:=1/a_N$ ; then $|r_N(x_N)-0|=a_N\ge1>\varepsilon$ . Thus the uniform quantifier condition in [L5] fails.

givenL1L5

2.1

By [L3] and [L4], endpoint values do not affect either piece, and splitting at $1/a_k$ when it lies in the interior, with the coincident-endpoint convention otherwise, gives $\int_0^1q_k=a_k(1/a_k)+0=1$ .

step 1.2L3L4algebra

3.1

Steps 1.2 and 2.1 give $\int_0^1r_k=1$ for every $k$ , whereas the integral of the zero function is $0$ .

step 1.2step 2.1L3L4

4.1

The sequence therefore converges pointwise to $0$ but its integrals do not converge to the integral of the limit, refuting the claim.

step 1.3step 3.1∎

Depends on

Used by

Nothing in the library uses this result yet.

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Sources

William Faris, Real Analysis: Part I, §13.2 (standard reference, not scraped)