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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 $k \in N$ put $a_{k} : = ι (k + 1)$ , the positive canonical natural in $R$ , and define

$r_{k} (x) : = {\begin{cases} a_{k}, & 0 < x \leq 1 / a_{k}, \\ 0, & x = 0 or 1 / a_{k} < x \leq 1. \end{cases}$

Then $r_{k} \to 0$ pointwise while $\int_{0}^{1} r_{k} = 1$ for every $k$ .

Facts & Assumptions

Given: The functions $r_{k}$ in the Statement, with $a_{k} = ι (k + 1) > 0$ .

[L1]

For every real $ε > 0$ there is $N \geq 1$ with $1 / ι (N) < ε$ ; canonical naturals increase and their positive reciprocals decrease (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , The canonical natural $ι (n) = n \cdot 1_{F}$ 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 $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ 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 $ε > 0$ there is an $N$ such that $k \geq N$ implies $∣ f_{k} (x) - f (x) ∣ < ε$ ; 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 $1 / ι (N) < x$ ; for $k \geq N$ , monotonicity of the canonical naturals gives $1 / a_{k} < x$ , hence $r_{k} (x) = 0$ . Thus $r_{k} \to 0$ pointwise.

L1L5choose

1.4

To see explicitly that the convergence is not uniform, take $ε : = 1 / 2$ . For every proposed $N \in N$ , choose $k : = N$ and $x_{N} : = 1 / a_{N}$ ; then $∣ r_{N} (x_{N}) - 0 ∣ = a_{N} \geq 1 > ε$ . 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 $\int_{0}^{1} q_{k} = a_{k} (1 / a_{k}) + 0 = 1$ .

step 1.2L3L4algebra

3.1

Steps 1.2 and 2.1 give $\int_{0}^{1} r_{k} = 1$ 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.

Dependency tree · two levels

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Sources

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