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FALSE: in the substitution theorem the continuity of $f$ may be weakened to integrability, $f \circ φ$ still being integrable

Statement

False claim: let $c < d$ be reals, let $φ : [c, d] \to R$ be differentiable at every point of $[c, d]$ with $φ^{'}$ integrable, and let $f$ be Riemann integrable on an interval $J$ containing $φ [[c, d]]$ . Then $f \circ φ$ is Riemann integrable on $[c, d]$ — so the hypothesis " $f$ is continuous" in Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ may be weakened to " $f$ is integrable" without the right-hand side $\int_{c}^{d} (f \circ φ) φ^{'}$ losing its meaning.

The claim is false. Let $S \subseteq [0, 1]$ be the Smith-Volterra-Cantor set (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals), which is compact, nowhere dense and not null (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero), and let

$d_{S} (u) : = \inf {∣ u - s ∣ : s \in S}, φ (x) : = \int_{0}^{x} d_{S} (x \in [0, 1]) .$

Then $φ$ is differentiable at every point of $[0, 1]$ with $φ^{'} = d_{S}$ continuous, hence integrable; $φ$ is strictly increasing; and $φ [S]$ has measure zero. Taking

$f : = 1_{φ [S]} on J : = [0, φ (1)]$

gives an integrable $f$ , because its discontinuity set is contained in the null closed set $φ [S]$ , while

$f \circ φ = 1_{S} on [0, 1],$

whose discontinuity set is exactly $S$ and is not null. So $f \circ φ$ is not Riemann integrable.

What this does and does not show. It shows that continuity of $f$ in Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ cannot simply be weakened to integrability: the composite in the right-hand side need not be integrable. It does not exhibit a pair for which both sides of the substitution identity exist and differ, and no such pair is claimed here.

Facts & Assumptions

Given: The Smith-Volterra-Cantor set $S \subseteq [0, 1]$ , the function $d_{S}$ and the function $φ$ above, a real $ε > 0$ and a natural number $N \geq 1$ .

[L1]

$S$ is a compact, nowhere dense subset of $[0, 1]$ ; it is nonempty; and no cover of $S$ by intervals has total length below $2^{- 1}$ , so $S$ does not have measure zero (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals, The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero, Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover), A subset of $R$ is compact if and only if it is closed and bounded, Open cover, subcover, compact subset of $R$ (every open cover has a finite subcover), and sequentially compact subset).

[L2]

Nowhere dense plus closed means empty interior: no nonempty open interval is contained in $S$ (Nowhere dense, meager (first category), residual, and second category subsets of $R$ , Interior, closure, boundary and exterior of a subset of $R$ , Open subset of $R$ (every point has a neighbourhood inside it), closed subset (complement open), and clopen).

[L3]

$u \mapsto d_{S} (u)$ is defined for every real (the set is nonempty and bounded below by $0$ ) and is $1$ -Lipschitz, hence continuous ( $∣ d (x, A) - d (y, A) ∣ \leq d (x, y)$ , so the distance to a fixed nonempty set is $1$ -Lipschitz, Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space, The absolute value makes $R$ a metric space: $d (x, y) = ∣ x - y ∣$ is a metric, its open balls are the intervals $(x - r, x + r)$ , and it is unbounded, Dictionary: for $A \subseteq R$ with the metric $d (x, y) = ∣ x - y ∣$ , continuity and uniform continuity of $f : A \to R$ agree with the metric-space notions, the Lipschitz and Hölder conditions are the metric ones instantiated, and a subset of $R$ is compact in the open-cover sense of $R$ exactly when it is a compact metric subspace, clauses 3 and 6, Greatest lower bound (infimum), Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L4]

A continuous function on a closed bounded interval with distinct endpoints is integrable there (A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, 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 $\int_{a}^{b} f$ ).

[L5]

First fundamental theorem: for $u$ integrable on $[p, q]$ and continuous at every point, $x \mapsto \int_{p}^{x} u$ is differentiable with derivative $u$ (The first fundamental theorem: if $f$ is integrable on $[a, b]$ and continuous at $c$ , then $F^{'} (c) = f (c)$ ; in particular a continuous $f$ has $F$ as a primitive, The integral function $F (x) : = \int_{a}^{x} f$ of an integrable $f$ , 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 $\int_{b}^{a} f : = - \int_{a}^{b} f$ ).

[L6]

A continuous $w \geq 0$ on $[p, q]$ with $p < q$ and $\int_{p}^{q} w = 0$ vanishes identically (A continuous $f \geq 0$ on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically $0$ ).

[L7]

If $κ \leq w \leq μ$ on $[p, q]$ with $w$ integrable then $κ (q - p) \leq \int_{p}^{q} w \leq μ (q - p)$ (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , 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 $\int_{a}^{b} c = c (b - a)$ ).

[L8]

The continuous image of a compact set is compact, and a compact subset of $R$ is closed and bounded (The image of a compact subset of $R$ under a continuous real function is compact, A subset of $R$ is compact if and only if it is closed and bounded).

[L9]

Lebesgue's criterion: a bounded function on $[p, q]$ is integrable if and only if its discontinuity set has measure zero; a set of content zero has measure zero (Lebesgue's criterion for Riemann integrability: a bounded $f$ on $[a, b]$ is Riemann integrable if and only if its set of discontinuities has measure zero, A set of content zero has measure zero, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover), Lower bound, bounded below, bounded set).

[L10]

Finite sums: monotonicity in the terms and $\sum_{i < N} λ = ι (N) λ$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clauses 2 and 4); $ι (N) \geq 1 > 0$ for $N \geq 1$ , and for every real $η > 0$ there is $N \geq 1$ with $1 / ι (N) < η$ (The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ , Every complete ordered field is Archimedean).

[L11]

Ordered-field arithmetic and suprema: a nonempty bounded set has a supremum and an infimum; $\min {s, t}$ is at most the average of $s$ and $t$ when $s + t$ is fixed; multiplying inequalities by positive reals preserves them; the order is total and transitive (Complete ordered field (least-upper-bound property), Ordered field, Greatest lower bound (infimum), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Injection, surjection, bijection).

Refutation

technique · direct

1.1givenL1L3L11

$d_{S} \geq 0$ everywhere, $d_{S} (u) = 0$ for $u \in S$ , and $d_{S} (u) > 0$ for $u \notin S$ : $S$ is closed by [L1], so some $ρ > 0$ has $(u - ρ, u + ρ) \cap S = \emptyset$ , whence $∣ u - s ∣ \geq ρ$ for every $s \in S$ and $d_{S} (u) \geq ρ$ .

2.1step 1.1L3L4L5

$d_{S}$ is continuous by [L3], hence integrable on every $[0, x]$ with $x > 0$ by [L4]; so $φ$ is defined on $[0, 1]$ , and by [L5] it is differentiable at every point of $[0, 1]$ with $φ^{'} = d_{S}$ , which is integrable by [L4].

3.1step 1.1step 2.1L2L5L6L7

$φ$ is strictly increasing, hence injective. For $0 \leq x < y \leq 1$ , $φ (y) - φ (x) = \int_{x}^{y} d_{S} \geq 0$ by [L5] and [L7]; if it were $0$ then [L6] would force $d_{S} \equiv 0$ on $[x, y]$ , so $[x, y] \subseteq S$ by step 1.1, contradicting [L2]. Hence $φ (x) < φ (y)$ . In particular $φ (0) = 0 < φ (1)$ .

3.2step 2.1L5L7L11

A quadratic contraction on $S$ . Let $x < y$ both lie in $S$ . For $u \in [x, y]$ one has $d_{S} (u) \leq \min {u - x, y - u} \leq (y - x) \cdot 2^{- 1}$ , since $x, y \in S$ ; so by [L5] and [L7], $0 \leq φ (y) - φ (x) = \int_{x}^{y} d_{S} \leq (y - x)^{2} \cdot 2^{- 1}$ .

4.1step 3.2L10L11construct

$φ [S]$ has content zero. Fix $N \geq 1$ and for $i < N$ put $J_{i} : = [ι (i) / ι (N), ι (i + 1) / ι (N)]$ , so the $J_{i}$ cover $[0, 1]$ and each has length $1 / ι (N)$ . If $S \cap J_{i} \neq \emptyset$ let $E_{i} : = φ [S \cap J_{i}]$ , a nonempty bounded set, and put $a_{i} : = \inf E_{i}$ , $b_{i} : = \sup E_{i}$ ; otherwise put $a_{i} : = b_{i} : = 0$ .

5.1step 3.2step 4.1L11

For $z, w \in E_{i}$ one has $∣ z - w ∣ \leq (1 / ι (N))^{2} \cdot 2^{- 1}$ by step 3.2, the two preimages lying in $S \cap J_{i}$ ; hence $b_{i} \leq a_{i} + (1 / ι (N))^{2} \cdot 2^{- 1}$ , since every $z \in E_{i}$ is at most $w + (1 / ι (N))^{2} 2^{- 1}$ for each fixed $w$ , and then $w \geq b_{i} - (1 / ι (N))^{2} 2^{- 1}$ for every $w$ .

6.1step 4.1step 5.1L10

Every point of $φ [S]$ lies in some $[a_{i}, b_{i}]$ , because every point of $S$ lies in some $J_{i}$ ; and $\sum_{i < N} (b_{i} - a_{i}) \leq ι (N) \cdot (1 / ι (N))^{2} \cdot 2^{- 1} = 1 / (2 ι (N))$ by [L10].

7.1step 6.1L9L10

Given $ε > 0$ , [L10] supplies $N \geq 1$ with $1 / (2 ι (N)) \leq ε$ ; so $φ [S]$ has content zero and therefore measure zero by [L9].

8.1step 2.1step 7.1L1L3L8L9

$f : = 1_{φ [S]}$ is integrable on $J = [0, φ (1)]$ . It is bounded, with values in ${0, 1}$ . $S$ is compact by [L1] and $φ$ is continuous by [L5] and [L3], so $φ [S]$ is compact, hence closed, by [L8]; therefore at every $z \in J ∖ φ [S]$ some neighbourhood misses $φ [S]$ and $f$ vanishes on it, so $f$ is continuous there. The discontinuity set of $f$ is thus contained in $φ [S]$ , which is null by step 7.1, so $f$ is integrable by [L9].

9.1step 3.1step 8.1

$f \circ φ = 1_{S}$ on $[0, 1]$ . For $x \in [0, 1]$ : if $x \in S$ then $φ (x) \in φ [S]$ and $f (φ (x)) = 1$ ; if $x \notin S$ then $φ (x) \notin φ [S]$ , since $φ$ is injective by step 3.1, and $f (φ (x)) = 0$ . Also $φ [[0, 1]] \subseteq J$ by step 3.1.

10.1step 9.1L1L2L11

$1_{S}$ is discontinuous at every point of $S$ . Let $x \in S$ and $ρ > 0$ ; the set $(x - ρ, x + ρ) \cap (0, 1)$ contains a nonempty open interval, which by [L2] is not contained in $S$ , so some $y$ in it has $1_{S} (y) = 0$ while $1_{S} (x) = 1$ ; no $δ$ works for $ε = 2^{- 1}$ . At $x \notin S$ the function vanishes on a neighbourhood, $S$ being closed, so it is continuous there.

11.1step 9.1step 10.1L1L9

The discontinuity set of $f \circ φ$ on $[0, 1]$ is therefore exactly $S$ , which is not null by [L1]; so $f \circ φ$ is bounded and not Riemann integrable, by [L9].

12.1step 2.1step 8.1step 9.1step 11.1∎

So $φ$ is differentiable on $[0, 1]$ with $φ^{'}$ integrable, $f$ is integrable on an interval containing $φ [[0, 1]]$ , and $f \circ φ$ is not integrable: the claim is false, and the continuity hypothesis on $f$ in Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ cannot be weakened to integrability.

Remarks

Why $φ$ has to be built and cannot be a familiar function. The set on which $f \circ φ$ misbehaves is $φ^{- 1}$ of a null set, and for a Lipschitz $φ$ with $φ^{'}$ bounded away from $0$ that preimage is again null. What makes the witness work is that $φ^{'} = d_{S}$ vanishes on the whole of $S$ , so $φ$ crushes a set of positive measure onto a null set while remaining injective; step 3.2 is the quantitative form of that crushing.
The substitution identity itself is not refuted here, and that is stated rather than glossed over. For this very pair one has $(f \circ φ) φ^{'} = 1_{S} \cdot d_{S}$ , which is the zero function because $d_{S}$ vanishes on $S$ ; so the right-hand side of the substitution formula does exist for this pair and equals $0$ , as does the left-hand side. What fails is the intermediate claim that the composite $f \circ φ$ is integrable, which is what a proof of the identity with $f$ merely integrable would have to establish, and which is what Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ 's continuity hypothesis delivers through If $f$ is integrable on $[a, b]$ with values in $[m, M]$ and $φ$ is continuous on $[m, M]$ , then $φ \circ f$ is integrable. No pair with both sides defined and unequal is exhibited anywhere on this page.
The fat Cantor set is doing exactly one job. It supplies a closed set with empty interior that is not null. Any such set would serve; $S$ is the published one, and The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero is what supplies all three properties without reproving them.

FALSE: in the substitution theorem the continuity of $f$ may be weakened to integrability, $f \circ φ$ still being integrable

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: in the substitution theorem the continuity of ff may be weakened to integrability, f∘φf∘φ still being integrable

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources

FALSE: in the substitution theorem the continuity of $f$ may be weakened to integrability, $f \circ φ$ still being integrable