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Integrable $φ$ and integrable $f$ with $φ \circ f$ not integrable: the order of the hypotheses in the composition theorem cannot be reversed

Statement refuted

False claim: if $f : [a, b] \to [m, M]$ is Riemann integrable and $φ : [m, M] \to R$ is Riemann integrable, then $φ \circ f$ is Riemann integrable on $[a, b]$ .

That is 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 with "continuous" weakened to "integrable" on the outer function, and it is false. On $[0, 1]$ take $f : = t$ , Thomae's function (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ ), whose values lie in $[0, 1]$ , and

$φ : [0, 1] \to R, φ (u) : = {\begin{cases} 0 & u = 0, \\ 1 & 0 < u \leq 1. \end{cases}$

Both are Riemann integrable, $t$ because its discontinuity set is at most countable and $φ$ because it is nondecreasing. But

$φ \circ t = 1_{Q} on [0, 1],$

the Dirichlet function, which is not Riemann integrable.

Exactly one hypothesis of the composition theorem fails, and it is named: $φ$ is not continuous, being discontinuous at $0$ . The theorem's hypothesis is continuous after integrable; here the outer function is merely integrable, and that is not enough.

Facts & Assumptions

Given: Thomae's function $t$ on $[0, 1]$ , with $t (x) = 1 / ι (q (x))$ at a rational $x$ of least denominator $q (x) \geq 1$ and $t (x) = 0$ at an irrational $x$ , and the function $φ$ above.

[L1]

$0 \leq t (x) \leq 1$ for every $x$ , and $t (x) > 0$ exactly at the rational $x$ , since $ι (q (x)) \geq 1 > 0$ (The Dirichlet function $1_{Q}$ , and Thomae's function $t$ with $t (x) = 1 / q$ at a rational $x = p / q$ in lowest terms with $q \geq 1$ and $t (x) = 0$ at every irrational $x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing, Ordered field).

[L2]

$t$ is continuous at every irrational and discontinuous at every rational, and $1_{Q}$ is continuous at no real (The Dirichlet function is continuous at no point of $R$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t (c)$ , claims 1 and 3, 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).

[L3]

$Q$ is countably infinite and every subset of an at most countable set is at most countable ( $Q$ is countably infinite, Every subset of an at most countable set is at most countable, Finite, countably infinite, countable, uncountable).

[L4]

A bounded function on $[p, q]$ whose set of discontinuities is at most countable is Riemann integrable there (A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable, Lower bound, bounded below, bounded set, 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]

A monotone function on a closed bounded interval with distinct endpoints is bounded and integrable there (A monotone function on $[a, b]$ is Riemann integrable: for the uniform partition into $N$ parts the upper minus lower sum telescopes to $∣ f (b) - f (a) ∣ (b - a) / ι (N)$ , Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of $R$ , with the dictionary to monotone sequences).

[L6]

A bounded function on $[p, q]$ is integrable if and only if its discontinuity set has measure zero; a nondegenerate interval does not have measure zero; an at most countable set does (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 sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero, Every at most countable subset of $R$ has measure zero, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L7]

Ordered-field arithmetic and intervals: the order is total and transitive (Ordered field, Complete ordered field (least-upper-bound property), Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct

1.1

$t$ is bounded on $[0, 1]$ with values in $[0, 1]$ , by [L1].

givenL1

1.2

The set of discontinuities of $t$ in $[0, 1]$ is $Q \cap [0, 1]$ by [L2], which is at most countable by [L3].

L2L3

1.3

$φ$ is nondecreasing on $[0, 1]$ : for $u \leq v$ the only case not giving $φ (u) = φ (v)$ is $u = 0 < v$ , where $φ (u) = 0 < 1 = φ (v)$ . So $φ$ is bounded and integrable on $[0, 1]$ by [L5].

givenL5L7

2.1

Hence $t$ is Riemann integrable on $[0, 1]$ by [L4].

step 1.1step 1.2L4

2.2

For $x \in [0, 1]$ : if $x$ is rational then $t (x) > 0$ by [L1], so $φ (t (x)) = 1$ ; if $x$ is irrational then $t (x) = 0$ , so $φ (t (x)) = 0$ . Hence $φ \circ t$ agrees with $1_{Q}$ at every point of $[0, 1]$ .

step 1.1givenL1

3.1

$1_{Q}$ is bounded on $[0, 1]$ and continuous at no point of it by [L2], so its discontinuity set is $[0, 1]$ , which does not have measure zero by [L6]; therefore $φ \circ t$ is not Riemann integrable on $[0, 1]$ by [L6].

step 2.2L2L6

4.1

So $t$ and $φ$ are integrable while $φ \circ t$ is not, and the claim is false. The hypothesis of 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 that fails here is the continuity of the outer function: $φ$ is discontinuous at $0$ , since $φ (0) = 0$ while $φ (u) = 1$ for every $u \in (0, 1]$ and every neighbourhood of $0$ in $[0, 1]$ contains such a $u$ .

step 2.1step 1.3step 3.1L7∎

Remarks

The order of composition is the whole point. The published theorem is continuous after integrable, and it is true; the reversal, integrable after integrable, is refuted here. A reader who remembers only "composition preserves integrability" will get this wrong in one direction out of two, and the direction that fails is the one this item names.
Nothing is claimed about the inner function being continuous. If $f$ is continuous and $φ$ merely integrable, the composite may still fail to be integrable; that case is not covered above and no claim is made about it here.
Thomae is the right inner function because its positivity set is exactly $Q$ . Any integrable $f$ that is positive exactly on a dense set of measure zero and zero elsewhere would do; Thomae is the published one, and The Dirichlet function is continuous at no point of $R$ , and Thomae's function is continuous at every irrational and at no rational, so its set of continuity points is exactly the set of irrationals and its oscillation at $c$ equals $t (c)$ is what supplies its continuity set without recomputing it.

Depends on

Used by

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Dependency tree · two levels

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Sources

Thomae's function (Wikipedia) (standard reference, not scraped)
Riemann integral (Wikipedia) (standard reference, not scraped)
Springer article on compositions of Riemann-integrable functions (standard reference, not scraped)

Integrable φφ and integrable ff with φ∘fφ∘f not integrable: the order of the hypotheses in the composition theorem cannot be reversed