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A function that is not Riemann integrable although $∣ f ∣$ is

Statement refuted

False claim: if $∣ f ∣$ is Riemann integrable on $[a, b]$ then so is $f$ ; that is, the first clause of If $f, g$ are integrable on $[a, b]$ then so are $∣ f ∣$ , $f^{2}$ , $f g$ , $\max (f, g)$ and $\min (f, g)$ , and $∣ \int_{a}^{b} f ∣ \leq \int_{a}^{b} ∣ f ∣$ has a converse.

Let $1_{Q}$ be the Dirichlet 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 $x$ ) and put

$f : [0, 1] \to R, f (x) : = 2 1_{Q} (x) - 1 = {\begin{cases} 1 & x rational, \\ - 1 & x irrational . \end{cases}$

Then $∣ f ∣$ is the constant function $1$ , integrable with $\int_{0}^{1} ∣ f ∣ = 1$ , while $f$ is not Riemann integrable on $[0, 1]$ : every lower Darboux sum of $f$ is $- 1$ and every upper Darboux sum is $1$ , so the lower and upper integrals are $- 1$ and $1$ .

Facts & Assumptions

Given: The function $f = 21_{Q} - 1$ on $[0, 1]$ , and a partition $P = (n, t)$ of $[0, 1]$ .

[L1]

$1_{Q} (x) = 1$ at a rational $x$ and $0$ at an irrational $x$ (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 $x$ ).

[L2]

Both $Q$ and the irrationals are dense in $R$ , so every nonempty open interval contains a rational and an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

[L3]

For a partition $P = (n, t)$ of $[0, 1]$ : $n \geq 1$ , $Δ_{i} = t_{i + 1} - t_{i} > 0$ , $\sum_{i < n} Δ_{i} = 1$ , and the open interval $(t_{i}, t_{i + 1})$ is nonempty (Partition of $[a, b]$ as a finite strictly increasing list $a = t_{0} < t_{1} < \dots < t_{n} = b$ , its subintervals and their lengths, its mesh, refinement, and the common refinement of two partitions, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

$L (u, P) = \sum_{i < n} m_{i} Δ_{i}$ and $U (u, P) = \sum_{i < n} M_{i} Δ_{i}$ with $m_{i} = \inf u [I_{i}]$ and $M_{i} = \sup u [I_{i}]$ ; a set with a least element has it as its infimum and with a greatest element has it as its supremum (For bounded $f$ on $[a, b]$ and a partition $P$ : the infimum $m_{i}$ and supremum $M_{i}$ of $f$ on the $i$ -th subinterval, and the lower and upper Darboux sums $L (f, P) = \sum_{i} m_{i} Δ_{i}$ and $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , Greatest lower bound (infimum), Maximum and minimum of a set).

[L5]

$\underline{\int_{0}^{1}} u = \sup_{P} L (u, P)$ and $\overline{\int_{0}^{1}} u = \inf_{P} U (u, P)$ ; $u$ is integrable exactly when the two agree (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 $\int_{a}^{b} f$ , Lower bound, bounded below, bounded set).

[L6]

Finite sums: scaling and $\sum_{i < n} λ Δ_{i} = λ \sum_{i < n} Δ_{i}$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, clause 2).

[L7]

$\int_{p}^{q} c = c (q - p)$ for a constant $c$ (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]

Absolute value and ordered-field arithmetic: $∣ 1 ∣ = ∣ - 1 ∣ = 1$ , and the order is total (Absolute value in an ordered field, Ordered field, Complete ordered field (least-upper-bound property)).

Counterexample

technique · direct

1.1

$∣ f ∣$ is the constant function $1$ on $[0, 1]$ : at a rational $x$ , $f (x) = 2 \cdot 1 - 1 = 1$ , and at an irrational $x$ , $f (x) = 2 \cdot 0 - 1 = - 1$ , and $∣ 1 ∣ = ∣ - 1 ∣ = 1$ by [L1] and [L8]. Hence $∣ f ∣$ is integrable with $\int_{0}^{1} ∣ f ∣ = 1$ by [L7].

givenL1L7L8

1.2

Let $P = (n, t)$ be any partition of $[0, 1]$ and let $i < n$ . The open interval $(t_{i}, t_{i + 1})$ is nonempty by [L3], so it contains a rational and an irrational by [L2]; both lie in $I_{i}$ , so $1 \in f [I_{i}]$ and $- 1 \in f [I_{i}]$ .

givenL2L3

2.1

$f$ is bounded, with values in ${- 1, 1}$ , so its Darboux sums are defined by [L4] and [L5].

step 1.1givenL4L5

3.1

Since $f [I_{i}] \subseteq {- 1, 1}$ and both values occur, $m_{i} = - 1$ and $M_{i} = 1$ by [L4].

step 2.1step 1.2L4

4.1

Hence $L (f, P) = \sum_{i < n} (- 1) Δ_{i} = - 1$ and $U (f, P) = \sum_{i < n} 1 \cdot Δ_{i} = 1$ , by [L4], [L6] and [L3].

step 3.1L3L4L6

5.1

That holds for every partition $P$ , so the set of lower sums is ${- 1}$ and the set of upper sums is ${1}$ ; by [L5], $\underline{\int_{0}^{1}} f = - 1 \neq 1 = \overline{\int_{0}^{1}} f$ and $f$ is not integrable.

step 4.1L5

6.1

So $∣ f ∣$ is integrable on $[0, 1]$ while $f$ is not, and the claim is false.

step 1.1step 5.1∎

Remarks

The Lebesgue criterion says the same thing. $f$ agrees with $21_{Q} - 1$ , and $1_{Q}$ is continuous at no point of $R$ (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 $t (c)$ , claim 1), so the discontinuity set of $f$ in $[0, 1]$ is the whole of $[0, 1]$ , which is not null (A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero); by 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 $f$ is not integrable. The direct computation above is given because it also locates both Darboux integrals, which the criterion does not.
Why no converse can be expected. Passing to $∣ f ∣$ destroys all sign information, and integrability is a statement about the oscillation of $f$ ; here $∣ f ∣$ has oscillation $0$ on every subinterval while $f$ has oscillation $2$ on every subinterval. The corollary's implication runs only in the direction the composition theorem provides, because $t \mapsto ∣ t ∣$ is continuous and no continuous $ψ$ satisfies $ψ (∣ t ∣) = t$ .
The same example separates $\int ∣ f ∣$ from $∣ \int f ∣$ in the strongest possible way: the right-hand side is not even defined.

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Sources

Riemann integral (Wikipedia) (standard reference, not scraped)
Thomae's function (Wikipedia) (standard reference, not scraped)

A function that is not Riemann integrable although ∣f∣∣f∣ is