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A function that is not Riemann integrable although $\lvert f\rvert$ 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 $\lvert f\rvert$ , $f^{2}$ , $fg$ , $\max(f,g)$ and $\min(f,g)$ , and $\bigl\lvert\int_a^b f\bigr\rvert \le \int_a^b\lvert f\rvert$ has a converse.

Let $\mathbf{1}_{\mathbb{Q}}$ be the Dirichlet function (The Dirichlet function $1_{\mathbb{Q}}$ , and Thomae's function $t$ with $t(x) = 1/q$ at a rational $x = p/q$ in lowest terms with $q \ge 1$ and $t(x) = 0$ at every irrational $x$ ) and put

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

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 $f = 2\mathbf{1}_{\mathbb{Q}} - 1$ on $[0,1]$ , and a partition $P = (n,t)$ of $[0,1]$ .

[L1]

$\mathbf{1}_{\mathbb{Q}}(x) = 1$ at a rational $x$ and $0$ at an irrational $x$ (The Dirichlet function $1_{\mathbb{Q}}$ , and Thomae's function $t$ with $t(x) = 1/q$ at a rational $x = p/q$ in lowest terms with $q \ge 1$ and $t(x) = 0$ at every irrational $x$ ).

[L2]

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

[L3]

For a partition $P = (n,t)$ of $[0,1]$ : $n \ge 1$ , $\Delta_i = t_{i+1}-t_i > 0$ , $\sum_{i<n}\Delta_i = 1$ , 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 $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

$L(u,P) = \sum_{i<n}m_i\Delta_i$ and $U(u,P) = \sum_{i<n}M_i\Delta_i$ 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 $L(f,P) = \sum_i m_i \Delta_i$ and $U(f,P) = \sum_i M_i \Delta_i$ , 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 $\sum_{i<n}\lambda\Delta_i = \lambda\sum_{i<n}\Delta_i$ (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 $m \le f \le M$ 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 $\int_a^b c = c(b-a)$ ).

[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 $x$ , $f(x) = 2\cdot1 - 1 = 1$ , and at an irrational $x$ , $f(x) = 2\cdot0-1 = -1$ , 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 $L(f,P) = \sum_{i<n}(-1)\Delta_i = -1$ and $U(f,P) = \sum_{i<n}1\cdot\Delta_i = 1$ , 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], $\underline{\int_0^1}f = -1 \ne 1 = \overline{\int_0^1}f$ 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 $2\mathbf{1}_{\mathbb{Q}} - 1$ , and $\mathbf{1}_{\mathbb{Q}}$ is continuous at no point of $\mathbb{R}$ (The Dirichlet function is continuous at no point of $\mathbb{R}$ , 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 $\psi$ satisfies $\psi(|t|) = t$ .
The same example separates $\int|f|$ from $\bigl|\int f\bigr|$ 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∣\lvert f\rvert∣f∣ is