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Thomae's integrand is discontinuous at every rational, yet its integral function is identically zero and differentiable everywhere

Example

Let $t$ be Thomae's function on $[0, 1]$ . Then $t$ is Riemann integrable and

$\int_{0}^{1} t = 0.$

Consequently its integral function $T (x) = \int_{0}^{x} t$ is identically zero and is differentiable at every point. At every irrational $x$ , $T^{'} (x) = t (x) = 0$ ; at every rational $x$ , $t$ is discontinuous and $T^{'} (x) = 0 \neq t (x)$ . Thus the derivative of an integral function may exist at every point even though the integrand is discontinuous on a dense set.

Facts & Assumptions

Given: Thomae's function $t$ , equal to $1 / q$ at a rational with least positive denominator $q$ and to $0$ at an irrational (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 $x$ ).

[L1]

Thomae's continuity points are exactly the irrationals (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 $t (c)$ ).

[L2]

The rationals are countable, and subsets of countable sets are countable ( $Q$ is countably infinite, Every subset of an at most countable set is at most countable).

[L3]

A bounded function with at most countably many discontinuities is Riemann integrable (A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable).

[L4]

Darboux upper sums bound the integral from above, and a nonnegative integrable function has nonnegative integral (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 $U (f, P) = \sum_{i} M_{i} Δ_{i}$ , 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 $\int_{a}^{b} f$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ).

[L5]

Integral functions and additivity give $T (y) - T (x) = \int_{x}^{y} t$ (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 $a, b, c$ ).

[L6]

The rationals and irrationals are both dense in the reals (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable).

Verification

technique · direct

1.1

The function satisfies $0 \leq t \leq 1$ , and by [L1] its discontinuity set is $Q \cap [0, 1]$ , which is countable by [L2]. Hence [L3] makes $t$ integrable.

givenL1L2L3

1.2

Fix $ε > 0$ and choose an integer $N \geq 1$ with $1 / N < ε / 2$ . The rationals in $[0, 1]$ with least denominator at most $N$ form a finite set, since each has a representation $p / q$ with $0 \leq p \leq q \leq N$ .

givenchoosealgebra

2.1

Choose finitely many intervals around that finite set with total length below $ε / 2$ . A partition containing their endpoints has upper contribution below $ε / 2$ on those intervals because $t \leq 1$ , and below $1 / N < ε / 2$ on their complement because every positive value there has denominator greater than $N$ . Thus it has upper sum below $ε$ .

step 1.2L4construct

3.1

Since $t \geq 0$ , [L4] and the arbitrarily small upper sums in step 2.1 force $\int_{0}^{1} t = 0$ . The same argument on every subinterval gives $\int_{x}^{y} t = 0$ .

step 2.1L4

4.1

By [L5], $T (y) - T (x) = 0$ for all $x, y$ , so $T$ is identically zero and $T^{'} = 0$ everywhere, with relative derivatives at $0$ and $1$ .

step 3.1L5

5.1

By [L6], the rationals and irrationals are both dense. Combining the definition of $t$ , [L1], and step 4.1 gives the claimed equality at irrationals and failure at rationals.

givenstep 4.1L1L6∎

Depends on

Used by

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Sources

Thomae's function (standard reference, not scraped)