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FALSE: a nonnegative Riemann integrable function on $[a, b]$ with $\int_{a}^{b} f = 0$ is identically zero

Statement

False claim: if $f : [a, b] \to R$ is Riemann integrable (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 $\int_{a}^{b} f$ ) with $f (x) \geq 0$ for every $x \in [a, b]$ and $\int_{a}^{b} f = 0$ , then $f (x) = 0$ for every $x \in [a, b]$ .

The claim is true under the additional hypothesis that $f$ is continuous, and that is the version worth remembering; without it the integral simply cannot see a function that is positive on a null set. Thomae's function on $[0, 1]$ (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 $x$ ) is nonnegative, integrable, has integral $0$ , and is positive at every rational point of $[0, 1]$ , of which there are infinitely many.

Facts & Assumptions

Given: Thomae's function restricted to $[0, 1]$ , that is $t : [0, 1] \to R$ with $t (x) = 1 / ι (q (x))$ at a rational $x$ with least denominator $q (x) \geq 1$ , and $t (x) = 0$ 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 $x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field).

[A1]

The false claim: a nonnegative Riemann integrable function on a closed bounded interval with distinct endpoints whose integral is $0$ vanishes identically.

[L1]

$t (x) = 1 / ι (q (x))$ with $ι (q (x)) \geq 1 > 0$ at a rational $x$ , and $t (x) = 0$ at an irrational $x$ ; hence $0 \leq t (x) \leq 1$ everywhere and $t (x) > 0$ at every rational $x$ (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 $x$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

[L2]

Thomae's function on $R$ is continuous at every irrational and discontinuous at every rational (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 $t (c)$ ); a restriction of a function continuous at a point of the smaller domain is continuous there, the same $δ$ serving a condition quantified over fewer points (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 $[a, b]$ with an at most countable set of discontinuities is Riemann integrable (A bounded function on $[a, b]$ whose set of discontinuities is at most countable is Riemann integrable, Lower bound, bounded below, bounded set).

[L5]

The irrationals are dense in $R$ , so every nonempty open interval contains an irrational (Both $Q$ and $R ∖ Q$ are dense in $R$ , and every nonempty open subset of $R$ is uncountable, The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points, Interior, closure, boundary and exterior of a subset of $R$ , The $ε$ -neighbourhood and the punctured $ε$ -neighbourhood of a point of $R$ ).

[L6]

For a partition $P = (n, t^{P})$ of $[0, 1]$ : $n \geq 1$ , $t_{i}^{P} < t_{i + 1}^{P}$ , $Δ_{i} > 0$ , and $I_{i} = [t_{i}^{P}, t_{i + 1}^{P}] \subseteq [0, 1]$ (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).

[L7]

$L (t, P) = \sum_{i < n} m_{i} Δ_{i}$ with $m_{i} = \inf t [I_{i}]$ ; $\underline{\int_{0}^{1}} t$ is the supremum of the lower sums, $\overline{\int_{0}^{1}} t$ the infimum of the upper sums, and the integral is their common value when they agree (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 $\int_{a}^{b} f$ ).

[L8]

A set with a least element has it as its infimum; the supremum of ${0}$ is $0$ (Greatest lower bound (infimum), Maximum and minimum of a set, Complete ordered field (least-upper-bound property)).

[L9]

Finite sums: $\sum_{i < n} 0 = 0$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L10]

Ordered-field arithmetic: the order is total and transitive; a reciprocal of a positive quantity is positive; $0 < 1$ (Order is preserved by adding a constant and by adding inequalities, Sign rules for products and monotonicity of multiplication, Ordered field, Complete ordered field (least-upper-bound property)). These order-arithmetic facts are stated by their sources for the strict order only; the nonstrict forms used below follow by adjoining the equality case, in which the two sides coincide.

Refutation

technique · direct

1.1

$t$ is nonnegative and bounded on $[0, 1]$ , with $0 \leq t (x) \leq 1$ for every $x$ , by [L1].

givenL1L10

1.2

$t$ is continuous at every irrational point of $[0, 1]$ by [L2], so its set of discontinuities in $[0, 1]$ is contained in $Q \cap [0, 1]$ , which is at most countable by [L3]; every subset of it is then at most countable by [L3] as well.

givenL2L3

1.3

Separately, and independently of everything below, $t$ does not vanish identically: $2^{- 1}$ is a rational point of $[0, 1]$ , so $t (2^{- 1}) > 0$ by [L1] and [L10].

L1L10

2.1

By [L4] applied to $[0, 1]$ , with $0 < 1$ , the function $t$ is Riemann integrable on $[0, 1]$ .

step 1.1step 1.2L4

2.2

Every lower Darboux sum of $t$ is $0$ : let $P$ be a partition of $[0, 1]$ and $i < n$ ; the open interval $(t_{i}^{P}, t_{i + 1}^{P})$ is nonempty by [L6] and contains an irrational $y$ by [L5], and $y \in I_{i} \subseteq [0, 1]$ , so $t (y) = 0$ by [L1]. Since $t \geq 0$ by [L1], the value $0$ is the least element of $t [I_{i}]$ and $m_{i} = 0$ by [L8]. Hence $L (t, P) = \sum_{i < n} 0 \cdot Δ_{i} = 0$ by [L7] and [L9].

step 1.1L1L5L6L7L8L9

3.1

The set of lower sums is therefore ${0}$ , so $\underline{\int_{0}^{1}} t = 0$ by [L8], and since $t$ is integrable by step 2.1 its integral is $\int_{0}^{1} t = 0$ by [L7].

step 2.1step 2.2L7L8

4.1

So $t$ is a nonnegative Riemann integrable function on $[0, 1]$ with $\int_{0}^{1} t = 0$ that is not identically zero; [A1] fails at $t$ and the claim is false.

step 1.1step 3.1step 1.3A1∎

Remarks

The correct statement, and the hypothesis that repairs it. If $f$ is continuous on $[a, b]$ , nonnegative and $\int_{a}^{b} f = 0$ , then $f \equiv 0$ : a point with $f (c) > 0$ would, by continuity, force $f > f (c) \cdot 2^{- 1}$ on a whole subinterval, and $\int_{a}^{b} f$ would then be at least $f (c) \cdot 2^{- 1}$ times the length of that subinterval, by 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 $\int_{a}^{b} c = c (b - a)$ applied there. That argument is not carried out here, because the additivity of the integral over subintervals is not available at this point in the reading order; what is asserted by this item is only that the hypothesis of continuity cannot simply be dropped.
The exceptional set is exactly the rationals. $t$ is positive precisely on $Q \cap [0, 1]$ , a countable dense set, which is null (Every at most countable subset of $R$ has measure zero). The integral is blind to a null set of positive values, and the Lebesgue criterion explains why: what governs integrability, and here the value, is the size of the set where the function misbehaves, measured by interval covers (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).
Every upper sum, by contrast, is positive. Each subinterval contains a rational, so $M_{i} > 0$ for every $i$ and $U (t, P) > 0$ for every partition $P$ ; the upper integral is nevertheless $0$ , since the infimum of a set of positive numbers may be $0$ . That is the one place where this example is worth pausing over, and it is the reason the lower sums, not the upper ones, are what step 2.2 computes.

Depends on

Used by

Dependency tree · two levels

91 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Thomae's function (Wikipedia) (standard reference, not scraped)
Riemann integral (Wikipedia) (standard reference, not scraped)
MAT 125B Discussion 3 (UC Davis) (standard reference, not scraped)

FALSE: a nonnegative Riemann integrable function on [a,b][a,b] with ∫abf=0∫ab​f=0 is identically zero