Zero Definite Integral of Nowhere Negative Function implies Zero Function

Theorem

Let $\closedint a b \subseteq \R$ be a closed real interval.

Let $h: \closedint a b \to \R$ be a continuous real function such that:

$\forall x \in \closedint a b: \map h x \ge 0$

Let:

$\ds \int_a^b \map h x \rd x = 0$

Then:

$\forall x \in \closedint a b: \map h x = 0$

Proof

Aiming for a contradiction, suppose that:

$\exists c \in \closedint a b: \map h c > 0$

As $h$ is continuous, there exists some closed real interval $\closedint r s \subseteq \closedint a b$ where $r < s$ such that:

$\exists \epsilon \in \R_{>0}: \forall x \in \closedint r s: \map h x > \dfrac {\map h c} 2$

From Sign of Function Matches Sign of Definite Integral:

$\ds \int_r^s \map h x \rd x > 0$

This makes a strictly positive contribution to the integral.

$h$ is still continuous on $\closedint a r$ and $\closedint s b$.

So, again from Sign of Function Matches Sign of Definite Integral:

$\forall x \in \closedint a r: \map f x \ge 0 \implies \ds \int_a^r \map f x \rd x \ge 0$

$\forall x \in \closedint s b: \map f x \ge 0 \implies \ds \int_s^b \map f x \rd x \ge 0$

Thus as:

$\ds \int_a^b \map h x \rd x = \int_a^r \map f x \rd x + \int_r^s \map f x \rd x + \int_s^b \map f x \rd x$

it follows that:

$\ds \int_a^b \map h x \rd x > 0$

But by hypothesis:

$\ds \int_a^b \map h x \rd x = 0$

Thus by contradiction, there can be no $c \in \closedint a b$ such that $\map h c > 0$.

Hence the result.

$\blacksquare$

Sources

• 1975: W.A. Sutherland: Introduction to Metric and Topological Spaces ... (previous) ... (next): $2$: Continuity generalized: metric spaces: $2.2$: Examples: Lemma $2.2.10$