Continuous Non-Negative Real Function with Zero Integral is Zero Function/Proof 1

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Theorem

Let $a, b$ be real numbers with $a < b$.

Let $f : \closedint a b \to \R$ be a continuous function.

Let:

$\map f x \ge 0$

for all $x \in \closedint a b$.

Let:

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


Then $\map f x = 0$ for all $x \in \closedint a b$.


Proof

From Definite Integral of Constant, if $\map f x = 0$ for all $x \in \closedint a b$, then:

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

We want to show that if:

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

then:

$\map f x = 0$ for all $x \in \closedint a b$.

Since $\map f x \ge 0$, by Relative Sizes of Definite Integrals:

$\ds \int_a^b \map f x \rd x \ge 0$

It therefore suffices to show that if:

$\map f x > 0$ for some $x \in \closedint a b$

then:

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

We split this into three cases:

$x = a$
$a < x < b$
$x = b$

since continuity on endpoints is defined slightly differently.

Consider the case:

$a < x < b$

By continuity at $x$ we have that there exists $\delta > 0$ such that:

for all $y \in \openint {x - \delta} {x + \delta}$ we have $\size {\map f y - \map f x} < \dfrac {\map f x} 2$

In particular for $y \in \openint {x - \delta} {x + \delta}$ we have:

$0 < \dfrac {\map f x} 2 < \map f y$

Pick $\delta$ sufficiently small so that:

$\openint {x - \delta} {x + \delta} \subseteq \closedint a b$

We then have:

\(\ds \int_a^b \map f y \rd y\) \(=\) \(\ds \int_a^{x - \delta} \map f y \rd y + \int_{x - \delta}^{x + \delta} \map f y \rd y + \int_{x + \delta}^b \map f y \rd y\) Sum of Integrals on Adjacent Intervals for Continuous Functions
\(\ds \) \(\ge\) \(\ds \int_{x - \delta}^{x + \delta} \map f y \rd y\) Relative Sizes of Definite Integrals gives $\ds \int_a^{x - \delta} \map f y \rd y + \int_{x + \delta}^b \map f y \rd y \ge 0$
\(\ds \) \(>\) \(\ds \delta \map f x\) Relative Sizes of Definite Integrals, Definite Integral of Constant
\(\ds \) \(>\) \(\ds 0\)

In the case $x = a$, by the definition of right continuity, there exists $\delta > 0$ such that:

for all $x \in \openint a {a + \delta}$ we have $\size {\map f x - \map f a} < \dfrac {\map f a} 2$

That is, there exists some $x \in \openint a b$ such that:

$\map f x > \dfrac {\map f a} 2 > 0$

So the former proof applies in this case.

The case $x = b$ follows similarly.

In the case $x = b$, by the definition of left continuity, there exists $\delta > 0$ such that:

for all $x \in \openint {b - \delta} b$ we have $\size {\map f x - \map f b} < \dfrac {\map f b} 2$

That is, there exists some $x \in \openint a b$ such that:

$\map f x > \dfrac {\map f b} 2 > 0$

So, again, the proof for the case $a < x < b$ applies.

We have covered all three cases, so we are done.

$\blacksquare$