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

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:

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.