Solution of A Nonnegative Continuous Function with Zero Integral

Suppose, seeking a contradiction, that $0<f(x_0)$ for some $x_0\in[a,b]$, and put $\varepsilon=f(x_0)/2$, so that $0<\varepsilon$ and $f(x_0)-\varepsilon=\varepsilon$.

Step 0: a global upper bound for $f$. Since $a<b$ and $f$ is continuous on $[a,b]$, Extreme Value Theorem on a Closed Real Interval gives a point $x_{\max}\in[a,b]$ with $f(x)\le f(x_{\max})$ for every $x\in[a,b]$. Put $M=f(x_{\max})$, so that

using also the hypothesis $0\le f(x)$. The same two inequalities then hold for every $x$ in any subset of $[a,b]$, in particular on every closed subinterval used below.

Step 1: a nondegenerate subinterval on which $f\ge\varepsilon$. Since $f$ is continuous at $x_0$, there is $\delta>0$ such that every $x\in[a,b]$ with $|x-x_0|<\delta$ satisfies $|f(x)-f(x_0)|<\varepsilon$ and hence

Put $\rho=\min\{\delta,\ b-a\}/2$. Then $0<\rho$, $\rho<\delta$ and $2\rho\le b-a$. Set

Then $a\le u\le b$ and $a\le v\le b$, and since $a\le x_0\le b$ we have $u\le x_0\le v$.

We check that $\rho\le v-u$, by considering which of the two arguments realises each of $u$ and $v$. If $u=a$ and $v=b$, then $v-u=b-a\ge 2\rho\ge\rho$. If $u=a$ and $v=x_0+\rho$, then $v-u=(x_0-a)+\rho\ge\rho$, since $a\le x_0$. If $u=x_0-\rho$ and $v=b$, then $v-u=(b-x_0)+\rho\ge\rho$, since $x_0\le b$. If $u=x_0-\rho$ and $v=x_0+\rho$, then $v-u=2\rho\ge\rho$. In every case $\rho\le v-u$, and since $0<\rho$ we get $u<v$.

Finally, $x_0-\rho\le u$ and $v\le x_0+\rho$, so every $x\in[u,v]$ satisfies $x_0-\rho\le x\le x_0+\rho$, that is, $|x-x_0|\le\rho<\delta$. As $[u,v]\subseteq[a,b]$, the display above applies and gives $\varepsilon<f(x)$, in particular

Step 2: splitting the integral. By A Continuous Function on a Closed Interval is Riemann Integrable §integrable, $f$ is Riemann integrable on $[a,b]$, so the additivity lemma may be applied to it. Applying Additivity of the Riemann Integral on Adjacent Intervals to the triple $a\le u\le b$ shows that the restrictions of $f$ to $[a,u]$ and to $[u,b]$ are Riemann integrable and that

Applying the same lemma to the restriction of $f$ to $[u,b]$ and the triple $u\le v\le b$ shows that the restrictions to $[u,v]$ and $[v,b]$ are Riemann integrable and that $\int_u^b f=\int_u^v f+\int_v^b f$. Combining,

Step 3: the outer two pieces are nonnegative. Consider $\int_a^u f$. If $a<u$, then $0\le f(t)\le M$ for every $t\in[a,u]$ by Step 0, so clause 2 of Uniform Partitions and Order Bounds for the Riemann Integral applied with these bounds $m=0$ and $M$ gives $0\le\int_a^u f(t)\,dt$. If $a=u$, then applying Additivity of the Riemann Integral on Adjacent Intervals to the triple $a\le a\le b$ gives $\int_a^b f=\int_a^a f+\int_a^b f$, whence $\int_a^a f=0$. In both cases $0\le\int_a^u f(t)\,dt$.

The same argument applies to $\int_v^b f$: if $v<b$ then $0\le f(t)\le M$ for every $t\in[v,b]$ by Step 0, and clause 2 of Uniform Partitions and Order Bounds for the Riemann Integral applied with these bounds gives $0\le\int_v^b f(t)\,dt$, and if $v=b$ then applying the additivity lemma to the triple $a\le b\le b$ gives $\int_a^b f=\int_a^b f+\int_b^b f$, whence $\int_b^b f=0$.

Step 4: the middle piece is strictly positive. By Step 1 we have $u<v$ and $\varepsilon\le f(t)$ for every $t\in[u,v]$, and $f(t)\le M$ there by Step 0; so clause 2 of Uniform Partitions and Order Bounds for the Riemann Integral applied with the bounds $m=\varepsilon$ and $M$ gives

Since $0<\varepsilon$ and $0<v-u$, the product $\varepsilon\,(v-u)$ is positive by Elementary Order Arithmetic in an Ordered Field.

Step 5: the contradiction. Combining Steps 2, 3 and 4,

so $0<0$, which is impossible.

Therefore there is no $x_0\in[a,b]$ with $0<f(x_0)$. Since the order of $\mathbb{R}$ is total, every $x\in[a,b]$ satisfies $f(x)\le 0$; combined with the hypothesis $0\le f(x)$ this gives $f(x)=0$ for every $x\in[a,b]$.