$f$ continuous on $[a,b]$, $f\geq 0$, $\int_{a}^{b}f=0$ then $f=0$

Question

I'm curious as to if anyone can provide some feedback on this proof.

If $f:[a,b]\rightarrow \mathbb{R}$ continuous, $f\geq 0$, $\int_{a}^{b}f=0$ then $f=0$.

My proof: Let's assume there exists some $c\in [a,b]$ such that $f(c)>0$. Then $f>0$ on an open set $c\in I\subset [a,b]$, this must be true because otherwise we would have a non-continuous piecewise function, $f(x)=0$ for all $x\in [a,b]-\{c\}$ and $f(c)=0$. Now let's build a lower Darboux sum greater than $0$.

$L(f,\sigma)=\sum_{i=0}^{n-1} m_i(f)\,\Delta x_i=m_k(f)\,\Delta x_i>0,$ where my partition is $\sigma=\{ a=x_0<x_1<\dots<x_n=b \}$ picked in a way such that $[x_k,x_{k+1}]\subset I$. Also $m_k(f)=\inf\{f(x):x\in [x_k,x_{k+1}]\}$. Now we can conclude that $\underline{\int_a^b}f=\sup \{\text{L}(f,\sigma):\sigma \in \ \text{the set of all partitions of} \ [a,b] \}>0$ but this contradicts the fact that $\underline{\int_a^b} f=\int_a^b f=0$. Therefore $f=0$.

What do you think? The book "A First Course in Real Analysis by Sterling Berberian" gives a hint to help the student with the proof and says: Assume $f(c)>0$ at some point c, argue that $f(x)\geq \frac{1}{2}f(c)$ on a subinterval of $[a,b]$ and construct a lower sum for $f$ that is $>0$.

As you can see I based most of my proof on that help, but not entirely. If you have any idea as to how to properly use the help please share it!.

Answer 1

Simpler idea

If $f(c)>0$ and $f$ is continuous, then $f(x)\ge\epsilon>0$ on a neighborhood $[u,v]$ of $c$. Then:

$$\int_a^b f\ge\int_u^v f\ge\epsilon(v-u)>0$$

In your proof using Darboux sums, you still need to prove $f(x)\ge\epsilon>0$ on a neighborhood of $c$ in order to show that the lower Darboux sum is positive (i.e. $m_k(f)>0$). Otherwise, even if $f(x)>0$ on $[x_k,x_{k+1}]$, you may still have $m_k(f)=0$. As in $\inf\{1,1/2,\ldots 1/n \ldots\}=0$, even though each element of the set is positive.

Answer 2

You point out that $f$ must be positive in some open neighborhood of $c.$ You can use the partition $a<x_1<x_2<b$ where $x_1,x_2$ are in that open neighborhood. For that partition, the lower sum is positive. If there is one partition for which the lower sum is positive, then the supremum of all lower sums is positive, thus the integral is positive.

I wouldn't say "this must be true because otherwise we would have a non-continuous piecewise function" for a couple of reasons. One of those is that "piecewise" is not a property of functions, but a property of a particular way of defining a function. The other is there is a simple way to say it that connects it with the definition of "continuous". Choose $\varepsilon = f(c)/2.$ Then there exists $\delta>0$ small enough so that for $c-\delta<x<c+\delta$ one has $f(x)> f(c)-\varepsilon > 0.$