Here's what I'm doing:

Prove that if $f: [a,b] \to [0,+\infty)$ is a continuous nonnegative function with $\int_{a}^{b} f(x) \ dx = 0$, then $f(x) = 0$ for all $x \in [a,b]$.

Proof Attempt:

Suppose that $f(x) \neq 0$ for some $x \in [a,b]$. Let $c$ be one of those points in $[a,b]$ where $f(c) \neq 0$. Let $\epsilon > 0$ be given. Since $f$ is Riemann Integrable, there exists a $\delta > 0$ such that for all partitions $P$ with partition norm $|P| < \delta$ and all associated evaluation sets $T$:

$$|R(f,P,T)-0| = |R(f,P,T)| < \epsilon$$

where $R(f,P,T)$ is shorthand for the Riemann Sum taken over some partition and evaluation set. Now, we know that $R(f,P,T) \geq 0$ and, in fact, if $c \in T$, then $R(f,P,T) > 0$.

So, then, pick $T$ above so that $c \in T$ and let $\epsilon = R(f,P,T)$. Then, we get:

$$R(f,P,T) < R(f,P,T)$$

which is impossible. Hence, $f(x) = 0$ for all $x \in [a,b]$. $\Box$

Does the proof above work? If it doesn't, then why? How can I fix it?


An usual and simple way would be to say, like you did, if there exists $c\in [a,b]$ such that $f(c)>0$ than by continuity, given $\epsilon =f(c)/2$, there exists $\delta >0$ for which for every $x\in (c-\delta,c+\delta)$, $f(x)\ge \frac{f(x)}{2}$. So we have this: $$\int_a^bf(x)dx \ge \int_{c-\delta}^{c+\delta}f(x)dx\ge \int_{c-\delta}^{c+\delta}\frac{f(c)}{2}dx=\frac{f(c)}{2}2\delta>0\mbox,$$ which goes against our hipoyhesys

  • $\begingroup$ Ah so my argument wasn't correct? Or was it just too complicated? $\endgroup$
    – Abhi
    Nov 28 '20 at 12:08
  • $\begingroup$ For what i followed it made sense to me, but you made it way to complicated for me $\endgroup$
    – Measure me
    Nov 28 '20 at 12:09
  • $\begingroup$ Ah okay I see. Your proof sounds a lot more clever than mine, thank you very much! $\endgroup$
    – Abhi
    Nov 28 '20 at 12:10

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