# Is my solution correct to prove that $f (x) = 0$ for all $x \in [a, b]$? [duplicate]

Possible Duplicate:
$f\geq 0$ and $\int_a^b f=0$ implies $f=0$ everywhere on $[a,b]$

Is this a correct solution?

• It is correct, and the usual way to show this problem. Jan 9, 2013 at 3:40
• Looks correct and pretty nice. Jan 9, 2013 at 3:41
• Was going to ask for the source, but then I realized this was from baby Rudin. Jan 9, 2013 at 3:45
• A nitpick, $\delta > 0$ and $\delta < \min(x_0-a, b-x_0)$ to be completely sure.
– user17762
Jan 9, 2013 at 3:46

Since $\int f = 0$, and $f$ is continuous, we have $\int f = \sup_{\pi \in {\cal P}} L(f,\pi) = 0$, where ${\cal P}$ is the set of partitions of $[a,b]$, and $L(f,\pi)$ is the Riemann sum. Since $[a,b]$ is compact, $f$ is uniformly continuous.
Let $\epsilon>0$. Choose $\delta>0$ so that if $|x-y| < \delta$, then $|f(x)-f(y) | < \epsilon$. Choose a partition $\pi$ with $\text{mesh } \pi < \delta$. Since $f \geq 0$, we have $L(f,\pi) = 0$, and hence on every subinterval $[x,y]$ of $\pi$, we have some $\xi$ such that $f(\xi) = 0$. Uniform continuity shows that $f(t) < \epsilon$ for all $t \in [x,y]$. Hence $f(t) < \epsilon$ for all $t \in [a,b]$. Since $\epsilon>0$ was arbitrary, we are finished.
• Great proof. One (little) thing I can comment on is that the subintervals of $\pi$ don't have to be of the form $[x,y]$ but can also have the form $(x,y]$ or $(x,y)$ etc. But this is no problem since the length of the intervals do not change by removing or adding single points.
• @André: A partition is of the form $\pi=(x_0=a,x_1,...,x_n=b)$ and the intervals formed from the partition are $[x_i, x_{i+1}]$, all of which are closed. Dec 17, 2014 at 16:24