If $f : [a,b] \to [0,M]$ is Riemann integrable and has integral zero then $f(x) = 0$ at every continuity point $x$ of $f$.
I don't know how to prove this though I have faint idea stated below. Can we use this idea?
Proof:
We are given that function is Riemann integrable and integral is zero.
$$\lim_{mesh \space P \to \space 0} R(f,P,T) = I$$
So,
Given $\epsilon > 0, \exists\delta>0, $ such that for any partition pair
$$mesh \space P < \delta \implies \left| R(f,P,T) – I\right| < \epsilon $$
If number of intervals are n then
$$ \left| \sum_{i=1}^n f(t_i)\Delta x_i \right| < \epsilon$$
Now we can take contradiction that function has value greater than 0 at some continuous points and then use this to contradict : If function is Riemann integrable, then it's set of discontinuities will be zero set.
But I am not able to formulate it. Give me hint. Thanks.
Best Answer
Prove by contradiction. Suppose that there exists $x_{0}\in[a,b]$ such that $f$ is continuous at $x_{0}$ and $f(x_{0})>0$. Let $c=\frac{1}{2}f(x_{0})>0$. Since $f$ is continuous at $x_{0}$ and $f(x_{0})>c$, there exists $\delta>0$ such that $f(x)>c$ for any $x\in[x_{0}-\delta,x_{0}+\delta]$. (We need to adjust the interval $[x_{0}-\delta,x_{0}+\delta]$ accordingly if $x_{0}=a$ or $x_{0}=b$.)
Since $f\geq0$, we have $$ \int_{a}^{b}f(x)\,dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}f(x)\,dx\geq\int_{x_{0}-\delta}^{x_{0}+\delta}c\,dx=2\delta c>0, $$ which is a contradiction.