Let there be an integrable, non-negative function $f$ in a range $[a,b]$. If the integral $\int_a^b f(x) \, dx$ equals $0$, prove that $f(x)=0$ for every $x$ for which $f$ is continuous.
I have already tried to solve this problem, but i'd like some usefull input.
My solution:
Let there be a number $c$ in the range $[a,b]$, for which $f(c)>0$.
$f$ is continuous, so there should be a positive $d>0$ near $c$, so that $f(x)>0$ for every $x$ within $(c-d,c+d)$.
Now let there be a partition $P=\{x_{0},x_{1},....,x_{n}\}$ of the interval $[c-d/2 , c+d/2]$
with $x_{0}=c-d/2$ and $x_{n}=c+d/2$.
Let be $m_{i}=\inf \{ f(x): x_{i-1}\leq x\leq x_{i}\}$ for $i=1,2,...,n$.
Therefore, $$L_{f,P}=\sum_{i=1}^{n} m_{i}(x_{i}-x_{i-1})>0\text{.}$$
However, we know that $\int_a^b f(x) \, dx=L_{f,P}=U_{f,P}=0$ and we have a contradiction.
As such, there is no $c$ in $[a,b]$ so that $f(c)>0$, and so $f(x)=0$ for every $x \in [a,b]$.
Was my assumption of using the interval $(c-d,c+d)$ correct?