I need to show that if $f: [0,1] \rightarrow \mathbb{R}$ is continuous, $f(x_0) = c > 0$ for some $x_0 \in [0, 1]$, then there is an interval $I$ containing $x_0$ such that $f(x) > c/2$ for every $x \in I$.
I know that $f$ is Riemann integrable. This is part of a larger question on proving $f(x) = 0$, where $f$ is a continuous function where $f(x) \geq 0$ and the integral of $f$ is $0$.
I am struggling to see the larger point here if it's clear to anyone else. Is a sufficient answer to the question the singleton point ${x_0}$ and as a result, the condition is satisfied. Unless we take the definition of an interval to need more than one point, which in that case, then I'm unsure.
To find $I$ you simply take $\varepsilon=c/2$ and apply the definition of continuity. The point is that, if $f$ is Riemann integrable with $f(x)\geq0$ and $f(x)>M$ on some open interval of length $L$, then you can show that the integral of $f$ is at least $LM$. Hence if $f$ has integral $0$, then such an interval should not exist.
EDIT: After reading your comment again, I realised that $I$ might be taken as $[x_0,x_0]$ which is completely useless. If you specify that $I$ must be open, this makes a lot more sense.