Let $f:[0,1] \rightarrow \mathbb{R}$ be Riemann integrable.Prove that, if $\int_{[0,1]}f \neq 0$ then there exists $\delta>0$ & a non-empty open interval $I \subseteq[0,1]$ s.t $|f|\geq \delta$ throughout $I$.
Any hint on how to proceed? How do I construct such an open interval?
On
Here is an elementary proof:
Take any partition $P=\{x_k\}^n_{k=1}$. Then, the Riemann sum $\sum^n_{k=1}f(x_k^*)\Delta x_k$ where $x_k^*$ is an $\textit{arbitrary}$ point in $[x_k,x_{k-1}]$ can always be chosen to equal zero because in every interval there is a point $x$ such that $f(x)=0$.
It follows that $\int^1_0f(x)dx=0$ if it exists.
I will propose an elementary proof that depends only on Riemann sums (although in comments there is more simple and elegant one).
First step is to consider $|f|$. Since $f$ is Riemann integrable then $|f|$ is also Riemann integrable and moreover $0 < |\int\limits_0^1 f(x)dx| \le \int\limits_0^1 |f(x)|dx$. So, $\int\limits_0^1 |f(x)|dx > 0$.
In the second step we prove proposition by contradiction. Assume that for all open intervals $I \subset [0,1]$ and for all $\delta > 0$ there exists $x \in I$ s.t. $|f(x)| < \delta$. Now we consider lower Riemann (Darboux) sum that correspond to a partition $0 = x_0 < x_1 < \dots < x_n = 1$: $$s = \sum\limits_{i = 1}^n (x_i - x_{i-1})\inf\limits_{x \in [x_{i-1}, x_i]}|f(x)|. $$ From our assumption we have that $\inf\limits_{x \in [x_{i-1}, x_i]}|f(x)| = 0$. Therefore $s = 0$. But lower Riemann sums converge to $\int\limits_0^1 |f(x)|dx$ as partition diameter tends to $0$. Therefore $\int\limits_0^1 |f(x)|dx = 0$. That's a contradiction.