Let $f$ be integrable on $[a,b]$. Suppose $f(x) \ge 0$ for all $x \in [a,b]$ and $\displaystyle\lim_{x\to x_0} f(x) = f(x_0)$ with $f(x_0) > 0$ for some $x_0 \in [a,b]$. Prove that $\int_a^b f > 0$.

Since $f$ is continuous at $x_0$, we know that for any given $\varepsilon > 0$, there exists a $\delta > 0$ such that $|f(x) - f(x_0)| < \varepsilon$ whenever $x \in [x_0 - \delta/2,x + \delta/2]$. Therefore, set $\varepsilon = f(x_0)/2 > 0$ and notice that $-f(x_0)/2 < f(x) - f(x_0)$ so $f(x_0)/2 < f(x)$ for $x$ in the aforementioned interval. Thus, $f(x_0)/2$ is a lower bound of the set $\{ f…