If $f$ is integrable in $[a,b]\rightarrow\mathbb R$ and $f(x)≥0$ except for a value $c$ such that $f(c)=-1$. Show that the integral of is positive.

Question

Given $f:[a,b] \rightarrow \mathbb{R} $ where $f(x)\geq 0$ for all $x \in [a,b]-\lbrace c\rbrace $ and $c \in (a,b) $ such that $f(c)=-1$. Prove that if $f$ is integrable in $[a,b]$ then $ \int_{a}^{b} f(x) dx \geq 0$ .

It has to be solved with Riemann-integration and it just says that $f$ is integrable.

The thing is that I don't really know how to work with the removable discontinuity in c, even if the function is integrable.

Answer 1

Let $\varepsilon \in (0, \min(c-a, b-c))$, so that $P_\varepsilon = \{a < c-\varepsilon < c+\varepsilon < b\}$ gives a partition of $[a,b]$.

Then the corresponding lower Riemann sum $L(P_\varepsilon, f)$ satisfies $$\begin{align*}\int_a^bf(x)\,dx &\geq L(P_\varepsilon, f) \\ &= (c-\varepsilon-a)\inf f([a,c-\varepsilon])+2\varepsilon\inf f([c-\varepsilon, c+\varepsilon]) + (b-c-\varepsilon)\inf f([c+\varepsilon, b]) \\ &\geq (c-\varepsilon-a)(0) + 2\varepsilon(-1)+(b-c-\varepsilon)(0) \\ &= -2\varepsilon.\end{align*}$$

That is, $\int_a^b f(x)\,dx \geq -2\varepsilon$.

Let $\varepsilon \to 0$ to see $\int_a^b f(x)\,dx \geq 0$.

Answer 2

If $f$ is Riemann integrable on $[a,b]$, and hence on every sub interval of it, you can write the integral on $[a,b]$ as the limit as $\epsilon$ goes to zero of $\int_a^{c-\epsilon}f(x)dx+\int_{c+\epsilon}^b f(x) dx$ Now, let $m$ and $n$ be the infimum of $f$ in both these intervals (and hence real positive numbers, or zero). Then, $\int_a^{c-\epsilon}f(x)dx+\int_{c+\epsilon}^b f(x) dx \geq m(c-a-\epsilon)+n(b-c-\epsilon)$. In the limit as $\epsilon$ goes to zero, $\int_a^b f(x)dx \geq m(c-a)+n(b-c)\geq0$