# Thread: Continuous and regulated functions

1. ## Continuous and regulated functions

(a) Let $g:[a,b] \to \mathbb{R}$ be a continuous function. Suppose that $g(x) \geq 0 ~\forall x \in [a,b]$ and $\int_{a}^{b} g = 0$.

Prove that $g(x)=0 \forall x \in [a,b]$.

Intuitively, this statement is obviously true. However, I can't seem to find a starting point for a formal proof. Where to start? Any hints?

(b) Give an example of a regulated function $h:[a,b] \to \mathbb{R}$ such that $h(x) \geq 0 ~ \forall x \in [a,b]$, $\int_{a}^{b} h = 0$, and there is a point $c\in [a,b]$ such that $h(c) \neq 0$.

My problem here is that I don't really understand the difference between continuous and regulated functions. I do know the formal definition of a regulated function, and the fact that a continuous function is regulated but do not entirely grasp the concept.
Also, it seems rather counterintuitive to assert that there would be a point $c$ such that $h(c) \neq 0$.

2. Originally Posted by nmatthies1
(a) Let $g:[a,b] \to \mathbb{R}$ be a continuous function. Suppose that $g(x) \geq 0 ~\forall x \in [a,b]$ and $\int_{a}^{b} g = 0$.

Prove that $g(x)=0 \forall x \in [a,b]$.

Intuitively, this statement is obviously true. However, I can't seem to find a starting point for a formal proof. Where to start? Any hints?

(b) Give an example of a regulated function $h:[a,b] \to \mathbb{R}$ such that $h(x) \geq 0 ~ \forall x \in [a,b]$, $\int_{a}^{b} h = 0$, and there is a point $c\in [a,b]$ such that $h(c) \neq 0$.

My problem here is that I don't really understand the difference between continuous and regulated functions. I do know the formal definition of a regulated function, and the fact that a continuous function is regulated but do not entirely grasp the concept.
Also, it seems rather counterintuitive to assert that there would be a point $c$ such that $h(c) \neq 0$.
For (a), proceed by contradiction and assume that there exists a point $c\in[a,b]$ s.t. $g(c)>0$. Then use the continuity of $g$ to show that there must exist $\delta>0$ s.t. $\forall x\in(c-\delta,c+\delta)$, $g(x)>0$, and derive a contradiction from that.

For (b), what about something as simple as:

$h(x)\left\{\begin{array}{lr}1:&x=(a+b)/2\\0:&else\end{array}\right\}$

The left and right limits always exist, so it's regulated, and certainly has an integral of 0.