1. ## Strict

Let $\displaystyle f: [a,b] \to \mathbb{R}$ be Riemann integrable on $\displaystyle [a,b]$. Suppose that $\displaystyle f(x) \geq 0$ for all $\displaystyle x \in [a,b]$ and that $\displaystyle f$ is continuous on $\displaystyle [a,b]$. If $\displaystyle f>0$ somewhere in $\displaystyle [a,b]$, prove that $\displaystyle \smallint_{a}^{b} f > 0$.

Proof. If we look at the most extreme case in which $\displaystyle f$ is $\displaystyle 0$ everywhere except at $\displaystyle c \in [a,b]$ then $\displaystyle \mathcal{R}(f,P) \geq 0$. E.g. for the "extreme
case" we define $\displaystyle f$ as
$\displaystyle f(x) = \begin{cases} 0 \ \ \ \ \text{if} \ x \neq c \\ L \ \ \ \ \text{if} \ x = c \end{cases}$

where $\displaystyle c \in [a,b]$ and $\displaystyle L$ is a constant. So $\displaystyle \smallint_{a}^{b} f \neq 0$. Thus $\displaystyle \smallint_{a}^{b} > 0$. $\displaystyle \diamond$

Is this correct?

If $\displaystyle f$ were not continuous then the above statement would be false, right?

2. The way you define f doesn't work since f is assumed to be continuous. Incidentally the integral of your function is 0.

The key idea here is that if f is continuous and f(x)>0 for some x in [a,b], then there exists an open interval (c,d) contained in [a,b] such that x is in (c,d) and f(u)>0 for all u in (c,d).

3. Originally Posted by Diamondlance
The way you define f doesn't work since f is assumed to be continuous. Incidentally the integral of your function is 0.

The key idea here is that if f is continuous and f(x)>0 for some x in [a,b], then there exists an open interval (c,d) contained in [a,b] such that x is in (c,d) and f(u)>0 for all u in (c,d).
So suppose $\displaystyle a<c<d<b$. Let $\displaystyle k$ be any non-zero number. Define $\displaystyle f: \mathbb{R} \to \mathbb{R}$ as follows: $\displaystyle f(x) = \begin{cases} 0 \ \ \ \text{if} \ x < c \\ k \ \ \ \text{if} \ c \leq x \leq d \\ 0 \ \ \ \text{if} \ d < x \end{cases}$

4. Your function still isn't continuous. Actually, you don't need to define an explicit f.

There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?

5. Originally Posted by Diamondlance
Your function still isn't continuous. Actually, you don't need to define an explicit f.

There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?
$\displaystyle \int_{c}^{d} f \leq \int_{a}^{b} f$ and $\displaystyle \int_{c}^{d} f > 0$.

6. I think you've got it !

7. The only technical point we haven't addressed is if x=a or x=b. But this isn't too tough to deal with: we still know f is continuous, so if x=a, there exists d in (a,b] such that f(u)>0 on [a,d), and the same argument then applies. The case where x=b is similar.

8. Originally Posted by Diamondlance
The only technical point we haven't addressed is if x=a or x=b. But this isn't too tough to deal with: we still know f is continuous, so if x=a, there exists d in (a,b] such that f(u)>0 on [a,d), and the same argument then applies. The case where x=b is similar.

And if $\displaystyle f$ is not continuous then the statement in post #1 does not hold.

9. Right, and you can use the f you came up with in post 1 to illustrate that.

10. Originally Posted by Diamondlance
Your function still isn't continuous. Actually, you don't need to define an explicit f.

There's an interval (c,d) contained in [a,b] such that f(u)>0 for all u in (c,d). Question: What can you say about the Riemann integral of f from c to d?

This is not enough. You have to come up with a piecewise defined function $\displaystyle g(x)$ such that $\displaystyle f(x) \geq g(x)$ aka post #3. E.g. an open interval $\displaystyle (c,d)$ contained in $\displaystyle [a,b]$ such that $\displaystyle f(u) > z_0/2$.