I read a good number of posts which deal with this specific topic, but none of them seem to answer my question. If this is a repost, I apologize.
Alright so here is the problem, followed by a theorem which I use, and then my attempted proof:
Exercise 7.4.4. Show that if $f(x)>0$ for all $x\in [a,b]$ and $f$ is integrable, then $\int_{a}^{b} f>0$.
The theorem (and exercise) is from Abbott's Understanding Analysis, and my proof uses Theorem 7.4.2(ii):
Theorem 7.4.2.(ii) Assume $f$ is integrable on $[a,b]$. If $m\leq f(x)\leq M$ for all $x\in [a,b]$ then $$m(b-a)\leq\int_a^b f(x) \leq M(b-a)$$
Now here is my proof:
Proof. Let $f(x)>0$ for all $x\in [a,b]$ and assume $f$ is integrable. Since $f$ is integrable, then we can let (but can we?)
$$m=\inf\{f(x_0):x_0 \in [a,b]\}\text{ and }M=\sup\{f(x_0):x_0\in [a,b]\}$$
Where both $m$ and $M$ are greater than zero since $f(x)>0$.
Now by Theorem 7.4.2.(ii), we can write $$m(b-a)\leq \int_{a}^{b} f\leq M(b-a)$$
And since $0<m(b-a)\leq M(b-a)$, then the integral must be greater than zero as well.
Q.E.D.
I would just like some verification/correction. My doubt is whether or not the values $m$ and $M$ necessarily have to exist, since Theorem 7.4.2.(ii) states that the inequality is only true IF there exists such values for $m$ and $M$. But since $f$ is integrable and $[a,b]$ is compact, then the function should attain a maximum and a minimum, correct?
That is not correct. Since Riemann integrability implies boundedness, both infimum and supremum exist, but if we let $f(x)=x$ for $x\in(0,1]$ and $f(0)=1$, unfortunately, $\inf_{x\in[0,1]}f(x)=0$ and the infimum is not attainable.
Elementary Way?
Assume the contrary that $\displaystyle\int_{a}^{b}f(x)dx=0$, then find a partition $P=\{x_{0},...,x_{n}\}$ such that $U(f,P)\leq b-a$. It cannot be the case that all $\sup_{x\in[x_{i-1},x_{i}]}f(x)>1$, $i=1,...,n$, so some $i$ is such that $\sup_{x\in[x_{i-1},x_{i}]}f(x)\leq 1$, let us write $\sup_{x\in[a_{1},b_{1}]}f(x)\leq 1$.
Now $\displaystyle\int_{a_{1}}^{b_{1}}f(x)dx=0$. Choose another partition $Q$ in $[a_{1},b_{1}]$ such that $U(f,Q)\leq\dfrac{b_{1}-a_{1}}{2}$, then by the same reasoning we have some subinterval $[a_{2},b_{2}]$ of $[a_{1},b_{1}]$ such that $\sup_{x\in[a_{2},b_{2}]}f(x)\leq\dfrac{1}{2}$.
Proceed in this way we have a decreasing chain $\{[a_{n},b_{n}]\}$ such that $\sup_{x\in[a_{n},b_{n}]}f(x)\leq\dfrac{1}{n}$.
Take $c\in\displaystyle\bigcap_{n=1}^{\infty}[a_{n},b_{n}]$, then $f(c)\leq\dfrac{1}{n}$ for all $n=1,2,...$, now taking $n\rightarrow\infty$, we have $f(c)=0$.