Let $f:[a,b]\to\mathbb R$ be Riemann integrable and $f>0$. Prove that $\int_a^bf>0$. (No Measure theory) Is the Riemann integral of a strictly positive function positive?
This is not a duplicate. I'm specifically interested in a proof not involving Measure Theory. The thread above uses the fact that $f$ has to be continuous at some point and as such that its intergral is bounded below by a positive value. 
The question I posted and that was marked as a duplicate is as follows:
Here's a link to it: Let $f:[a,b]\to\mathbb R$ be Riemann integrable and $f>0$. Prove that $\int_a^bf>0$. (Without Measure theory)
"I've been struggling with this for a while, and I have a couple of leads that kind of got me nowhere:
At first I thought that if $f$ is continuous somewhere then the integral will be $>0$. So, if the integral was $0$ then that would mean it would need to be nowhere continuous. That seemed unlikely to me, but I couldn't prove the existence of a point at which it is continuous.
For the integral to be $0$ it would necessitate that for any sub interval of $[a,b]$ the function's infimum would have to be $0$. Also seems weird for $f>0$. Again, got me nowhere.
Thank you!"
If you feel this proof that I'm looking for is impossible to produce, please state why you think so. I have reasons to believe this is feasible (One being as it appeared as a a starred question in my homework assignment when we have only just started learning about integration a few weeks ago).
Edits: Wasn't aware that I could file a claim to re-open a post. I'll keep that in mind.
Also, I feel that the contradiction should come from assuming that $f$ is nowhere continuous, and then somehow show that there exists $x\in[a,b]$ such that $f(x)=0$, contradicting the definition of $f$.
 A: Given $x\in[a,b]$ and $n\in\mathbb{N}$, define 
$$\omega_n(x):=\sup\{f(y):y\in[a,b]\cap[x-\frac{1}{n},x+\frac{1}{n}]\}-\inf\{f(y):y\in[a,b]\cap[x-\frac{1}{n},x+\frac{1}{n}]\}.$$
For every $x\in[a,b]$, $\omega_n(x)$ is non-increasing w.r.t. $n$, so we can define
$$w(x):=\lim_{n\to\infty}w_n(x).$$
It is easy to see that $f$ is continuous at $x$ if and only if $w(x)=0$.
We claim that(in fact, the set below is of full Lebesgue measure in $[a,b]$) 
$$\{x\in[a,b]:\omega(x)=0\} \text{ is nonempty}.$$
If this is proved, then $f$ has some continuous point, and hence from $f>0$ we will see $\int_a^bf(x)dx>0$. 
Let us prove the claim. Denote $O_f:=\sup_{x\in[a,b]}f(x)-\inf_{x\in[a,b]}f(x)$. We may assume that $f$ is not constant, i.e. $O_f>0$. For every $n\in\mathbb{N}$, denote
$$E_n:=\{x\in[a,b]:\omega(x)\le \frac{O_f}{n}\}.$$
If we can find a sequence of decreasing closed intervals $\{I_n\subset[a,b]\}_{n\ge 1}$, such that $E_n\supset I_n$, $\forall n$, then we are done, because
$$\{x\in[a,b]:\omega(x)=0\}=\cap_{n=1}^\infty E_n\supset\cap_{n=1}^\infty I_n\ne\emptyset. $$
Let us construct $I_n\subset[a,b]$ inductively. By definition, $E_1=[a,b]$, so we can choose $I_1=[a,b]$. Once $I_n:=[a_n,b_n]$ is constructed, then since $f$ is Riemann integrable on $I_n$, for every $\epsilon>0$, there exists $m\in\mathbb{N}$, such that for the partition defined by $t_i=a_n+\frac{i}{m}(b_n-a_n)$, $i=0,1,\dots,m$, the difference between upper Darboux sum and lower Darboux sum is
$$\frac{b_n-a_n}{m}\sum_{i=1}^m (\sup_{x\in[t_{i-1},t_i]}f(x)-\inf_{x\in[t_{i-1},t_i]}f(x))\le\epsilon.$$
In particular, there exists $1\le i\le m$, such that
$$\sup_{x\in[t_{i-1},t_i]}f(x)-\inf_{x\in[t_{i-1},t_i]}f(x)\le \frac{\epsilon}{b_n-a_n}.$$
Note that when $x\in(t_{i-1},t_i)$, the left hand side of the inequality above is no less than $\omega(x)$. Therefore, if we choose $\epsilon=\frac{(b_n-a_n)O_f}{n+1}$ and $m$ accordingly, then $\omega(x)\le\frac{O_f}{n+1}$, $\forall x\in(t_{i-1},t_i)$. As a result, if we choose $I_{n+1}$ to be any closed subinterval of $(t_{i-1},t_i)$, then $I_{n+1}\subset E_{n+1}$, which completes the induction. 
