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 $&gt;0$. 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 $f&gt;0$. Again, got me nowhere.

I should mention that I'm aware that it's possible to prove this by defining "Measure" and all that, but I don't want to go there. I'm wondering if there are more elementary tools to show the above.

Thank you!
Sketch: For a bounded function $f$, we define the oscillation of $f$ at $a$ by $$\mathrm{osc}f(a) = \limsup_{x\to a} f(x) - \liminf_{x\to a} f(x). $$ Then it is easy to show that $f$ is continuous at $a$ if and only if $\mathrm{osc}f(a) = 0$.

Now let us given $\epsilon &gt; 0$ and $\delta &gt; 0$. Then we can find a partition $P$ such that $U(f,P)-L(f,P) &lt; \epsilon$. Thus the set $S = S(\epsilon, \delta)$ denotes the collections of subintervals $I$, formed by the points of $P$, where $\sup_{I} f - \inf_{I} f &gt; \delta$ satisfies

$$ \epsilon &gt; U(f,P)-L(f,P) &gt; \sum_{I\in S} \delta |I| $$

Now, choose $(\epsilon_n, \delta_n)$ such that $\delta_n \downarrow 0$ and

$$ l := \sum_{n=1}^{\infty} \frac{\epsilon_n}{\delta_n} &lt; b-a. $$

Choose $r &gt;1$ such that $rl &lt; b-a$. For each closed interval $I \in S(\epsilon_n, \delta_n)$, we choose an open interval $J$ containing $I$ such that $|J| = r|I|$. Finally, let $\mathcal{U}$ be the family formed by collecting all such open intervals $J$. If $f$ has no point of continuity, then each $x \in [a, b]$ is a point where $f$ has positive oscillation and hence lies in some $J\in \mathcal{U}$. Thus $\mathcal{U}$ is an open cover of $[a,b]$, and hence it has a finite subcover. But the sum of the length of the open intervals in that subcover cannot exceed $rl &lt; b-a$, a contradiction! Therefore $f$ must have a point of continuity.

Of course, this proof also hides the idea of measure, though indirectly and elusively.
What does it mean for it to be Riemann integrable? Why $should$ it be positive?  The Riemann sum is defined as 
$$\sum_{i=0}^{n-1} f(t_i)(x_{i+1}-x_i) : t_i \in (x_{i+1},x_i)$$
over a partition $a=x_0 &lt;x_1 &lt; x_2 &lt; \cdots &lt; x_n = b$

and what is the Riemann integral?  In essence, we let $n\to \infty$ (our partition becomes very fine).  So essentially your question is logically equivalent to asking why the above sum is always positive, well that's because $f(t_i)&gt;0$, so naturally the sum is too... and the integral version just gives us the best solution (but it will always be positive for the same reason).