Show that if $f : [a, b] \to (0, \infty)$ is integrable, then $\int_a^b f(x)dx \gt 0$. I can do it in case f is continuous in c:
Let $m= \frac{f}{2}$, there is $\epsilon \gt 0$ such that $f \gt m$ for all  $x \in [c−\epsilon,c+\epsilon]$ by continuity at the point $c$, we can take partitions that contain the points $c−\epsilon$ and $c+\epsilon$, so there is $s$ such that $t_{s−1}=c−\epsilon$, $t_s=c+\epsilon$,
$$m_s = \inf_{f∈[c-\epsilon,c+\epsilon]}f \ge m \gt 0$$
for the smallest is the largest of the lower dimensions, so $$s(f,P) = \sum_{k=1}^{s−1}m_k \Delta t_{k−1} + m_s \Delta t_{s−1}+ \sum_{k=s+1}^{n} m_k \Delta t_{k−1} \ge m(c+ \epsilon - c + \epsilon)=2m\epsilon$$ as $f $ is integrable we have
$$\int_a^bf = \sup s (f,p) \ge s(f,p) \ge 2m\epsilon \gt 0$$
so the integral is positive.
But without this hypothesis I'm kind of lost.
Thank's in advance for any help.
 A: If you are a little familiar with measure theory it is quite easy. Consider the subsets $A_0=f^{-1}([1/2,\infty)), A_m=f^{-1}([\frac{1}{2}^{m+1},\frac{1}{2}^m))\subset[a,b]$. You know that $\bigcup^\infty_{m=0} A_m=[a,b]$. That means that there must be a $\tilde{m}$ for which $A_\tilde{m}$ is not a null-set, so it has Lesbegue measure $\neq0$. Therefore we have that
\begin{equation}
\int_a^b f(x)dx>\mathcal{L}(A_\tilde{m})\frac{1}{2}^{m+1}>0
\end{equation}
A: There is a theorem due to Lebesgue which says that a function $f:[a,b] \rightarrow \mathbb{R}$ is (Riemann) integrable if and only the set of discontinuity points of $f$ has Lebesgue measure zero.
Now let $f:[a,b] \rightarrow (0, \infty)$ be a Riemann integrable function. By the theorem above $f$ has a continuity point $c \in (a,b)$. Therefore, since $f(c)>0$, there exists some $\varepsilon>0$ such that $(c-\varepsilon, c+\varepsilon) \subset (a,b)$ and $f(x)\ge \frac{f(c)}{2}$ for any $x \in (c-\varepsilon, c+\varepsilon)$. Thus
$$\int_a^b f(x) dx \ge \int_{c-\varepsilon}^{c+\varepsilon} f(x) dx \ge 2\cdot\varepsilon\cdot \frac{f(c)}{2} = \varepsilon f(c) >0$$
