# 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.

• Depends on your definition of integrable. Are you using the defintiion upper Riemann sum = lower Riemann sum, the definition that uses partitions or something a bit more measure theoretic?
– Rdrr
Commented Jun 18, 2021 at 18:24
• But Lebesgue's criterion says that if $f$ is Riemann integrable, then $f$ is continuous a.e. Thus you have the existence of your continuity point $c.$
– zhw.
Commented Jun 18, 2021 at 18:31
• By the way, if $f$ is continuous, there's an easier argument: since $[a, b]$ is compact, $f$ attains a minimum value $m$. Clearly $m > 0$ and $\int f >= m(b - a)$. Commented Jun 18, 2021 at 18:42
• @Rdrr defintiion upper Riemann sum = lower Riemann sum Commented Jun 18, 2021 at 21:32
• You can simply argue then that both the upper and lower Riemann sums are positive. This should be somewhat clear since they are both of the form $\sum_{i=1}^n c_i(x_i-x_{i-1})$, where $c_i=f(y_i)$ for some point $y_i \in [x_{i-1},x_i]$, and $c_i >0$.
– Rdrr
Commented Jun 21, 2021 at 12:52

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 $$\int_a^b f(x) dx \ge \int_{c-\varepsilon}^{c+\varepsilon} f(x) dx \ge 2\cdot\varepsilon\cdot \frac{f(c)}{2} = \varepsilon f(c) >0$$
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 $$$$\int_a^b f(x)dx>\mathcal{L}(A_\tilde{m})\frac{1}{2}^{m+1}>0$$$$