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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.

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    $\begingroup$ 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? $\endgroup$
    – Rdrr
    Commented Jun 18, 2021 at 18:24
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    $\begingroup$ 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 $c.$ $\endgroup$
    – zhw.
    Commented Jun 18, 2021 at 18:31
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    $\begingroup$ 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 $\int f >= m(b - a)$. $\endgroup$
    – Hew Wolff
    Commented Jun 18, 2021 at 18:42
  • $\begingroup$ @Rdrr defintiion upper Riemann sum = lower Riemann sum $\endgroup$
    – Andre
    Commented Jun 18, 2021 at 21:32
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    $\begingroup$ 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 $c_i >0$. $\endgroup$
    – Rdrr
    Commented Jun 21, 2021 at 12:52

2 Answers 2

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

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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}

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    $\begingroup$ Are you missing a tilde in the last line exponent? Also, this might be overkill if this is a question in a first Analysis course :) $\endgroup$ Commented Jun 18, 2021 at 18:57
  • $\begingroup$ @BenjaminWang is correct, measure theory is to much for me at the moment. But thanks anyway. $\endgroup$
    – Andre
    Commented Jun 18, 2021 at 21:43
  • $\begingroup$ You are taking for granted that Riemann integrable implies Lebesgue measurable, not to mention the connection between the associated integrals. $\endgroup$ Commented Jun 21, 2021 at 15:07

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