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Let $f : [a, b] \to [0,\infty)$ be an integrable function. Prove that if there exists $c \in [a, b]$ such that $f(c) \gt 0$ and $f$ is continuous in $c$, then $\int_a^b f \gt 0$.

My attempt:

There is $\delta \gt 0$ such that $x \in[c−\delta,c+\delta] \to f \gt 0$, by the continuity of, therefore $$ \int_a^bf= \int_a^{c- \delta} f + \int_{c-\delta}^{c+\delta} f + \int_{c+\delta}^b \gt 0$$ 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.

Thank's in advance for any help.

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  • $\begingroup$ What is the use of your first $\epsilon$? Also denoting twice $\epsilon$ for different purposes is confusing. $\endgroup$ – mathcounterexamples.net Jun 10 at 19:01
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If $f(c)>0$ and $f$ continuous in $c$, then from the definition of continuity $f(x)>\epsilon>$ for some $x\in[c-\delta,c+\delta]$, with $\delta>0$

Thus:

$$\int_a^bf(x)dx=\int_a^{c-\delta}f(x)dx+\int_{c-\delta}^{c+\delta}f(x)dx+\int_{c+\delta}^{b}f(x)dx\ge0+2\delta\epsilon+0>0$$

The corner cases $c=a$ and $c=b$ can be solved with the same idea (one-sided).

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Since f is integrable and non-negative, the integral of f must be 0 or greater. Since f(c) > 0 and continuous at c, there exists a small ball centered on c such that f is positive within this ball. Therefore, the integral of f over this ball is positive. Therefore, the integral of f from a to b, which must be greater than the integral of f over this ball, must be positive.

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