# Let $f : [a, b] \to [0,\infty)$ be an integrable function…

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.

• What is the use of your first $\epsilon$? Also denoting twice $\epsilon$ for different purposes is confusing. – mathcounterexamples.net Jun 10 at 19:01

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 $$\delta>0$$
$$\int_a^bf(x)dx=\int_a^{c-\delta}f(x)dx+\int_{c-\delta}^{c+\delta}f(x)dx+\int_{c+\delta}^{b}f(x)dx\ge0+2\delta\epsilon+0>0$$
The corner cases $$c=a$$ and $$c=b$$ can be solved with the same idea (one-sided).