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Given $f$ is Rimann integrable on $[a,b]$, and $f(x)$ is positive at $\forall x \in [a,b]$. Using Heine-Borel's finite covering theorem to prove $\int_a^b f(x) \,\mathrm{d} x>0$.

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Consider $\inf _ {x \in [a, b]} f (x)$. Also, put some effort into your questions. –  Karolis Juodelė Nov 4 '12 at 16:28
Consider the Riemann function $f(x)=1/q$, when $x=p/q$; and $f(x)=0$ when $x$ isn't a quotient number. We could find that $g(x)=f(x)+\epsilon$ is Riemann integrable on $[0,1]$ and $\inf_{x\in[a,b]} f(x)=0$, while $f(x)>0$ on $[0,1]$. So, I just don't know what @Karolis Juodelė want to say! –  sdfjldh Nov 5 '12 at 1:24
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1 Answer

I have a proof using Closed interval theorem:

Because $\int_a^b f$ is exists, the upper sum $$\sum M_i\Delta x_i \to \int_a^b f$$ when $\lambda=\max_i\Delta x_i\to0$. If the $\int_a^b f =0$ we have an interval $[a_1,b_1]:=[x_{j},x_{j+1}]$, on it $$\sup_{x\in[a_1,b_1]} f(x)=M_i<1.$$

Due to the $f$ is still integrable on $[x_j,x_{j+1}]$ we could find the second interval $[a_2,b_2]\subsetneqq[a_1,b_1]$, for which $$\sup_{x\in[a_2,b_2]}f(x)<1/2.$$

By induction we have $[a_n,b_n]_{n\in\mathbb N}$ satisfied:

  1. $[a_n,b_n]\subsetneqq[a_{n+1},b_{n+1}]$;
  2. $\sup_{x\in[a_n,b_n]}<1/n$;
  3. $b_n-a_n\to0$.

So $\lim b_n=\lim a_n=x_0$, and $f(x_0)=0$, it's a contradiction!

But, how to prove the proposition by finite covering theorem?

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