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 $\int_a^b f(x) \,\mathrm{d} x>0$.
Prove \int_a^b f(x)\,\mathrm dx>0 if f(x)>0 at $\forall x\in [a,b]$ and Riemann integrable (via finite covering theorem).
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calculus
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0Consider 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! – 2012-11-05
1 Answers
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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:
- $[a_n,b_n]\subsetneqq[a_{n+1},b_{n+1}]$;
- $\sup_{x\in[a_n,b_n]}<1/n$;
- $b_n-a_n\to0$.
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?