# Prove that if $f$ is continuous at $x_0\in [a,b]$ and $f(x_0)\neq 0,$ then $\sup\limits_{P}L(|f|,P)>0$

Can you help me check if this proof is correct? If not, kindly provide a better proof

Prove that if $$f$$ is continuous at $$x_0\in [a,b]$$ and $$f(x_0)\neq 0,$$ then $$\sup\limits_{P}L(|f|,P)>0$$

Suppose $$a. For $$\epsilon=|f(x_0)|/2,$$ there exists $$\delta>0$$ such that $$|f(x)|>\dfrac{1}{2}|f(x_0)|,\;\;\text{whenever}\;\;x\in(x_0-\delta,x_0+\delta).$$ Choose a uniform partition $$P_n$$, for each $$n,$$ such that $$a=x_0 Hence, \begin{align}\sup\limits_{P_n}L(|f|,P_n)&= \lim\limits_{n\to \infty}\sum^{n}_{j=1}|f(t_j)|(x_j-x_{j-1})\\&\geq \dfrac{1}{2}|f(x_0)|\lim\limits_{n\to \infty}\sum^{n}_{j=1}(x_j-x_{j-1})\\&\geq \dfrac{1}{2}|f(x_0)|\lim\limits_{n\to \infty}(x_n-x_{0})\\&= \dfrac{1}{2}|f(x_0)|\lim\limits_{n\to \infty}(b-a)\\&= \dfrac{1}{2}|f(x_0)|(b-a)\\&>0\end{align}

• Please put the statement you're trying to prove in the question itself, not just as the title. Without it, your post is hard to understand. Jan 15 '19 at 22:59
• @Michael Burr: I'll do that! Jan 15 '19 at 23:02

No, it is not correct. That inequality containg the second $$\geqslant$$ doesn't hold; you seem to be assuming here that each $$m_j$$ is greater than or equal to $$\frac12\bigl\lvert f(x_0)\bigr\rvert$$, but that is not true.
Note that you only have to proved an example of one partition $$P_0$$ such that $$L\bigl(\lvert f\rvert,P_0\bigr)>0$$. Then it will follow automatically that $$\displaystyle\sup_PL\bigl(\lvert f\rvert,P\bigr)>0$$.
• Why do you think that $\bigl\lvert f(t_j)\bigr\rvert\geqslant\frac12\bigl\lvert f(x_0)\bigr\rvert$ for each $j$? There is no reason for that. Jan 15 '19 at 23:07
• If you take $a_1,b_1\in[a,b]$ such that $a<a_1<x_0<b_1<b$ and that $(\forall x\in[a_1,b_1]):\bigl\lvert f(x)\bigr\rvert\geqslant\frac12\bigl\lvert f(x_0)\bigr\vert$ and if $P_0=\{a,a_1,b_1,b\}$, then$$L\bigl(\lvert f\rvert,P_0\bigr)=\frac12\bigl\lvert f(x_0)\bigr\vert(b_1-a_1)>0.$$ Jan 15 '19 at 23:12