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 $a<b$. 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 $$a=x_0<x_1<\cdots<x_n=b\;\;\text{and}\;\;x_j-x_{j-1}=(b-a)/n,\;\;j\in \{1,2,\cdots,n\}.$$ 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}

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    $\begingroup$ 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. $\endgroup$ Jan 15 '19 at 22:59
  • $\begingroup$ @Michael Burr: I'll do that! $\endgroup$ 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 $\displaystyle\sup_PL\bigl(\lvert f\rvert,P\bigr)>0$.

  • $\begingroup$ Okay, let me edit it. $\endgroup$ Jan 15 '19 at 23:04
  • $\begingroup$ Please, how about now? $\endgroup$ Jan 15 '19 at 23:05
  • $\begingroup$ 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. $\endgroup$ Jan 15 '19 at 23:07
  • $\begingroup$ So, what should I do? $\endgroup$ Jan 15 '19 at 23:10
  • $\begingroup$ 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$$L\bigl(\lvert f\rvert,P_0\bigr)=\frac12\bigl\lvert f(x_0)\bigr\vert(b_1-a_1)>0.$$ $\endgroup$ Jan 15 '19 at 23:12

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