# Continous Function on [a, b] $\implies$ Riemann Integrable without use of uniform continuity

Is my proof (without using uniform continuity) correct?

$f$ continuous on a compact set $\implies$ $\exists \ M,m, \ f(M)\geq f(x) , \ f(m) \leq(x) \ \forall x \in[a, b]$

$M_i:=\sup\{f(x):t_{i-1}\leq x\leq t_i\}$ and $m_i:=\inf\{f(x):t_{i-1}\leq x\leq t_i\}$

Let $P=\{t_0, \cdots , t_n\}$ be a partition of $[a, b]$ such that $t_i-t_{i-1}<\delta$ and let $\delta < \dfrac{\epsilon}{n(f(M)-f(m))}$

\begin{align}&U(f, P)-L(f, P)\\=& \sum (M_i-m_i)(t_i-t_{i-1}) \\\leq& (f(M)-f(m)) \sum (t_i-t_{i-1}) \\\leq& (f(M)-f(m))\cdot\delta n \\<&\epsilon\end{align}

• Your proof is incorrect, because $\delta$ cannot be dependent on $n$. $\delta$ can only depend on $\epsilon$. – 5xum Apr 13 '18 at 9:11
• Note that $\sum (t_i-t_{i-1}) = b - a$ always holds. – Martin R Apr 13 '18 at 9:11
• @5xum Your comment is a full answer. I suggest that you post it as such. – José Carlos Santos Apr 13 '18 at 9:11
• It might also make you suspicious that you did not use the continuity of $f$ at all, only the boundedness. – Martin R Apr 13 '18 at 9:14

Your estimates on $\delta$ are inconsistent, because the standalone fact that $t_{i+1}-t_i<\delta$ for all $i$ guarantees that $n\cdot\delta>b-a$.