$\textbf{The Problem:}$ Let $f:[a,b]\to\mathbb R$ be a continuous function. Prove that $f$ is Riemann integrable on $[a,b]$.

$\textbf{My Thoughts and Attempt:}$ Since $[a,b]$ is compact, $f$ is also uniformly continuous on it, so there is a $\delta(\varepsilon)>0$ such that for all $\varepsilon>0$ and all $x,y\in[a,b]$ with $\vert x-y\vert\leqslant\delta(\varepsilon)$ it follows that $\vert f(x)-f(y)\vert<\varepsilon.$ So let $\varepsilon>0$ be given and consider the partition $$\mathcal P=\{x_0=a,x_1=a+\delta(\varepsilon), x_2=a+2\delta(\varepsilon),\dots,a+(n-1)\delta(\varepsilon),x_n=b\},$$ where $n\in\mathbb N$ is such that $\vert x_j-x_{j-1}\vert\leqslant\delta(\varepsilon)$ implies $\color{red}{\vert f(x)-f(y)\vert<\displaystyle\frac{\varepsilon}{n}}$. Then the uniform continuity of $f$ guarantees that the values $$M_j=\sup\limits_{x\in[x_{j-1},x_j]}f(x)\quad\text{and}\quad m_j=\inf\limits_{x\in[x_{j-1},x_j]}f(x)$$ are attained and that $\color{red}{M_j-m_j<\displaystyle\frac{\varepsilon}{n}}$. So we see that \begin{align*} \mathcal U(f,\mathcal P)-\mathcal L(f,\mathcal P)&=\sum^{n}_{j=1}(x_{j}-x_{j-1})\cdot(M_j-m_j)\\ &\leqslant\delta(\varepsilon)\cdot\sum^{n}_{j=1}\frac{\varepsilon}{n}\\ &=\delta(\varepsilon)\cdot\varepsilon. \end{align*} And it follows that $f$ is Riemann integrable on $[a,b]$.

$\textbf{Note:}$ The notation $\delta(\varepsilon)$ means that $\delta$ can only depend on $\varepsilon.$

$\textbf{My Concern:}$ My concern is that I am not sure if the statements in $\color{red}{\text{red}}$ are correct. Also, if there are any other mistakes, please feel free to point them out.

Thank you for your time.


No it should be $M_{j} - m_j \leq \varepsilon.$ You need just proceed like following \begin{align*} \mathcal U(f,\mathcal P)-\mathcal L(f,\mathcal P)&=\sum^{n}_{j=1}(x_{j}-x_{j-1})\cdot(M_j-m_j)\\ &\leqslant\varepsilon\sum^{n}_{j=1} (x_{j}-x_{j-1})\\ &= (b-a )\varepsilon. \end{align*}


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