Theorem says that if $f:[a,b]\longrightarrow \mathbb{R}$ is a continuous function, then it's integrable in Riemann's sense. Here's my proof:
Proof
Let $\mathfrak{P}=\{x_0,x_1,…x_n\}$ be a subdivision of $[a,b]$. Let denote $m_k=\inf_{x\in[x_{k-1},x_k]}f(x)$, $M_k=\sup_{x\in[x_{k-1},x_k]}f(x)$ and a diameter as $\max_{k=0,1…,n}(x_{k+1}-x_k)$. Then, the lower Darboux sum is $s(f,\mathfrak{P})=\sum_{k=1}^n m_k(x_k-x_{k-1})$ and upper sum is $S(f,\mathfrak{P})=\sum_{k=1}^n M_k(x_k-x_{k-1})$.
We know that $f\in\mathfrak{R}([a,b])\iff \forall_{\epsilon >0}\exists_{\mathfrak{P}}:S(f,\mathfrak{P})-s(f,\mathfrak{P})<\epsilon$
Every funtion continuous on closed interval is also uniformly continuous, so $$\forall_{\epsilon >0}\exists_{\delta >0}\forall_{x_1,x_2\in [a,b]}:|x_1-x_2|<\delta \Rightarrow |f(x_1)-f(x_2)|<\frac{\epsilon}{b-a}$$
This means that for every subdivision $\mathfrak{P}$ of a diameter smaller than $\delta$ we get $$S(f,\mathfrak{P})-s(f,\mathfrak{P})=\sum_{k=1}^n(M_k-m_k)(x_k-x_{k-1})\leq \frac{\epsilon}{b-a}(b-a)=\epsilon$$ which ends the proof.
Is it understandable and, what's more important, correct?
Best Answer
Yes, it is correct, and very clear. The dirty job is actually proving that if $f$ is a continuous function defined in a closed interval, then $f$ is uniformly continuous. But if you already have this result, then there are no issues.