We can use the following:
Lemma. If $f$ is continuous on $[a,b]$ then for any $\epsilon > 0$ there exists $\delta > 0$ such that for any partition $P$ of $[a,b]$ with $\|P\| < \delta$ and any refinement $R$ of $P$, we have $|S(f,P) - S(f,R)| < \epsilon$.
Applying the lemma, we can show that if $f$ is continuous, then the Cauchy criterion is satisfied. That is, for any $\epsilon >0$ there exists $\delta > 0$ such that if $P$ and $Q$ are any partitions satisfying $\|P\|, \|Q\| < \delta$, then $|S(f,P) - S(f,Q)| < \epsilon.$
To see this, let $R = P \cup Q$ be a common refinement and take $\delta$ as specified in the lemma such that if $\|P\|, \|Q\| < \delta$, we have $|S(f,P) - S(f,R)| < \epsilon/2$ and $|S(f,Q) - S(f,R)| < \epsilon/2$. Whence, it follows that
$$|S(f,P) - S(f,Q)| \leqslant |S(f,P) - S(f,R)| + |S(f,Q) - S(f,R)| < \epsilon/2 + \epsilon/2 = \epsilon.$$
It remains to prove the lemma.
Since $[a,b]$ is compact, $f$ is uniformly continuous and for any $\epsilon > 0$ there exists $\delta >0$ such that if $|x - y| < \delta$, then $|f(x) - f(y)| < \epsilon/(b-a)$. Suppose $\|P\| < \delta$ and $R$ is a refinement of $P$. Any subinterval $[x_{j-1}, x_{j}]$ of $P$ can be decomposed as the union of subintervals of $R$,
$$[x_{j-1},x_{j}] = \bigcup_{k=1}^{n_j}[y_{j,k-1}, y_{j,k}],$$
and
$$\begin{align}\left|S(f,P) - S(f,R)\right| &= \left|\sum_{j=1}^n f(\xi_j)(x_j - x_{j-1}) - \sum_{j=1}^n \sum_{k=1}^{n_j}f(\eta_{j,k})(y_{j,k} - y_{j,k-1})\right| \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}|f(\xi_j) - f(\eta_{j,k})|(y_{j,k} - y_{j,k-1}) \\ &\leqslant \sum_{j=1}^n \sum_{k=1}^{n_j}\frac{\epsilon}{b-a}(y_{j,k} - y_{j,k-1}) \\ &= \epsilon\end{align}$$
I want to show using ONLY this definition that The Dirichlet function $f : [0, 1] → \mathbb R$, defined by
Best Answer
I believe this is the original definition of the Riemann integral. To show that the Dirichlet function is not Riemann integrable according to this definition, we have to show that $\forall B\in\mathbb{R}, \exists \epsilon >0 : \forall \delta >0, \exists P\in\mathcal{P} : |P|<\delta$ and there exists a Riemann sum $S(P,f)$ such that $|S(P,f)-B|\ge \epsilon$ where $\mathcal{P}$ is the set of all partitions of $[0, 1]$.
For simplicity, let us just consider the case where $B\in [0, 1]$ since $0\le f\le 1$. Let $\epsilon=1/2$. Let $\delta>0$ be arbitrary, $n\in\mathbb{N}$ be large enough such that $1/n<\delta$ and let $P=(0, 1/n, 2/n, \dots, (n-1)/n, 1)$. By construction, $|P|=1/n<\delta$ and $S(P,f)\in \{0, 1/n, 2/n, \dots, (n-1)/n, 1\}$ for any choice of Riemann sum $S(P,f)$ because the intermediate points can either be rational or irrational. Now, for some $k=0, 1,\dots, n$, $|B-k/n|\ge 1/2$ and so on $[(j-1)/n, j/n]$, we can pick the intermediate point $\xi_j$ to be rational if $j\le k$ and irrational if $j>k$. This way, $S(P,f)=k/n$ and $$|S(P,f)-B|=|B-k/n|\ge 1/2$$