I am trying to prove
If $f(x) = 0$ for all irrational $x, f(x) = 1$ for all rational $x$, prove that $f \notin \mathscr{R}$ on $[a, b]$ for any $a<b$.
My attempt:
Suppose $f(x) = 0$ for all irrational $x$ and $f(x) = 1$ for all rational $x$. Let $[a, b]$ be an arbitrary chosen real-valued interval. To show that the Dirichlet function is not Riemann-integrable on $[a, b]$, by the discussion on Page $121$, it suffices to show that $\int_a^{-b} f(x)\,dx \ne \int_a^{-b} f(x)\,dx$, where $\int_a^{-b} f(x)\,dx$ and $\int_{-a}^b f(x)\,dx$ are lower and upper Riemann integrals of $f$ over $[a, b]$, respectively. Note that
\begin{equation}\tag{4.1}
\int_{a}^{-b} f(x)\,dx = \inf U(P, f) \quad and \quad \int_{-a}^b f(x)\,dx = \sup L(P, f)
\end{equation}
where the $\sup$ and the $\inf$ in (4.1) are taken over all partitions $P$ of $[a, b]$. Now, corresponding to any partition $P$ of $[a, b]$, due to the orientation of the Dirichlet function (and the density of rationals in $\mathbb{R}$), we have
\begin{align*}
U(P, f) &= \sum_{i=1}^n M_i \; \Delta x_i = \underbrace{\Delta x_i+\Delta x_i+\dots+\Delta x_i}_{n \textrm{ summands}} = n \Delta x_i \quad \textrm{where } n\ge 1 \\
L(P, f) &= \sum_{i=1}^n m_i \; \Delta x_i = 0
\end{align*}
where $M_i = \sup f(x)$ and $m_i = \inf f(x)$ for $x_{i-1}\le x \le x_i$. Therefore,
\begin{equation*}
\int_{a}^{-b} f(x)\,dx = \inf \{n \Delta x_i, n \Delta x_i, \dots\} \ne \int_{-a}^{b} f(x)\,dx = \sup \{0, 0, \dots\} = 0
\end{equation*}
since $n \Delta x_i \ne 0$ by construction and $n \ge 1$.
Can someone please verify the proof above and suggest how it might be improved? This is my very first proof pertaining to integrals, so it is quite possible that there are some major mistakes in it.
Best Answer
The only mistake is $\sum M_i\Delta x_i=b-a$, not $n\Delta x_i$. (It is $\Delta x_1+\Delta x_2+...+\Delta x_n$). Otherwise your proof is OK.