Proof that the Riemann integral of the given function is 0 From "An Introduction to Lebesgue Integration and Fourier Series" by Howard J. Wilcox and David L. Myers:

1.1 Definition: A partition $P$ of a closed interval $[a, b]$ is a finite sequence $(x_{0}, x_{1}, \ldots, x_{n})$ such that $a = x_{0} < x_{1} < \ldots < x_{n} = b$. The norm of $P$, denoted $\left|\left|P\right|\right|$, is defined by $\left|\left|P\right|\right| = \max_{1 \leq i \leq n} (x_{i} - x_{i-1})$.
1.2 Definition: Let $P = (x_{0}, \ldots, x_{n})$ be a partition of $[a, b]$, and let $f$ be defined on $[a, b]$. For each $i = 1, \ldots, n$, let $x_{i}*$ be an arbitrary point in the interval $[x_{i-1}, x_{i}]$. Then any sum of the form $R(f, P) = \sum_{i=1}^{n} f(x_{i}*)(x_{i} - x_{i-1})$ is called a Riemann sum of $f$ relative to $P$.
1.3 Definition: A function $f$ is Riemann integrable on $[a, b]$ if there is a real number $R$ such that for any $\epsilon > 0$, there exists a $\delta > 0$ such that for any partition $P$ of $[a, b]$ satisfying $\left|\left|P\right|\right| < \delta$, and for any Riemann sum $R(f, P)$ of $f$ relative to $P$, we have $\left|R(f,P) - R\right| < \epsilon$.

If $R$ exists then $\int_{a}^{b} f(x) dx = R$.

Exercise 5.6: Let $f(x) = 0$ for $x \neq 1/n$, $n = 1,2,3, \ldots$, and let $f(1/n) = 1$. Show that $\int_{0}^{1} f(x) dx = 0$.

Is there a solution to this exercise that only uses the given definitions? I am having trouble finding an equation to relate the size of the norm to the maximum value of the Riemann sum.
 A: Write the integral $\int_{0}^{1} =\int_{0}^{\frac{1}{n}}+\int_{\frac{1}{n}}^{\frac{1}{n-1}} +... +\int_{\frac{1}{2}}^{1}$
Now, $\|P\|=0.5<0.6$ and note that $f(x_i*)=0$ for any interval in the above partition (by definition $f(x)=0$ for all $x\ne \frac{1}{2}, ...,\frac{1}{n}$), which gives that $R(f,P)=0$ and so you can take $R=0$
A: $\newcommand{\abs}[1]{\left| #1 \right|}$
$\newcommand{\norm}[1]{\left|\left| #1 \right|\right|}$
Let $R = 0$.
Let $\epsilon > 0$.
Let $k > 0$ be such that $1/k < \epsilon / 2$.
Let $\delta_{1} > 0$ be such that for any partition $P = (x_{0}, \ldots, x_{n})$ of $[0, 1]$ such that $\norm{P} < \delta_{1}$ there exists an $x \in P$ such that $1/(k+1) < x < 1/k$.
If $P = (x_{0}, \ldots, x_{n})$ is a partition of $[0, 1]$ and $x \in P$ is such that $1/(k+1) < x < 1/k$ then there are at most $2k - 1$ subintervals of $P$ in $[x, 1]$ that contain a number of the form $1/n$. Let $p = 2k - 1$.
Let $\delta_{2} = \epsilon / 2p$.
Let $\delta = \min(\delta_{1}, \delta_{2})$.
Let $P = (x_{0}, \ldots, x_{n})$ be a partition of $[0, 1]$ such that $\norm{P} < \delta$.
For each $i = 1, \ldots, n$, let $x_{i}^{*}$ be an arbitrary point in the interval $[x_{i-1}, x_{i}]$.
Let $R(f,P) = \sum_{i=1}^{n} f(x_{i}^{*})(x_{i} - x_{i-1})$. Then $R(f,P)$ is a Riemann sum of $f$ relative to $P$.
Let $m$ be such that $x_{m} \in P$ and $1/(k+1) < x_{m} < 1/k$.
Then
\begin{align*}
\abs{R(f,P) - R}
&= \abs{\sum_{i=1}^{n} f(x_{i}^{*})(x_{i} - x_{i-1}) - 0} \\
&= \abs{\sum_{i=1}^{n} f(x_{i}^{*})(x_{i} - x_{i-1})} \\
&= \abs{\sum_{i=1}^{m} f(x_{i}^{*})(x_{i} - x_{i-1}) + \sum_{i=m+1}^{n} f(x_{i}^{*})(x_{i} - x_{i-1})} \\
&\leq \abs{\sum_{i=1}^{m} f(x_{i}^{*})(x_{i} - x_{i-1})} + \abs{\sum_{i=m+1}^{n} f(x_{i}^{*})(x_{i} - x_{i-1})} \\
&\leq \sum_{i=1}^{m} \abs{f(x_{i}^{*})(x_{i} - x_{i-1})} + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})(x_{i} - x_{i-1})} \\
&= \sum_{i=1}^{m} \abs{f(x_{i}^{*})} \abs{(x_{i} - x_{i-1})} + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} \abs{(x_{i} - x_{i-1})} \\
&= \sum_{i=1}^{m} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&\leq \sum_{i=1}^{m} 1 \cdot (x_{i} - x_{i-1}) + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&= \sum_{i=1}^{m} (x_{i} - x_{i-1}) + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&= x_{m} - x_{0} + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&= x_{m} - 0 + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&= x_{m} + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&< 1/k + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&< \epsilon / 2 + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} (x_{i} - x_{i-1}) \\
&\leq \epsilon / 2 + \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} \norm{P} \\
&= \epsilon / 2 + \norm{P} \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} \\
&< \epsilon / 2 + \delta \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} \\
&\leq \epsilon / 2 + \delta_{2} \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} \\
&= \epsilon / 2 + \epsilon / 2p \sum_{i=m+1}^{n} \abs{f(x_{i}^{*})} \\
&\leq \epsilon / 2 + \epsilon / 2p \cdot p \\
&= \epsilon / 2 + \epsilon / 2 \\
&= \epsilon.
\end{align*}
