# 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.

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 $$R=0$$

$$\newcommand{\abs}{\left| #1 \right|}$$ $$\newcommand{\norm}{\left|\left| #1 \right|\right|}$$

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*}