Let $f:[a,b]\rightarrow\mathbb{R}$. A tagged partition, $\mathcal{P}$ of $[a,b]$ is a set of ordered pairs defined as $$\mathcal{P}:=\{([x_{k−1},x_k]),t_k)\}^n_{k=0},$$ where $a=x_0<...<x_n=b$ and the "tags" $t_k∈[x_{k−1},x_k]$, where $\mathbb{P}_{[a,b]}$ is the set of all tagged partitions over $[a,b]$. $$\|\mathcal{P}\|:=\sup\{x_k-x_{k-1}|1\leq k\leq n\}$$ is the mesh of the partition. The Riemann sum of $f$ over $[a,b]$ w.r.t $\mathcal{P}$ is defined as $$S(f,\mathcal{P}):=\sum\limits^n_{k=1}f(t_k)(x_k−x_{k−1})$$ and $f$ is said to be Riemann integrable with $\int_a^bf=L$ iff $$(\forall\epsilon>0)(\exists\delta>0)(\forall \mathcal{P}\in\mathbb{P}_{[a,b]})\bigg(\|\mathcal{P}\|<\delta\Rightarrow |S(f,\mathcal{P})-L|<\epsilon\bigg)$$

I am having understanding a part of the following proof which says that if $f$ is Riemann integrable on $[a,b]$ then $f$ must be bounded on $[a,b]$. The proof (directly from the text) is as follows:

If $f$ is unbounded. For every $n\in\mathbb{N}$ divide the interval into $n$ parts. Hence, for every $n$, $f$ is unbounded on at least one of these n parts. Call it $I_n$ Now, let $\epsilon>0$ be given. Consider an arbitrary $\delta>0$. Let $\mathcal{P}$ be a tagged partition such that $\|\mathcal{P}\|<\delta$ and $(I_n,t_n)\in\mathcal{P}$, where $t_n$ is taken so as to satisfy $|f(t_n)|>n\epsilon$. Thus we have that $$|S(f,\mathcal{P})−L|>\epsilon\space \space \space\space (\dagger) .$$A contradiction to the fact that f is Riemann integrable.

My trouble is trying to get $(\dagger)$. I know we can choose $|f(t_n)|$ as big as we want, so how do we arrive on the lower bound of $n\epsilon$? I am not sure how split up the $|S(f,\mathcal{P})-L|$ to get the inequality, I was thinking the reverse triangle but get nowhere. So any help will be really appreciated and needed in understanding how to get $(\dagger)$. Also if there is a mistake in the proof please let me know and how to correct it, otherwise if you have another proof please let me know

Thanks in advance


The proof you cite seems dubious.

For an alternative, we can follow your idea of using the reverse triangle inequality to show

$$|S(f,P) - L| = \left|f(t_n)(x_n - x_{n-1}) + \sum_{k \neq n}f(t_k)(x_k - x_{k-1}) - L \right| \\ \geqslant |f(t_n)|(x_n - x_{n-1}) - \left|\sum_{k \neq n}f(t_k)(x_k - x_{k-1}) - L \right|$$

Since $f$ is unbounded on $I_n$, choose $t_n$ such that

$$|f(t_n)| > \frac{\epsilon + \left|\sum_{k \neq n}f(t_k)(x_k - x_{k-1}) - L \right|}{x_n - x_{n-1}},$$

and it follows that

$$|S(f,P) - L| > \epsilon.$$

Thus, when $f$ is unbounded, it is impossible to find $L$ such that for every $\epsilon > 0$ and sufficiently fine partitions, the condition $|S(f,P) - L| < \epsilon$ holds. We can always select the tags so that the inequality is violated.

  • $\begingroup$ I don't think the proof covers the case $a = b$. $\endgroup$ – Björn Lindqvist Jul 28 '18 at 17:50
  • $\begingroup$ @BjörnLindqvist:If you read the question, it is understood that $a < b$ since a reference is made to a partition $a = x_0 < \ldots < x_n = b$. If you allow for a partition to include a single degenerate interval $[a,a]$ which has length $0$, then you can extend the Riemann definition of integral to get $\int_a^a f(x) \, dx = 0$ when $f(a)$ is finite (since lower and upper sums must be $0$). The question is about proving that boundedness is a necessary condition. In the case of $[a,a]$, unboundedness can only mean $f(a) = \pm \infty$ if you allow functions with extended real values. $\endgroup$ – RRL Jul 28 '18 at 18:55
  • $\begingroup$ If we allow $f(a) = \pm \infty$ then either the Riemann integral is undefined or again $0$ if we adopt the convention $0 \cdot \infty = \infty$ as with Lebesgue integrals, and this is not typically done in the Riemann theory. Of course, the Lebesgue integral is always $0$ when taken over a set of measure $0$. $\endgroup$ – RRL Jul 28 '18 at 18:58

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