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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

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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.

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  • $\begingroup$ 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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