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Let $f:[a,b]\to\mathbb{R}$ and $L\in\mathbb{R}$

We say that $f$ is Integrable on $[a,b]$ if and only if for every $\varepsilon>0$, there exists $\delta>0$ such that for every partition $\mathcal{\dot{P}}$ satisfying $\|\mathcal{\dot{P}}\|<\delta$, we have that $|S(f,\mathcal{\dot{P}})-L|<\varepsilon$

If $f:[a,b]\to\mathbb{R}$ is Riemann integrable, then $f$ is bounded over $[a,b]$

Proof

Assume if possible that f is unbounded. For every $n\in\mathbb{N}$ divide the interval $[a,b]$ into $n$ parts. Hence, for every $n\in\mathbb{N}$, $f$ is unbounded on at least one of these n parts. Call it $I_n$.

Now, let $\varepsilon>0$ be given. Consider an arbitrary $\delta>0$. Let $\mathcal{\dot{P}}$ be a tagged partition such that $\|\mathcal{\dot{P}}\|<\delta$ and $(I_n,t_n)\in\mathcal{\dot{P}}$, where $t_n$ is taken so as to satisfy $|f(t_n)|>n\varepsilon$. Thus we have that $|S(f,\mathcal{\dot{P}})-L|>\varepsilon$. But as $\delta>0$ is arbitrary, we have a contradiction to the fact that $f$ is Riemann integrable.

Hence, $f$ is bounded.

The problem I have is that I can intuitively see why $|S(f,\mathcal{\dot{P}})-L|>\varepsilon$, but I can't show that this is true; so can anyone please help fill in the details for me, any help is greatly appreciated.

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Assume that $f$ is integrable with integral $L$, and assume at the same time that $f$ is unbounded. There is a $\delta>0$ such that for every $\mathcal{\dot{P}}$ with $\|\mathcal{\dot{P}}\|<\delta$ one has $$|S(f,\mathcal{\dot{P}})-L|<1\ .\tag{1}$$ Consider now a tagged partition $\mathcal{\dot{P}}$ of $[a,b]$ into $N$ equal parts of length $\delta':={b-a\over N}<\delta$. In at least one of these parts, say in $I_k$, the function $f$ is unbounded. Let $\tau$ be the sampling point used in $I_k$. Replacing $\tau$ by another sampling point $\tau'$ where $|f(\tau')|$ is large, and leaving everything else as it is, we obtain a tagged partition $\mathcal{\dot{P}}'$ with $\|\mathcal{\dot{P}}'\|<\delta$, and we can make $$|S(f,\mathcal{\dot{P}}')-S(f,\mathcal{\dot{P}})|=|f(\tau')-f(\tau)|\>\delta'>2\ $$ by a suitable choice of $\tau'$. It follows that $S(f,\mathcal{\dot{P}})$ and $S(f,\mathcal{\dot{P}}')$ cannot both satisfy $(1)$.

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  • $\begingroup$ Thank you I really appreciate your feedback but I am still not sure, on why this works. I know $|f(\tau')|$ can be made as big as we want so what is your lower bound for it for you to obtain your inequality, is it $|f(\tau')|>(L+f(\tau)) / \delta$? And how exactly does your inequality contradict (1) as there no $L$ explicitly in it? I guessing it must be very obvious but I am not seeing it so would appreciate it if you could explain further? Danke. $\endgroup$ – Carsten Mar 5 '15 at 18:46
  • $\begingroup$ @Carsten: See my edit. $\endgroup$ – Christian Blatter Mar 5 '15 at 19:14

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