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This question already has an answer here:

I see these two:

  1. How to show that a Riemann integrable function is bounded

  2. If a function $f(x)$ is Riemann integrable on $[a,b]$, is $f(x)$ bounded on $[a,b]$?

The first uses a very different definition of Riemann integrable functions. The second post offers casual intuition, not a formal proof.

The definitions I'm working with:

A Riemann Sum is defined for a partition $\mathcal{P}$ of $[a,b]$ as:

\begin{align*} \mathcal{R}(f, \mathcal{P}) &= \sum\limits_{j=1}^k f(s_j) \Delta_j \\ \end{align*}

The function is Riemann integrable if Riemann sums converge to a number $\ell$ as the mesh sizes of the partitions approach zero. A function $f$ is Riemann integrable if for any $\epsilon > 0$, there must exist some $\delta > 0$ and some partition $\mathcal{P}$ such that:

\begin{align*} m(\mathcal{P}) < \delta &\implies |\mathcal{R}(f, \mathcal{P}) - \ell| < \epsilon \\ \end{align*}

Intuitively, if $f$ is unbounded, it looks like that Riemann sum will not converge, but I can't see how to formally demonstrate that.

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marked as duplicate by Paramanand Singh real-analysis Nov 15 '18 at 3:09

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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How are you choosing the $s_j$? The definition in Stephen Abbott's book allows us to choose the $s_j$ freely so long as $\Delta_j<\delta$.

Let $P=\{x_i\}_1^n$ with $x_i<x_{i+1}$ and $\Delta_i<\delta$ for all $i$. Then if $f$ is bounded on $[x_i,x_{i+1}]$, then it is bounded on $[a,b]$, hence if it is unbounded, then there exists $[x_i,x_{i+1}]$ such that $f$ is unbounded on $[x_i,x_{i+1}]$. Suppose $f$ is unbounded and choose an $i$ such that $f$ is unbounded on $[x_i,x_{i+1}]$. For simplicity we assume it is unbounded above (the case for unbounded below is similar).

Now fix $s_j\in[x_j,x_{j+1}]$ for all $j\not=i$. Then we can make $$\sum_{j=1}^nf(s_j)\Delta_j$$ as large as we wish by varying $s_i$. For if $Q>0$, then choose $s_i$ such that $$f(s_i)>\dfrac{Q-\sum_{\substack{j=1\\j\not=i}}^nf(s_j)\Delta_j}{\Delta_i}.$$ Then $$\sum_{j=1}^nf(s_j)\Delta_j>\sum_{\substack{j=1\\j\not=i}}^nf(s_j)\Delta_j+\dfrac{Q-\sum_{\substack{j=1\\j\not=i}}^nf(s_j)\Delta_j}{\Delta_i}\Delta_i=Q.$$ This shows for any $\delta$ that there exists no real number $A$ such that for every tagged partition $(P,\{s_k\})$ with $\Delta_k<\delta$ we find $\mathcal{R}(f,P)\in(A-\epsilon,A+\epsilon)$. Contrapositively if $f$ is Riemann integrable, then it is bounded.

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Assuming $f$ is not bounded we get that for every $K>0$ and every Partition $\mathcal{P}$ of $[a,b]$ we find $x\in [a,b]$ such that for every $\Delta_j$ of this partition we have $f(x)\cdot \Delta_j>K$ so $f$ is not Riemann integrable.

So we conclude that if $f$ is Riemann integrable, it is bounded.

That should be all you need to show.

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