# Show that Riemann integrable function $f$ on $[a,b]$ must be a bounded function. [duplicate]

I see these two:

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.

## marked as duplicate by Paramanand Singh real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Nov 15 '18 at 3:09

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 $$x_i and $$\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$$ is Riemann integrable, then it is bounded.
Assuming $$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.