Suppose $f$ is Riemann integrable over $[a,b]$. The proof I am reading that shows that $f$ must be bounded, just says if $f$ is unbounded then there is some point $x$, such that for any $n\in\mathbb{N}$ and $\epsilon>0$ if we take any partition divded into $n$ parts we have $|f(x)|>n\epsilon$ and so if we take $x$ as a tag in the partition then $$|S(f,\mathcal{P})-L|>\epsilon\space\space\space\space\space\space (\dagger),$$ where $S(f,\mathcal{P})$ is the Riemann sum of $f$ with respect to the partition $\mathcal{P}$ and $L=\int_a^bf$.

My trouble is trying to get $(\dagger)$, I am not sure how split up the Riemann sum to get the inequality, I was thinking the reverse triangle but get nowhere. So any help will be really appreciated and needed.

Thanks in advance

  • 2
    $\begingroup$ This is direct from the definition (one could even say that boundedness is a part of the definition of Riemann integrability). To wit, if $f$ is Riemann integrable, then $f$ is bounded from above and from below by some step functions. Every step function is bounded hence $f$ is bounded. The "proof" in your post seems unnecessarily convoluted (in addition, it is frankly suspect). $\endgroup$ – Did Apr 5 '17 at 12:23

You have not told us what your working definition of Riemann integrability is.

If integrability is defined in terms of lower and upper sums, and $f$ is unbounded above, then every single upper sum is undefined (or $=\infty$, if you prefer). Basta; no further epsilontics needed.

If integrability is defined in terms of quantities like $\|\Delta f\|_{I_j}:=\sup_{x, \ y\in I_j}|f(x)-f(y)|$ then again for every partition at least one of these quantities is $=\infty$, hence you can never make $\sum_j\|\Delta f\|_{I_j}\>|I_j|<\epsilon$. But this is required before you can even think of computing "general Riemann sums" $\sum_{j=1}^N f(\tau_j)\>|I_j|$ as approximations to the intended integral.

  • $\begingroup$ My suspicion is that a 'tagged partition' is a sequence $a = t_0< t_1 < ... < t_n = b$ together with values $x_i \in (t_{i-1}, t_i)$, and there's some notion of the sum associated to a tagged partition, a partial order on tagged partitions, and a limit as the mesh of the partition goes to zero. (If this is right, I think it's probably pedagogically a bad way to go...even proving that $f(x) = x$ is integrable on $[0, 1]$ could take a week of class time!) $\endgroup$ – John Hughes Apr 5 '17 at 14:36
  • $\begingroup$ Yes, that is right, we do however prove that this "classical" definition is equivalent to the Darboux definition where $f$ is taken to be bounded, I do have a proof (not proof by contradiction) that $f$ must be bounded in E.D. Bloch's Real numbers and real analysis bit it is a bit long winded so I found this proof which seems much shorter but my trouble is showing $(\dagger)$. So this proof is from the Real Analysis wikibook (en.wikibooks.org/wiki/Real_Analysis/Riemann_integration) so I apologise for not being clear in my 1st post. $\endgroup$ – user152874 Apr 5 '17 at 15:10
  • $\begingroup$ So any help with showing $(\dagger)$ will really be appreciated. I am guessing it is probably quiet simple and easier statement are given more detail in the text. But I just can't see it. Thanks again $\endgroup$ – user152874 Apr 5 '17 at 15:12

This seems wrong to me. I might be misunderstanding the use of "tag", however.

Look at $$ f(x) = \frac{1}{x-1} + \frac{1}{x+1} $$ on the interval $(-1, 1)$.

The points $+1$ and $-1$ are examples of the point $x$ with the necessary property (unboundedness near $x$). But if you take the $k$-interval equipartition with its $k+1$ breakpoints evenly spaced from $-1$ to $1$, and pick $k$ equispaced points between $-1$ and $1$ as the "samples" or "tags", you find that $S(f, P) = 0$. That's at least consistent with $L = 0$ (even through the function is not, in fact, Riemann integrable.

In short: the claim that unboundedness messes things up is correct. The alleged proof of this claim appears wrong.

  • $\begingroup$ This the proof from the text: If $f$ is unbounded. For every $n$ 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$. A contradiction to the fact that $f$ is Riemann integrable. So what are the details to show $(\dagger)$? Thanks $\endgroup$ – user152874 Apr 5 '17 at 13:11
  • $\begingroup$ What does $\|P\|$ denote? The "mesh" of $P$, i.e., the length of the largest interval in $P$? And what is $L$? It seems to be unbound in this quotation at least. Is it the purported value of the integral? The sentence "Thus we have that ..." seems to hide a great deal (and continuing to look at my example, I don't actually believe it's true). $\endgroup$ – John Hughes Apr 5 '17 at 13:51
  • $\begingroup$ For $f$ unbounded, what's the definition of the Upper Sum of $f$ with respect to a partition $P$? In my favorite definition, we let $M_i$ be the max of $f$ on the $i$th subinterval of $P$, and the upper sum involves $M_i (t_{i+1} - t_i)$, but that only makes sense because the definition restricts to bounded functions at the start. $\endgroup$ – John Hughes Apr 5 '17 at 14:01
  • $\begingroup$ A tagged partition of $[a,b]$ is defined as $\{([x_{k-1},x_k]),t_k)\}_0^n$, where $a=x_0<...<x_n=b$ and the "tags" $t_k\in[x_{k-1},x_k]$, with $\mathbb{P}_{[a,b]}$ is the set of all tagged partitions over $[a,b]$. Yes $\|\mathcal{P}\|$ is the mesh of the partition. The Riemann sum of $f$ over $[a,b]$ w.r.t $\mathcal{P}$ is defined as $S(f,\mathcal{P}):=\sum_1^n f(t_k)(x_k-x_{k-1})$ an $f$ is Riemann integrable with $\int_a^bf=L$ iff $$(\forall\epsilon>0)(\exists\delta>0)(\forall \mathcal{P}\in\mathbb{P}_{[a,b]})(\|\mathcal{P}\|<\delta\Rightarrow |S(f,\mathcal{P})-L|<\epsilon)$$ $\endgroup$ – user152874 Apr 5 '17 at 15:23
  • $\begingroup$ In that case, the proof is certainly wrong, for $I_n$ has length $(b-a)/n$, and for $\delta = (b-a)/(2n)$, it's impossible for $\|P\| = \delta$ and for $I_n \in P$. I'm growing to have grave doubts about your text... $\endgroup$ – John Hughes Apr 5 '17 at 16:30

In comments, you wrote the following (which I've slightly polished):

A tagged partition of $[a,b]$ is a sequence
$\{([x_{k−1},x_{k}]),t_k)\}_{k=0}^n$, where $a=x_0<...<x_n=b$ and the "tags" $t_k$ satisfy $$ t_k\in [x_{k−1},x_k] $$ for $k = 1, \ldots n$.

$ \Bbb P[a,b]$ denotes the set of all tagged partitions over $[a,b]$.

$\|P\| = \max_{k=1}^n (x_k - x_{k-1})$ is called the mesh of the partition.

The Riemann sum of $f$ over $[a,b]$ w.r.t a partition $P$ is defined as $$ S(f,P):=\sum_1^n f(t_k)(x_k−x_{k−1}). $$

Finally, $f$ is Riemann integrable with $\int_a^b f = L$ iff $$ \forall \epsilon > 0~ \exists \delta > 0 ~\forall P \in \Bbb P[a,b], \|P\| < \delta \implies |S(f,P)−L|<\epsilon. $$

You want to prove that if $f$ is unbounded, $f$ is not integrable.

Proof: By contradiction: Suppose $f$ is unbounded, but that it's also integrable with integral $L$.

Pick $\epsilon = 1$. Then there's a $\delta$ with the property that for any tagged partition $P$ with mesh less than $\delta$, $ |S(f, P) - L | < 1$. Let $P$, with split points $x_i$ and tags $t_i$, be such a partition** . Let $n$ be the number of tags in $P$.

I'll exhibit a different partition, $Q$ with the same mesh that does not have $$ |S(f, Q) - L| < 1. $$

Now $f$ must be unbounded on interval $[x_{j-1}, x_j]$ for some $j$. (If not, it's bounded on a finite collection of intervals, hence bounded.) Note that $j$ here is the index of just one interval. (I'll use "$k$" for a generic interval index.)

Therefore, for any $M$, there's a point $s \in [x_{j-1}, x_j]$ with $|f(s)| > M$.

I'll choose $$ M = f(t_j) + \frac{10}{x_j - x_{j-1}}. $$ Let $s_j$ be the point with $|f(s_j)| > M$.

Now, let $Q$ be the partition with the same split-points as $P$, and the same tags as $P$, except replace the single tag $t_j$ with $s_j$. Let's compare $S(f, Q)$ to $S(f, P)$: $$ S(f, P) - S(f, Q) = \sum_{k=1}^n f(t_k)(x_k−x_{k−1}) - \left[ \sum_{k=1}^{j-1} f(t_k)(x_k−x_{k−1}) + f(s_j) (x_j - x_{j-1}) \sum_{k = j+1}^n f(t_k)(x_k−x_{k−1}) \right] $$ All terms here cancel except for the $j$th, so \begin{align} |S(f, Q) - S(f, P) | &= | f(t_j)(x_k−x_{k−1}) - f(s_j) (x_j - x_{j-1}) |\\ &= |(f(t_j) - f(s_j))(x_j−x_{j−1}) | \\ &> |(\frac{10}{x_j - x_{j-1}})(x_j−x_{j−1}) | \\ &= 10 \end{align} Now since $|S(f, P) - L | < 1$ but $|S(f, Q) - S(f, P)| > 10$, the triangle inequality tells us that $|S(f, Q) - L | > 9$, and hence is not less than one.

That's a contradiction.

  • $\begingroup$ Thanks, this really helps Also I have already re-posted it following your suggestion so plesae ignore that as it will be a duplicate and I will delete it shortly. $\endgroup$ – user152874 Apr 5 '17 at 17:50

You never shared with us your definition of the Riemann integral. In Rudin, for example, boundedness of $f$ on $[a,b]$ is part of the definition.

I suppose you are letting $f:[a,b]\to \mathbb R,$ with no boundedness assumed, and are using the $\epsilon-\delta$ definition with Riemann sums. So $\int_a^b f =L$ means for every $\epsilon>0$ there is $\delta >0$ such that if $P$ is a partition with mesh size $<\delta,$ then

$$\tag 1 |\sum_{P}f(c_k)\Delta x_k -L| <\epsilon$$

for all choices of $c_k \in I_k.$ (Here the $I_k$ are the subintervals induced by $P,$ and $\Delta x_k$ is the length of $I_k.$)

Suppose $f$ is unbounded on $[a,b].$ WLOG $f$ is not bounded above. Take any partition $P$ of $[a,b].$ Then $f$ is not bounded above on some $I_{k_0}.$ This implies there is a sequence of points $d_m\in I_{k_0}$ such that $f(d_m) \to \infty.$ In the subintervals $I_k, k\ne k_0,$ take $c_k$ to be the left end point of $I_k.$ We'll keep the $c_k$'s fixed while letting $d_m$ vary. We get

$$\tag 2 \lim_{m\to \infty} \left (\sum_{k\ne k_0} f(c_k)\Delta x_k + f(d_m)\Delta x_{k_0}\right ) = \infty.$$

Now all Riemann sums in $(2)$ take place with respect to the partition $P.$ And this can be done for any $P.$ This destroys any hope that something like $(1)$ could hold.


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