0
$\begingroup$

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

$\endgroup$
  • 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
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.