Let $I$ b a generalized rectangle in $\Bbb R^n$

Suppose the bounded function $f:I\to \Bbb R$ assumes the value $0$ except at a single point $x \in I$

Show that $f$ is integrable and $\int_{I} f=0$

I think that

Let $\epsilon >0$

Let's choose a partition $$\vert \vert P\vert\vert ^n< \frac{\epsilon}{2\vert f(x)\vert} $$

and x contained in exactly one generalized rectangle in $P$.

I need to calculate $L(f,P)$ and $U(f,P)$.

How to calculate $L$ and $U$? Please help me for the evaluation. Thank you.

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    $\begingroup$ Thank you for starting a bounty. :) $\endgroup$ – B11b Jul 30 '13 at 13:45
  • $\begingroup$ If you're dissatisfied with the provided answer, you should unaccept it, unless there's a time limit I don't know about and that's not possible. $\endgroup$ – Kevin Arlin Jul 30 '13 at 13:49
  • $\begingroup$ Question: Why wouldn't you use the lebesgue integral? That would make it a trivial task as $\lambda^n(\{x\}) = 0\ \forall\ x\in \mathbb{R}^n$ $\endgroup$ – AlexR Aug 5 '13 at 20:16

Assume that $f\geq 0$; if not, since it is only non-zero at $x$, you can replace $f$ by $-f$.

Then every (generalized) rectangle in your partition that does not contain $x$ has both upper and lower sum 0. For the rectangle that contains $x$, note that all $y$ in the rectangle satisfy $$ 0\leq f(y)\leq f(x), $$ and so $x$ must actually be maximum on the rectangle. Hence the lower sum is 0, and the upper sum is $f(x)\cdot V_P$, where $V_P$ is the volume of the box containing $x$.

Hence for any partition $P$ in which $x$ lies in only one box, we have $$ L(f,P)=0\qquad\text{and}\qquad U(f,P)=f(x)\cdot V_{P}. $$

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    $\begingroup$ Well, thank you:) but what is the value of $V_p$? $\endgroup$ – Bstr Jul 27 '13 at 13:39
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    $\begingroup$ All that really matters is that it tends to $0$ as your partition gets finer and finer. So, you can certainly make it smaller than $\frac{\epsilon}{\lvert f(x)\rvert}$, as you suggested. $\endgroup$ – Nick Peterson Jul 27 '13 at 13:47

Since $f$ assumes the value $0$ except that at a single point $x\in I$, $f(x)$ is non-zero.

And this point $x$ falls in one generalized rectangle in the partition.

The fineness of partition is important because the size of the rectangle has measure controlled by this.

Then, In order to keep the fineness of partition, let choose small enough volume $V_p$ of the relevant part, which contains of $x$.

And let's assume that $f$ is positive.

Then, the lower sum is zero since the value of $f$ is zero except at $x$.

i.e $L(f,P)=0$

But, the upper sum has just one nonzero contribution. And the contribution is at most $f(x)\cdot V_p$

So, $U(f,P)=f(x)\cdot V_p$

Let choose $\epsilon >0$ such that $V_p< \frac{\epsilon}{f(x)}$

Then, $$U(f,P)-L(f,P)=f(x) V_p -0< f(x) \frac{\epsilon}{f(x)}=\epsilon$$

Thus, f is integrable.

As for $\int_I f$,

as the partition $P$ becomes finer and the upper and lower sums converge to zero, the integral of $f$ is zero.

Note that this answer is the combination of N. R. Peterson's answer and M. Bennet's good and wide explanation.


This is a commentary on what @NicholasR.Peterson has written, in the hope that it will help others to understand the steps of the proof. However it is not a proper answer, it was just getting unwieldy for a comment.

What is happening here is that as the partition gets finer and finer, the sum over most of $I$ is zero - the upper and lower sums there are easy.

$f(x)$ is non-zero at just one point $x$. And this point falls in one [generalised] "rectangle" in the partition. Here we can't do anything about $f(x)$, which is fixed, but we can make the rectangles in the partition as small as we like. The size of the rectangle has measure controlled by the fineness of the partition. So if the volume of the relevant part is at most $V_P$ (the fineness of the partition), there will be a contribution of at most $f(x)V_P$ to the upper or lower sum.

There is a technical issue about the sign of $f(x)$ which will affect whether it contributes to the upper sum or the lower sum. Nicholas has fixed $f(x)$ to be positive so it affects the upper sum. The lower sum for this part is still zero, because the value of $f$ at points other than $x$ is zero.

The lower sum is made up of lots of zeros. The upper sum has just one non-zero contribution. We just need to show that we can make that contribution as small as we like - less than any small positive number $\epsilon$ - and with $f(x)$ fixed we do that by choosing $V_P$ to be small enough.

So in the limit, as the partition becomes finer, upper and lower sums converge to zero, the integral is defined, and is equal to zero.

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    $\begingroup$ Thank you so much, I understand Peterson's answer thanks to you. I am grateful of you. You are a so kind and helpful person. Thanks a lot! Dear Mark Bennet. By the way, I wrote what I understand Peterson's answer with the help of your wide explanation. Is this my writing enough and satisfactory to prove this question? $\endgroup$ – B11b Jul 30 '13 at 16:17
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    $\begingroup$ I wish everyone wrote an answer explanatorily just like yours :) $\endgroup$ – B11b Jul 30 '13 at 16:22
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    $\begingroup$ This is worth understanding, because there are some important proofs in measure theory involving functions which are well-behaved most of the time. The trick is then to fit the badly behaved parts into small enough boxes to get the proof to go through. $\endgroup$ – Mark Bennet Jul 30 '13 at 17:14

We know that $f$ is integrable in $I$ if and only if for every $\epsilon>0$ there exist a partition $P$ such that:

$$ |L(f,P)-U(f,P)|<\epsilon $$

now, recall that a partition $P$ is a set of non-overlapping generalized rectangles whose union is I. From the definition: $$ L(f,P)=\sum_{I_n\in P}\inf\{f(x):x\in I_n\}vol(I_n)\\ U(f,P)=\sum_{I_n\in P}\sup\{f(x):x\in I_n\}vol(I_n) $$ where $vol(I)$ is the volume of a particular generalized rectangle in $\mathbb{R}^n$: $$ I=[a_1,b_1]\times[a_2,b_2]\times...\times[a_n,b_n]\\ vol(I)=(b_1-a_1)(b_2-a_2)...(b_n-a_n) $$ in this particular lets fix $\epsilon>0$, and pick a partition such that: $$ ||P||<\frac{\epsilon}{|f(x)|}, $$ (here I am supposing that $||P||=\max\{vol(I_n):I_n\in P\}$, if you use $||P||$ as the max length of the unidimensional intervals you should use $||P||^n$) as the function takes the $0$ value at every point inside $I$ except for $x$, supposing that $x\in I_{n_0}$ we have: $$ L(f,P)=\inf\{f(x):x\in I_{n_0}\}vol(I_{n_0})=0\\ U(f,P)=\sup\{f(x):x\in I_{n_0}\}vol(I_{n_0})=f(x)vol(I_{n_0}) $$ hence: $$ |L(f,P)-U(f,P)|=|f(x)vol(I_{n_0})|\\ =|f(x)||vol(I_{n_0})|\\ \le |f(x)|||P||\\ <|f(x)|\frac{\epsilon}{|f(x)|} =\epsilon $$ so $f$ is integrable. Now, to calculate its value we can take either the $\sup$ of $L(f,P)$ over all the possible partitions, or the $\inf$ of $U(f,P)$ over all the possible partitions. As we now know that $f$ is integrable this two values are equal. So, for a fixed $\epsilon>0$ we have: $$ \int_If=\sup\{L(f,P):P\mbox{ is a partition of }I\}\\= \sup\{0:P\mbox{ is a partition of }I\}\\=0 $$ so the value of the integral of $f$ over $I$ is $0$.

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