# Riemann integral of characteristic function.

1. Jan 22, 2013

### stripes

1. The problem statement, all variables and given/known data

The characteristic function of a set E is given by χe = 1 if x is in E, and χe = 0 if x is not in E. Let N be a natural number, and {an, bn} from n=1 to N, be any real numbers. Use the definition of the integral (Riemann) to show that $\int \sum b_{n} X_{ \left\{ a_{n} \right\} } (x) dx = 0$. The limits on the integral are from x=a to b, the limits on the summation is from n=1 to N.

2. Relevant equations

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3. The attempt at a solution

I've tried writing out the integral and simplifying it and I came to the conclusion that the integral is (b1 + b2 + ... + bn)x whenever x is equal to an and is zero otherwise. It's probably wrong but that's what I got.

Then I started by using the definition. I have a hard enough time proving the Riemann integral of regular functions using the definition. This seems impossible. Using the definition, I wouldn't even know where to start. Any help to get me started would be appreciated. Thanks in advance.
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2. Relevant equations

3. The attempt at a solution

2. Jan 22, 2013

### Dick

It looks to me like you are summing n functions each of which are nonzero at a single point. Can't you think of a way of defining a series of partitions so both the upper sums and lower sums converge to zero?

3. Jan 23, 2013

### stripes

Okay so I see that we're just summing a bunch of points from 1 to N. But I don't understand how the Riemann integral could be zero of the sum of a set of points that are nonzero, and I also don't quite understand the limits of integration. a and b are in no way related to the sequence of numbers, are they? That being said, if a and b can be any real numbers, does this mean we need to show the area under the curve is always zero no matter which interval I choose? The only function I can think of is the constant zero. But the sequence {an, bn} can be any real numbers, which means that the sum will be nonzero. This means that the integral will be nonzero, doesn't it? Unless we're talking about the Cauchy principle value of a sinusoidal function...but that's not what we're doing here.

Basically I don't see how the sum of these functions that are nonzero at only one point can end up having an area underneath it of zero. The sum will just be a constant for any natural number N, and the integral of a nonzero function will be nonzero.

Define a series of partitions whose upper and lower sums converge to zero. Since the series is finite, we could choose an alternating series like (-1)^n if n = 2k for some natural number k.

4. Jan 23, 2013

### jbunniii

You're not just summing points. You are summing characteristic functions, each of which is zero everywhere except at one point. So the sum of these characteristic functions will be a new function which is zero everywhere except at (up to) N points.
Probably the values $a_n$ are assumed to satisfy $a \leq a_n \leq b$, so that the function's discontinuities all lie within the interval of integration. But even if this is not the case, the integral will still be zero.
Yes, unless the problem statement stipulates that $a \leq a_n \leq b$ for each $n$.
Surely that is not the only possibility. What if you have
$$f(x) = \begin{cases} 1 & \textrm{if }x = 0 \\ 0 & \textrm{if }x \neq 0\end{cases}$$
What are the following integrals in this case?
$$\int_{-1}^{1}f(x) dx$$
$$\int_{0}^{1}f(x) dx$$
$$\int_{1}^{2}f(x)dx$$
$$\int_{a}^{b}f(x) dx$$
I claim that they will all be zero. Your function is similar to this one, except it has up to N "spikes" instead of just one.

5. Jan 23, 2013

### stripes

I wasn't thinking. I kept thinking of intervals centered at these spikes, but I didn't think that, like the Dirac delta function, the integral is just zero at a spike. Anyways, am I right when I say

$f(x) = \sum ^{N}_{n=1} b_{n} X_{ \left \{a_{n} \right\}} (x) = \begin{cases} \sum ^{N}_{n=1} b_{n} & \textrm{if }x = a_{n} \\ 0 & \textrm{if } x \neq a_{n} \end{cases}$?

I don't think this is right, but it looks like I'm going in the right direction? Or maybe not.

Edit: nope this is definitely wrong.

Last edited: Jan 23, 2013
6. Jan 23, 2013

### stripes

Wait, my function would be:

$f(x) = \sum ^{N}_{n=1} b_{n} X_{ \left \{a_{n} \right\}} (x) = \begin{cases} b_{1} & \textrm{if }x = a_{1} \\ b_{2} & \textrm{if }x = a_{2} \\ ... \\ b_{n} & \textrm{if }x = a_{n} \\ 0 & \textrm{if } x \neq a_{n} \end{cases}$?

Or no...

7. Jan 23, 2013

### Dick

Yes. That's it. Now think of finite width boxes of height bi around each ai. Can you integrate that? Then let the boxes get narrower and narrower.

8. Jan 23, 2013

### jbunniii

Yes, that's right, except the last line should read "for all other $x$" instead of "if $x \neq a_n$".

9. Jan 23, 2013

### stripes

$\int ^{b}_{a} f(x) dx = \int^{b}_{a} \begin{cases} b_{1} & \textrm{if }x = a_{1} \\ b_{2} & \textrm{if }x = a_{2} \\ ... \\ b_{n} & \textrm{if }x = a_{n} \\ 0 & \textrm{if } x \neq a_{n} \end{cases} dx = lim_{n \rightarrow \infty} \sum^{N}_{i=1} f(a_{i}) \frac{ a_{i} - a_{i-1} }{n} = lim_{n \rightarrow \infty} \sum^{N}_{i=1} b_{i} \frac{ a_{i} - a_{i-1} }{n}$

Well it's not correct because I made the assumption that the integral is zero, and then found a series that converges to zero to "make it work". If I go by the definition of the Riemann integral, how do I introduce that n in the denominator, if that's even right at all?

10. Jan 23, 2013

### jbunniii

It might be helpful to write the definition you are using for the Riemann integral. Another suggestion: try proving that the integral is zero for the simpler case I mentioned earlier:
$$f(x) = \begin{cases} 1 & \textrm{if }x = 0 \\ 0 & \textrm{if }x \neq 0\end{cases}$$
Then you can use linearity to extend this result to the case where there are N spikes instead of 1.

11. Jan 24, 2013

### stripes

Here is the definition we are using. I will consider your other suggestion but I should be able to prove the integral without using linearity, so if you could possibly help me with that, I would really appreciate it.

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12. Jan 24, 2013

### Dick

Try proving it for the simple f(x) that jbunniii gave you first. Hint, pick A=0. And you really only have to worry about the segment of the partition that contains x=0.