# Prove Integrability of a Discontinuous Function

1. Dec 11, 2008

### Lazerlike42

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

Let f(x)= { 1 if x=$$\frac{1}{n}$$ for some n$$\in$$ the natural numbers,
or 0 otherwise}

Prove f is integrable on [0,1], and evaluate the integral.

2. Relevant equations

This is using Riemann Integrability. I know that the method of providing the solution is supposed to be by application of the following theorem:

"A bounded function f is integrable on [a,b] if and only if $$\forall \epsilon > 0$$, there is a partition $$P_{\epsilon}$$ of [a,b] where U(f, $$P_{\epsilon}$$) - L(f,$$P_{\epsilon} < \epsilon$$"

U($$P_{\epsilon}$$) and L($$P_{\epsilon}$$) are of course the upper and lower sums.

3. The attempt at a solution

I know that L($$P_{\epsilon}$$) is 0 everywhere, because for any sub-interval no matter how small there is some value of x where x /= $$\frac{1}{n}$$. I also know that the integral must equal 0, as the lower and upper sums must be equal in an integrable function.

What I am supposed to do is to find some partition of [a,b] where the upper sum is less than epsilon, but I can't figure out how to do that.

Given an epsilon, if I select some x0 where x0=$$\frac{1}{m}$$ for some m and x0 < $$\epsilon$$, then it follows that the upper sum over [0,x0] < $$\epsilon$$. The issue is then the upper sum on [x0, 1].

I have also noticed this fact: Suppose my x0=$$\frac{1}{500}$$. Then if I move up to $$\frac{2}{500}$$, that's actually $$\frac{1}{250}$$. Similarly, $$\frac{4}{500}=\frac{1}{125}$$, $$\frac{5}{500}=\frac{1}{100}$$, and $$\frac{10}{500}={\frac{1}{50}$$. That means that between $$\frac{2}{500}$$ and $$\frac{1}{50}$$, there are only 4 values equal to some $$\frac{1}{n}$$. The upper sum over that would then be simply 4 * $$\frac{1}{500}$$. Somewhere in all of that, I feel like there is a way to partition [x0, 1] so that it's less than a given epsilon, but I can't figure it out.

2. Dec 11, 2008

### Dick

For each integer M>0, pick an interval around each x=1/n of size 1/2^(n+M). You can do that, right? What happens as M->infinity?

3. Dec 11, 2008

### Lazerlike42

Ok, here is the solution I came up with. Any comments are appreciated.

By the given theorem, a bounded function f is integrable on [a,b] if and only if $$\forall \epsilon > 0 \exists$$ some partition $$P_{\epsilon}$$ of [a,b] such that $$U(f,P_{\epsilon})-L(f,P_{\epsilon}) < \epsilon$$.

Clearly the function f(x) is bounded, having only two possible values: 0 and 1.

So Consider $$L(f,P_{\epsilon})$$ for an arbitrary partition. Because the irrationals are dense on $$\Re$$, for any sub-interval $$[x_{k},x_{k-1}] \exists x_{0}$$ such that $$x_{0}\neq\frac{1}{n}$$ for any n$$\in$$ the naturals. Therefore, the $$L(f,P_{\epsilon})=0 \forall\epsilon$$

Then by the theorem, f is integrable if $$\forall \epsilon >0 \exists$$ a partition $$P_{\epsilon}$$ such that $$U(f,P_{\epsilon}) < \epsilon$$

Choose some $$x_{1}$$such that $$x_{1}=\frac{1}{m}$$ for some $$m\in$$ the naturals and $$x_{1} < \frac{\epsilon}{2}$$. We must produce a partition $$P_{\epsilon}$$ such that $$U(f,P_{\epsilon}) < \epsilon$$. For now, consider $$P_{\epsilon}$$ only over the interval [0,x1], and define $$P_{\epsilon}=$${$${0,x_{1}$$}. On [0,x1], $$U(f,P_{\epsilon})=1(x_{1}-0)=x_{1}$$

Now consider [x1,1]. In [x1,1] there are only m numbers of the form $$\frac{1}{n}$$ for some n: x1 (which is, as we chose it, $$\frac{1}{m}$$), $$\frac{1}{m-1}$$, $$\frac{1}{m-2}$$...1. Define $$P_{\epsilon}$$ over [x1,1] so as to satisfy the property that each number of the form $$\frac{1}{n}$$ (call them $$\frac{1}{n_{k}}$$) is in a subinterval as follows:

[$$\frac{1}{n_{k}}-\frac{1}{2m^{2}}, \frac{1}{n_{k}}+\frac{1}{2m^{2}}$$] (except for x1 and 1, the intervals for which need be defined only on the right and left sides, respectively). So for example, if x1=$$\frac{1}{5}$$, ensure that $$\frac{1}{4}$$ is in the sub-interval [0.23,0.27].

More specifically, $$P_{\epsilon}=$${$$0, x_{1}, \frac{1+m}{m^{2}}, \frac{1}{m-1}-\frac{1}{2m^{2}},\frac{1}{m-1}+\frac{1}{2m^{2}},\frac{1}{m-2}-\frac{1}{2m^{2}}...1-\frac{1}{2m^{2}},1$$}

Then over [x1,1], $$U(f,P_{\epsilon})$$=

$$1( \frac{1+m}{m^{2}}-\frac{1}{m})+0(\frac{1}{m-1}-\frac{1}{2m^{2}}-[ \frac{1+m}{m^{2}}])+1(\frac{1}{m-1}+\frac{1}{2m^{2}}-[\frac{1}{m-1}-\frac{1}{2m^{2}}])+0(\frac{1}{m-2}-\frac{1}{2m^{2}}-[\frac{1}{m-1}+\frac{1}{2m^{2}}]...+1(1-[1-\frac{1}{2,^{2}}])$$

=$$\frac{1}{2m^{2}}+\frac{1}{m^{2}}+\frac{1}{m^{2}}+...+\frac{1}{2m^{2}}$$

$$=(m-1)(\frac{1}{m^{2}})=\frac{1}{m}-\frac{1}{m^{2}} = x_{1}-\frac{1}{m^{2}}$$

$$<x_{1}<\frac{\epsilon}{2}$$

Also, over [0,x1], $$U(f,P_{\epsilon}) = x_{1} < \frac{\epsilon}{2}$$, as we saw above.

Then $$U(f,P_{\epsilon})$$ over [0,1]=$$U(f,P_{\epsilon})$$ over [0,x1]+$$U(f,P_{\epsilon})$$ over [x1,1]

$$<\frac{\epsilon}{2}+\frac{\epsilon}{2}$$
$$<\epsilon$$

Therefore, f is integrable over [0,1]

4. Dec 11, 2008

### Lazerlike42

Dick, I saw your hint when I had just finished posting my solution. Thanks! It looks like I had the same general idea as you hinted at, but I'm not sure if what I did is precisely the same.

5. Dec 11, 2008

### Dick

It's way simpler than you think, I think. Just pick a value of M based on epsilon. You don't need stuff like the density of irrationals. Sorry, but I can't read your post. Too tired.

6. Dec 11, 2008

### Lazerlike42

That's ok. I'm satisfied with what I've got so far... I see what you're saying. There are obviously a few ways to do it (I can think of a couple of others right now which involve approaching it from other angles, like countability), but the professor gave us a hint and I based my work off of that... I'm also too tired to bother thinking about whether your hint fits in with hers, but I'll take another look in the morning.

7. Dec 11, 2008

### Dick

Fair enough. Like you, I can only read about 6 lines before I doze off.