Prove Integrability of a Discontinuous Function

[Lazerlike42]

Homework Statement

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.

Homework 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.

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.

Thanks in advance.

 

[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?

 

[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]

 

[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.

 

[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.

 

[Lazerlike42]

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.

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.

 

[Dick]

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