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 &gt; 0 \exists some partition P_{\epsilon} of [a,b] such that U(f,P_{\epsilon})-L(f,P_{\epsilon}) &lt; \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 &gt;0 \exists a partition P_{\epsilon} such that U(f,P_{\epsilon}) &lt; \epsilon



Choose some x_{1}such that x_{1}=\frac{1}{m} for some m\in the naturals and x_{1} &lt; \frac{\epsilon}{2}. We must produce a partition P_{\epsilon} such that U(f,P_{\epsilon}) &lt; \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}}

&lt;x_{1}&lt;\frac{\epsilon}{2}

Also, over [0,x₁], U(f,P_{\epsilon}) = x_{1} &lt; \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]

&lt;\frac{\epsilon}{2}+\frac{\epsilon}{2}
&lt;\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.