1. ## function integrable?

Let $\displaystyle n$ be a fixed integer $\displaystyle >0;$ consider the function

$\displaystyle f_{n}(x) = \{0, if (x = \frac{1}{n}, \frac{2}{n},...,\frac{n-1}{n}), 1, otherwise.\}$
Prove directly from the definition of integrability that $\displaystyle f_{n}(x)$ is integrable on $\displaystyle [0,1].$

All I've got so far is that: $\displaystyle M_{i} = (\frac{i}{n}); m_{i} = (\frac{i-1}{n}); \Delta x_{i} = \frac{1}{n}.$

Any help? Thanks

2. Originally Posted by Anonymous1
Let $\displaystyle n$ be a fixed integer $\displaystyle >0;$ consider the function

$\displaystyle f_{n}(x) = \{0, if (x = \frac{1}{n}, \frac{2}{n},...,\frac{n-1}{n}), 1, otherwise.\}$

Prove directly from the definition of integrability that $\displaystyle f_{n}(x)$ is integrable on $\displaystyle [0,1].$

All I've got so far is that: $\displaystyle M_{i} = (\frac{i}{n}); m_{i} = (\frac{i-1}{n}); \Delta x_{i} = \frac{1}{n}.$

Any help? Thanks
$\displaystyle f_n(x)=\left\{\begin{array}{lr}0:&x=\frac{k}{n},~k \in\mathbb{N},~1\leq k<n\\1:&otherwise\end{array}\right\}$

One thing that should be obvious is that $\displaystyle U(P,f)=1$, since for all intervals in any partition you choose, $\displaystyle M_i$ will always equal $\displaystyle 1$. Proving that $\displaystyle L(P,f)=1$ is a matter of choosing a clever partition. I don't think the "standard" partition (i.e. the one you're using) will work. You want a partition such that the total length of the $\displaystyle n-1$ intervals containing the $\displaystyle \frac{k}{n}$ can be made as small as desired.

3. I don't think this function is Riemann integrable. Anyone disagree?

 Wait yes it is, since it only has a finite number of discontinuities.

4. Originally Posted by Anonymous1
I don't think this function is Riemann integrable. Anyone disagree?
I definitely disagree!
Because I know that $\displaystyle \left( {\forall n} \right)\left[ {f_n } \right]$ has only $\displaystyle n-1$ discontinuities.
Therefore, it is Riemann intergrable.

5. Hey just came back to look at this problem.

So, when I use the above partition I get $\displaystyle \frac{1}{2}$ as the limit for both the upper and lower sum. This seems to violate my intuitive answer of $\displaystyle 1.$

So my question is: can I just say, for some $\displaystyle N > \frac{n-1}{n}$ the sequence will get and stay at $\displaystyle 1,$ hence the limit of the upper sum is $\displaystyle 1$ which implies the value of the integral is $\displaystyle 1.$

Any faulty logic going on here?

Thanks