1. ## function integrable?

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

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

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

Any help? Thanks

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

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

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

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

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

One thing that should be obvious is that $U(P,f)=1$, since for all intervals in any partition you choose, $M_i$ will always equal $1$. Proving that $L(P,f)=1$ 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 $n-1$ intervals containing the $\frac{k}{n}$ 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 $\left( {\forall n} \right)\left[ {f_n } \right]$ has only $n-1$ 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 $\frac{1}{2}$ as the limit for both the upper and lower sum. This seems to violate my intuitive answer of $1.$

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

Any faulty logic going on here?

Thanks