Prove Integrability of f(x)| 0 to 1 Inequality

[FallArk]

Prove that the function
$$f(x) = 1+x, 0 \le x \le 1$$, x rational
$$f(x) = 1-x, 0 \le x \le 1$$, x irrational (they are one function, I just don't know how to use the LATEX code properly)
is not integrable on $$[0,1]$$
I don't know where to start, I tried to evalute the lower and upper Riemann sum but it does not seem to work.
I have learned some of the criteria for integrability such as Riemann's Criterion, Common Limit Criterion and Null Partitions Criterion. However, I am not familiar with them enough to use it. Any help?

 

[I like Serena]

Prove that the function
$$f(x) = 1+x, 0 \le x \le 1$$, x rational
$$f(x) = 1-x, 0 \le x \le 1$$, x irrational (they are one function, I just don't know how to use the LATEX code properly)
is not integrable on $$[0,1]$$
I don't know where to start, I tried to evalute the lower and upper Riemann sum but it does not seem to work.
I have learned some of the criteria for integrability such as Riemann's Criterion, Common Limit Criterion and Null Partitions Criterion. However, I am not familiar with them enough to use it. Any help?

Hey FallArk,

The $\LaTeX$ code would be as follows (click Reply With Quote to see what it looks like):
$$
f(x) =
\begin{cases}
1+x, & 0 \le x \le 1, & x \text{ rational} \\
1-x, & 0 \le x \le 1, & x \text{ irrational}
\end{cases}
$$
Now let's pick the standard partition.
And let's try to find an upper bound for the lower Riemann sum with points that are irrational.
Similarly let's find a lower bound for the upper Riemann sum with points that are rational.
What are their limits? (Wondering)

 

[FallArk]

What would be the standard partition? $$P= \left\{[{x}_{0},{x}_{1}],...,[{x}_{i-1},{x}_{i}],...,[{x}_{n-1},{x}_{n}]\right\}$$ ?
I think 1-x where x is irrational would be the lower sum, and it should be bounded above by 1, since 0 is not irrational. And 1+x would be the upper sum bounded by 1 as well.

 

[I like Serena]

What would be the standard partition? $$P= \left\{[{x}_{0},{x}_{1}],...,[{x}_{i-1},{x}_{i}],...,[{x}_{n-1},{x}_{n}]\right\}$$ ?
I think 1-x where x is irrational would be the lower sum, and it should be bounded above by 1, since 0 is not irrational. And 1+x would be the upper sum bounded by 1 as well.

The standard partition is where we divide the interval into equally sized intervals:
$$P_n=\left\{[0,\frac 1n], ..., [\frac{n-1}n,1]\right\}$$
In the interval $[\frac {i-1}n, \frac in]$ an upper bound of the corresponding term of the lower sum would be $1-\frac{i}n$... (Thinking)

 

[FallArk]

The standard partition is where we divide the interval into equally sized intervals:
$$P_n=\left\{[0,\frac 1n], ..., [\frac{n-1}n,1]\right\}$$
In the interval $[\frac {i-1}n, \frac in]$ an upper bound of the corresponding term of the lower sum would be $1-\frac{i}n$... (Thinking)

Then the lower bound of the upper sum would be $$1+\frac{i-1}{n}$$
After that do I simply evaluate the sums?

 

[I like Serena]

Then the lower bound of the upper sum would be $$1+\frac{i-1}{n}$$
After that do I simply evaluate the sums?

Yep. (Nod)

 

[FallArk]

Yep. (Nod)

Like this?
$$\lim_{{n}\to{\infty}}L\left(f,{P}_{n}\right) = \lim_{{n}\to{\infty}} \left(1-\frac{i}{n}\right)\cdot\frac{1}{n}$$

 

[I like Serena]

Like this?
$$\lim_{{n}\to{\infty}}L\left(f,{P}_{n}\right) = \lim_{{n}\to{\infty}} \left(1-\frac{i}{n}\right)\cdot\frac{1}{n}$$

We need to sum first.
That is only the contribution of the last interval.

 

[FallArk]

We need to sum first.
That is only the contribution of the last interval.

$$L\left(f,{P}_{n}\right) = \sum_{i=1}^{n}\left(1-\frac{i}{n}\right)\cdot\frac{i}{n}$$
then take the limit

 

[I like Serena]

$$L\left(f,{P}_{n}\right) = \sum_{i=1}^{n}\left(1-\frac{i}{n}\right)\cdot\frac{i}{n}$$
then take the limit

It should be:
$$L\left(f,{P}_{n}\right) \le \sum_{i=1}^{n}\left(1-\frac{i}{n}\right)\cdot\frac{1}{n}$$
and:
$$U\left(f,{P}_{n}\right) \ge \sum_{i=1}^{n}\left(1+\frac{i-1}{n}\right)\cdot\frac{1}{n}$$

If we take the limits, we should find that those are different.
Therefore the function is not Riemann integrable.