Proving Non-Integrable

You use the Definite Integral Darboux Sum Definition or Definite Integral Darboux Sum Epsilon Reformed Definition to prove non-integrability.

Example

Consider the Dirichlet Function: $f=\{0\text{ if }x\in Q,1\text{ if }x\notin Q\}$ Then, to show non-integrability, we want to show that the upper darboux sum is not equa lower darboux sum.

1. Let $a,b\in \mathbb{R}$ be arbitrary and suppose $a<b$
2. Let $P=\{x_i{\}}_{i=0}^n$ be an arbitrary partition
3. Then, the lower Darboux sum = $L(f,p)={\sum }_{k=1}^n{m}_i(x_i-x_{i-1})$
   1. $L(f,p)={\sum }_{k=1}^n(0)(x_i-x_{i-1})$ as the minimum is always 0, as there is always an irrational in a set by Density
   2. $L(f,p)=(0)(-x_0+x_n)$ by Linearity dist prop and since ${\sum }_{k=1}^n(x_i-x_{i-1})$ is a Telescoping Series
   3. $L(f,p)=0$
4. Then, the upper Darboux sum = $U(f,p)={\sum }_{k=1}^n{M}_i(x_i-x_{i+1})$
   1. $U(f,p)={\sum }_{k=1}^n(1)(x_i-x_{i-1})$ as the maximum is always 1, as there is always a rational in a real set by Density
   2. $U(f,p)=(1)(-x_0+x_n)$ by Linearity dist prop and since ${\sum }_{k=1}^n(x_i-x_{i-1})$ is a Telescoping Series
   3. $U(f,p)=(-x_0+x_n)$
   4. $U(f,p)=(-a+b)$ by defn of partiton
5. Note that $0\ne (-a+b)$, as $a<b⟹-a+b\ne 0$, thus $U(f,p)\ne L(f,p)$, thus the f(x) is not Integrable

Example with Integral Reformulation

1. Let $f(x)=\{\text{2 if x∈Q},\text{0 if x∈Q}\}$
2. Proving $f$ is not integrable on $[0,1]$
3. We can show this by proving the negation of the Integral Reformulation definition. In other words, lets prove that $\exists ϵ>0,\forall$ partitions $P$ on $[0,1]$ s.t $U(f,p)-L(f,p)\ge ϵ$
4. Choose $ϵ=2$
5. Let $P$ be an arbitrary partition of $[0,1]$
6. Note that $M_1=sup\{f(x)|x\in [x_{i-1},x_i]\}$
   1. $M_1=1$
7. Note that $m_1=inf\{f(x)|x\in [x_{i-1},x_i]\}$
   1. $m_1=0$
8. $U(f,p)=2$
9. $L(f,p)=0$
10. $U(f,p)-L(f,p)=2-0\le 2=ϵ$ $\square$