Question

Let𝛼(𝑥)=𝑥 and define𝑓 as𝑓(𝑥)=1 if𝑥 is rational and𝑓(𝑥)=0 if𝑥 is irrational. Find 𝐼(𝑓, 𝛼) and 𝐼(𝑓, 𝛼). 

Answer 1

By definition;

$\displaystyle \alpha(x)=x \text{ and }\\ f:[a,b]\rightarrow\R \text{ is defined as } f(x)=\begin{cases}1&\text{if }x\text{ is rational}\\0&\text{if }x\text{ is irrational}\end{cases}$

Now, for every partition P of $\displaystyle [a,b],$

$\displaystyle M_k(f) = \sup \{f(x) : x \in [x_{k-1}, x_k]\}=1\\ m_k(f) = \inf \{f(x) : x\in[x_{k-1}, x_k]\}=0$, since every subinterval contains both rational and irrational numbers.

Thus,

$\displaystyle U(P, f, a) =\sum^n_{k=1} M_k(f) \Delta \alpha_k=1\\\text{and}$

$\displaystyle L(P, f, \alpha) =\sum^n_{k=1} m_k(f) \Delta \alpha_k=0\ \ \forall\ \text{partition }P.\\ \text{Where }\Delta\alpha_k=\Delta\alpha(x_{k})=\alpha_k-\alpha_{k-1}.$

Hence, it follows that we have;

$\displaystyle \underline{I}(f,\alpha)=\underline{\int}^b_af\ dx=0\\ \quad\\ \overline{I}(f,\alpha)=\overline{\int}^b_af\ dx=1$