Question

Show that the function f:[0,4] by fx=0 where x is rational and fx=2 where x is irrational is not remann integrable but is lesberg integrable

Answer 1

Solution:

Given function is same as, f is 0 at every rational number and 2 at every irrational number.

If P={I1​,I2​,…,In​} is a partition of [0,4], then

$M_{k}=\sup _{I_{k}} f=2, \quad m_{k}=\inf _{I_{k}}=0$  

since every interval of non-zero length contains both rational and irrational numbers. It follows that

$U(f ; P)=2, \quad L(f ; P)=0$  

for every partition P of [0,1], so U(f)=1 and L(f)=0 are not equal.

Hence, this function is not Riemann integrable.

Further, the given function is clearly Lebesgue-integrable as being equal to 0 almost everywhere.

Let $x \in[a, b] \cap \mathbb{Q}$ . Then f(x)=2. Pick a sequence of irrational numbers $\left(x_{n}\right)_{n \in \mathbb{N}}\ with\ x_{n} \rightarrow x$ . Then $f\left(x_{n}\right)=0 \rightarrow 1=f(x)$ , so f is not continuous in x.

Let $x \in[a, b] \backslash \mathbb{Q}$ . Then f(x)=0. Pick a sequence of rational numbers $\left(x_{n}\right)_{n \in \mathbb{N}}\ with\ x_{n} \rightarrow x$ . Then $f\left(x_{n}\right)=2 \rightarrow 0=f(x)$ , so f is not continuous in x.

The above argument shows that f is discontinuous in all the points of [a, b], which is a set of Lebesgue measure $b-a \neq 0$ , therefore f is lebesgue integrable on [a, b].

Hence, proved.