# Check lebesgue integrable or not...

Consider the given question:

Let \begin{align} f(x) = \begin{cases} x, &\text{if} \ x\in [0,1]\cap\mathbb{Q}, \\ x^2, &\text{if} \ x\in[0,1]\setminus\mathbb{Q} \end{cases} \end{align} Then

(a) $$f$$ is Riemann integrable and Lebesgue integrable on $$[0,1]$$

(b) $$f$$ is Riemann integrable but not Lebesgue integrable on $$[0,1]$$

(c) $$f$$ is not Riemann integrable but Lebesgue integrable on $$[0,1]$$

(d) $$f$$ is neither Riemann integrable nor Lebesgue integrable on $$[0,1]$$

Clearly the function is not Riemann integrable, being continuous at only two points i.e. $$0$$ and $$1$$.. But is there a way to check whether it is Lebesgue integrable or not...

• what have you tried? What are basic facts about arithmetic of measurable functions? Nov 20, 2023 at 15:27
• Hint $f(x)=x^2\boldsymbol 1_{[0,1]}(x)$ a.e.
– Surb
Nov 20, 2023 at 15:43

$$f$$ is Lebesgue integrable. Since $$f$$ is bounded, we only need to show $$f$$ is measurable. For any $$a\in[0,1]$$, $$f^{-1}[-\infty,a)=\big(\mathbb{Q}\cap[0,a)\big)\cup\big((\mathbb{R}-\mathbb{Q})\cap[0,\sqrt a)\big).$$ The RHS is clearly Lebesgue measurable.