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...

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

1 Answer 1


$f$ 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.


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