Discussion
Consider the functions \(f, g : [0, 1] \to \mathbf{R}\) defined by :
\[ f(x) = \begin{cases} 1 & \text{if } x \text{ is rational} \\ 0 & \text{if } x \text{ is irrational} \end{cases} \]
and
\[ g(x) = \begin{cases} 1/x & \text{if } 0 \lt x \le 1 \\ 0 & \text{if } x = 0 \end{cases} \]
then :
Answer: D
\(f\) is Lebesgue integrable and \(g\) is not Lebesgue integrable.
\(f\) is the Dirichlet function. It equals 0 except on the rationals, a countable set of measure zero, so it is Lebesgue integrable with integral 0. (It is not Riemann integrable, since every upper sum is 1 and every lower sum 0.)
For \(g\), \(\int_{\epsilon}^{1} dx/x = -\log\epsilon \to \infty\), so \(\int_{[0,1]} g = \infty\) and \(g\) is not Lebesgue integrable. That rules out the options calling \(g\) integrable, and option A, which also wrongly denies that \(f\) is integrable.
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