Answer to Question #284347 in Real Analysis for Reens

Question

Are the statement true or false?give reason for your answers: the function f defined on r by f(x) ={0 x is rational 2 is irrational} Is integrable in the interval [2,3]

Answer 1

$\text{The answer is false and we give the reason in the following proof.}\\ \textbf{Proof}\\ \text{Given the function } \displaystyle f(x) = \begin{cases} 0 & \text{if $x$ is rational} \\ 2 & \text{if $x$ is irrational} \end{cases} \\ \text{if }[a,b] \subset [2,3] \text{ with $a < b$ then } \inf_{[a,b]} f = 0 \text{ and } \sup_{[a,b]} f = 2 \text{ since $[a,b]$ contains both rational and irrational numbers.} \\ \text{Thus } L(f, P,[2,3]) = 0 \text{ and } U(f,P,[2,3]) = 2 \text{ for every partition $P$ of $[2,3]$.}\\ \text{Hence we have that } L(f, [2,3]) = 0 \text{ and } U(f,[2,3]) = 2. \\ \text{Since } L(f, [2,3]) \ne U(f,[2,3]) \text{ we conclude that $f$ is not Riemann integrable.}$