Question

If \(f(x)=1\) when \(x\in Q\) and \(f(x)=-1\) when \(x\in R-Q\), then

• A. \(f\) is not bounded on \([a,b]\)
• B. Lower and upper Riemann integrals do not exist
• C. Lower and upper Riemann integrals exist but are not equal
• D. \(f\) is Riemann integrable on \([a,b]\)

Hint:

For functions defined differently on rationals and irrationals, use density of rationals and irrationals in every interval.

Answer 1

The Correct Option is C

Concept:
The function is defined as \[ f(x)= \begin{cases} 1, & x\in Q,\\ -1, & x\in R-Q. \end{cases} \] This is a modified Dirichlet-type function. Both rational and irrational numbers are dense in every interval.

Step 1: Check boundedness. The function takes only two values: \[ 1 \] and \[ -1 \] So it is bounded on every interval.

Step 2: Find supremum and infimum on any subinterval. In every subinterval of \([a,b]\), there are rational numbers and irrational numbers. Therefore, on every subinterval, \[ \sup f=1 \] and \[ \inf f=-1 \]

Step 3: Lower and upper sums. The lower sums are based on infimum, so they involve \(-1\). The upper sums are based on supremum, so they involve \(1\). Thus the lower and upper Riemann integrals exist, but they are not equal.

Step 4: Final conclusion. Since lower and upper integrals are not equal, the function is not Riemann integrable. \[ \boxed{\text{Lower and upper Riemann integrals exist but are not equal}} \]