Discussion
Consider the signed measure \(\nu\) on [0, 2] defined by \(\nu(A) = \int_{A} f(x)\, dx\)
where
\[ f(x) = \begin{cases} 1 & \text{if } x \in [0, 1) \\ -1 & \text{if } x \in [1, 2]. \end{cases} \]
Then \(\nu\left(\frac{1}{2}, \frac{3}{2}\right)\) is :
Answer: C
\(\nu(\frac{1}{2}, \frac{3}{2}) = \) 0.
Split the interval at 1, where \(f\) changes sign:
\[ \begin{aligned} \nu\left(\tfrac{1}{2}, \tfrac{3}{2}\right) &= \int_{1/2}^{1} 1\, dx + \int_{1}^{3/2} (-1)\, dx \\ &= \tfrac{1}{2} - \tfrac{1}{2} = 0 \end{aligned} \]
The positive and negative parts cancel. The Lebesgue measure of \((0, 1)\) is 1, so A and B are the same wrong value, which would ignore the negative part, and \(-1/2\) counts only the negative part.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development