Question

Consider the sequence of Lebesgue measurable functions \(f_n: \mathbb{R} \to \mathbb{R}\) given by
\[ f_n(x) = \begin{cases} n^2(x-n), & \text{if } x \in [n, n + \frac{1}{n^2}] \\ 0, & \text{otherwise} \end{cases} \]
For a measurable subset E of \(\mathbb{R}\), denote m(E) to be the Lebesgue measure of E.
Which of the following statements is/are correct?

• A. \(\sup_{x \in \mathbb{R}}|f_n(x)| \to 0\) as \(n \to \infty\)
• B. \(\int_{\mathbb{R}} |f_n(x)|dx \to 0\) as \(n \to \infty\)
• C. \(m(\{x \in \mathbb{R} : |f_n(x)|>\frac{1}{2}\}) \to 0\) as \(n \to \infty\)
• D. \(m(\{x \in \mathbb{R} : |f_n(x)|>0\}) \to 0\) as \(n \to \infty\)

Hint:

This is a classic example of a sequence of functions that converges to 0 in \(L^1\) and in measure, but does not converge uniformly. The "bump" keeps the same height (preventing uniform convergence) but gets narrower and moves away (allowing integral and measure to go to zero).

Answer 1

The Correct Option is A, B, C

Step 1: Understanding the Concept:
We are given a sequence of functions and asked to analyze their convergence properties in different senses: uniform convergence (related to sup norm), convergence in \(L^1\) norm, and convergence in measure. Each function is a linear "ramp" on a shrinking interval that moves to infinity.
Step 3: Detailed Explanation:
The function \(f_n(x)\) is non-zero only on the interval \([n, n + 1/n^2]\). On this interval, \(x-n \ge 0\), so \(f_n(x) \ge 0\). The function increases linearly from \(f_n(n)=0\) to \(f_n(n+1/n^2) = n^2((n+1/n^2)-n) = 1\).
(A) \(\sup_{x \in \mathbb{R}}|f_n(x)| \to 0\) as \(n \to \infty\): The supremum of \(f_n(x)\) is its maximum value, which is attained at the right endpoint of the interval.
\[ \sup_{x \in \mathbb{R}}|f_n(x)| = f_n(n + 1/n^2) = 1 \] This supremum is 1 for all \(n \ge 1\). The limit as \(n \to \infty\) is 1, not 0. Thus, the sequence does not converge uniformly to 0. (A) is FALSE.
(B) \(\int_{\mathbb{R}} |f_n(x)|dx \to 0\) as \(n \to \infty\):
The integral is the area under the graph of \(f_n(x)\). The graph is a right-angled triangle with base \(1/n^2\) and height 1.
\[ \int_{\mathbb{R}} |f_n(x)|dx = \int_n^{n+1/n^2} n^2(x-n) dx = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \cdot \frac{1}{n^2} \cdot 1 = \frac{1}{2n^2} \] As \(n \to \infty\), \(\frac{1}{2n^2} \to 0\). Thus, the sequence converges to 0 in \(L^1\). (B) is TRUE.
(C) \(m(\{x \in \mathbb{R} : |f_n(x)|>\tfrac{1}{2}\}) \to 0\) as \(n \to \infty\):
We need to find the measure of the set where \(n^2(x-n)>1/2\).
\[ x-n > \tfrac{1}{2n^2} \implies x > n + \tfrac{1}{2n^2} \] The set is the interval \((n + \tfrac{1}{2n^2}, n + \tfrac{1}{n^2}]\).
The measure of this set is its length: \((n + \tfrac{1}{n^2}) - (n + \tfrac{1}{2n^2}) = \tfrac{1}{2n^2}\).
As \(n \to \infty\), this measure \(\tfrac{1}{2n^2} \to 0\). (C) is TRUE.
(D) \(m(\{x \in \mathbb{R} : |f_n(x)|>0\}) \to 0\) as \(n \to \infty\):
The set where \(|f_n(x)|>0\) is the support of the function, excluding the left endpoint. This is the interval \((n, n+1/n^2]\).
The measure of this set is its length: \((n+1/n^2) - n = 1/n^2\).
As \(n \to \infty\), this measure \(\tfrac{1}{n^2} \to 0\). (D) is TRUE.
Step 4: Final Answer:
The correct statements are (B), (C), and (D).