The statement is true.

Here's why:

The given problem is in the context of measure theory and $L^1(\mu)$ spaces, which deal with integrable functions with respect to a measure $\mu$. Let’s break it down:

1. Setup: $(X, \mathcal{A}, \mu)$ is a measure space, and $\{u_n\}$ is an increasing sequence of non-negative functions in $L^1(\mu)$. Specifically, $u_1 \leq u_2 \leq u_3 \leq \cdots$, and $\sup_n u_n = 0$.

2. Condition: Since $\sup_n u_n = 0$, it implies that the functions $u_n$ approach zero pointwise almost everywhere (with respect to the measure $\mu$).

3. Conclusion: We are asked to determine if $\lim_{n \to \infty} \int u_n \, d\mu = 0.$ This is indeed true due to the monotone convergence theorem in measure theory.

   • By the assumption that $u_n$ is an increasing sequence and that $\sup_n u_n = 0$, we know that $u_n \to 0$ pointwise almost everywhere.
   • Since the functions $u_n$ are non-negative and integrable in $L^1(\mu)$, and the limit of the sequence is 0, the monotone convergence theorem guarantees that $\lim_{n \to \infty} \int u_n \, d\mu = \int \lim_{n \to \infty} u_n \, d\mu = \int 0 \, d\mu = 0.$

Thus, the limit of the integrals is indeed zero, confirming that the statement is true.

Would you like more details or have any questions?

Here are five related questions for further exploration:

1. What is the monotone convergence theorem in measure theory?
2. How does the Dominated Convergence Theorem differ from the Monotone Convergence Theorem?
3. Can a sequence of functions in $L^1(\mu)$ converge pointwise but not converge in the $L^1$-norm?
4. How do we prove that $\sup_n u_n = 0$ leads to pointwise convergence to zero?
5. Can the result still hold if the sequence $u_n$ were not increasing?

Tip: When dealing with sequences of functions, always check the conditions (like monotonicity or boundedness) to decide which convergence theorem to apply.