Doubt in understanding why $L^1$ is not reflexive

Question

I was reading Brezis

I donot understand

1) why author considered 2 cases highlighted

2) Also I do not understand how $\int u_n \chi_j=1$ and $\int u \chi_j=1$ also $\int u \chi_j\to 0$

Any Help will be appreciated

Answer 1

The second case is a contrary to the first case in some case, the second case considers that the measure has strictly greater than zero lower bound, where the first case considers the measure have sets that tend to zero.

$\displaystyle\int u_{n}\chi_{j}=\int\dfrac{\chi_{n}}{\|\chi_{n}\|}\chi_{j}=\int\dfrac{\chi_{n}}{\|\chi_{n}\|}=\dfrac{\|\chi_{n}\|}{\|\chi_{n}\|}=1$. Note that $(\omega_{i})$ is decreasing, then $\chi_{n}\chi_{j}=\chi_{n}$ for $n\geq j$.

The fact that $\displaystyle\int u\chi_{j}\rightarrow 0$ follows by Monotone Convergence Theorem.

While $\displaystyle\int u\chi_{j}=\lim_{n\rightarrow\infty}\int u_{n}\chi_{j}$ follows by the weak convergence in the topology $\sigma(L^{1},L^{\infty})$ because $\chi_{j}\in L^{\infty}$, and the pairing is exactly the integrals.