Convergence in measure is strictly weaker than pointwise convergence.
To see this, take $(X, \mathcal{A} , \mu)$ = $([0, 1], \mathcal{B}[0, 1], λ|[0,1])$ and set:
$u_n(x) := \textbf{1}_{[i2^{−k},(i+1)2^{−k}]}(x)$, where $n = i + 2^k, 0\leq i < 2^k$.
This is a sequence of rectangular pulses of width $2^{−k}$ moving in $2^k$ steps through $[0, 1]$, jump back to $x = 0$, halve their width and start moving again. Obviously $u_n \longmapsto 0$ in measure.
What about the pointwise convergence?
The book is reasoning by saying that the liminf = 0 and limsup = 1, so the limit doesn't converge anywhere. My doubt is on the liminf. Why should it be 0?
My reasoning was: fixing an x on the interval $[0,1]$, infinitely often there will be a rectangle that make my indicator function taking value 1.
But I don't know what to say on the liminf.
For any $x$ and any $k>1$ there is always an $i$ such that $[2^{k}x]\neq i, [2^{k}x]\neq i+1$. Let $n_k=i+2^{k}$. Then $x \notin [i2^{−k},(i+1)2^{−k}]$ so $u_{n_k}(x)=0$. Thus, infinitely many terms of the sequence $(u_n(x))$ are $0$ and this implies that $\lim \inf u_n(x)=0$.