Help with proof that convergence in measure is equivalent to Cauchy convergence in measure if the space is sigma-finite. Let $(X,\mathcal{A},\mu)$ be a $\sigma-$finite measure space and $(u_n)\subset \mathcal{M}(\mathcal{A})$. Show that $u_n \to u$ (as $n\to \infty$) in measure if and only if $u_n - u_k \to 0$ (as $n,k\to \infty$) in measure.
Here, convergence in measure is defined as 
$$\forall \epsilon>0 \forall A \in \mathcal{A}, \mu(A)<\infty: \lim_{n\to \infty} \mu(\{|u_n-u|>\epsilon\} \cap A)=0.$$
I have trouble understanding the solution for the only if direction. 
In the solution below, why does the uniform convergence of $\sum_j (u_{N_{j+1}}-u_{N_j})$ imply that $\lim_j u_{N_j}$ exists uniformly?
And how can we conclude that $\lim_j u_{N_j} 1_{A_l} = u^{(l)}1_{A_l}$ exists almost everywhere for some $u^{(l)}$ since $\mu(E_i^* \cap A_l)<2 \cdot 2^{-i}$?
Finally, I can't understand how we get (*) from the fact that for each $A_m$ we have an a.e. limit $u^{(m)}$. Here, $A_l \cap A_m$ should be $A_m$ if $m\le l$, so $u^{(l)} = u^{(m)}$ a.e. on $A_l$, but how do we extend this limit to the almost whole of $X$, from having the a.e. limits on $A_m$?



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*Recall that a series $\sum_{j=1}^{+\infty}g_j$ is uniformly convergence on a set $E$ if the sequence $\left(\sum_{j=1}^{n}g_j\right)_{n\geqslant 1}$ converges uniformly on $E$. Apply this to $g_j=u_{N_{j+1}}-u_{N_j}$.

*Let $B_i$ be the set $E_i^*\cap A_l$. Since $\sum_i \mu\left(B_i\right)$ is finite, for almost all $x\in A_l$, there exists an integer $I(x)$ such that for $i\geqslant I(x)$, $x$ does not belong to $B_i$. Therefore, for such an $i$, $x\in A_l\setminus E_i$ and we can use the previous convergence.

*An other way to view the things is to define $u(x)$ by $u^{(l)}(x)$ if $x\in A_l\setminus A_{l-1}$.
