Showing a certain set is measurable Let $(f_n: n>0)$ be a sequence of measurable real-valued functions on some measurable space $X.$ I want to show the set $A=\{x : \displaystyle{\sum_{n > 0}} |f_n(x)| \hspace{0.2cm}\textrm{is convergent} \}$ is measurable. I understand that I need to write $A$ as something like countable unions of intersections of unions of unions (or whatever number of them needed in whatever order). But I get really confused when there is union and intersection operations done after each other 4+ times in a row. Rather then the answer, if you could shed light on the thinking process that will be helpful.
 A: Denote $S_n= \sum_{m=1}^n |f_m(x)|$. We know that $S_n$ converges if and only if it is Cauchy sequence i.e. 
when $$\forall \epsilon ~ \exists N~ \forall m,k \geq N, ~|S_k-S_m|<\epsilon.$$
So, $$A= \bigcap_{n=1}^\infty \bigcup_{N=1}^{\infty}\bigcap_{k,m\geq N}\left \{ |S_k-S_m   |<\frac{1}{n}\right\}$$ 
A: The series $\sum_j|f_j(x)|$ is convergent when, for every $n\in\mathbb N $, there exists $K(n)\in\mathbb N$ such that $g_K(x)=\sum_{j\geq K}|f_j(x)|<\tfrac1n$ (this works because $\mathbb R$ is a metric space). I'm fairly sure that at this stage you have proven then limits of measurable functions are measurable; so each $g_K$ is measurable. Because $g_K$ is measurable, the set $\{x:\ g_K(x)<\tfrac1n\}$ (usually denoted $\{g_K<\tfrac1n\}$) is measurable. 
Now we are looking for those $x$ such that $x\in \{g_{K(n)}<\tfrac1n\}$ for all $n$. That is, the set 
$$
\bigcap_{n\in\mathbb N}\{g_{K(n)}<\tfrac1n\}.
$$
To make the dependence on this function $K(n)$, we may take those $x$ in $$\bigcup_{k\in\mathbb N}\{g_k<\tfrac1n\}.$$
Doing this for every $n$ means taking
$$\tag1\bigcap_{n\in\mathbb N}\bigcup_{k\in\mathbb N}\{g_k<\tfrac1n\}.$$
Let us confirm that this works. Suppose that $\sum_n|f_n(x)|$ converges. Fix $n$: then there exists $k$ with $|g_k(x)|<\tfrac1n$, so $x\in \bigcup_{k\in\mathbb N}\{g_k<\tfrac1n\}$; as we can do this for all $n$, we get $x\in\bigcap_{n\in\mathbb N}\bigcup_{k\in\mathbb N}\{g_k<\tfrac1n\}$.  Conversely, if $x\in \bigcap_{n\in\mathbb N}\bigcup_{k\in\mathbb N}\{g_k<\tfrac1n\}$, then for every $n$ we have that there exists $k$ such that $g_k(x)<\tfrac1n$, which is precisely that the series converges. 
A: Let $A_{m,n} = \{ x | \sum_{k=1}^n |f_k(x)| \le m \}$. Since the $f_k$ are measurable it
follows that $A_{m,n}$ is measurable.
Hence the $\cap_n A_{m,n} = \{ x | \sum_{k=1}^\infty |f_k(x)| \le m  \}$ are mesaurable
and so $\cup_m \cap_n A_{m,n} = \{ x | \sum_{k=1}^\infty |f_k(x)| \text{ is finite}  \}$
is measurable.
