TITLE: About a convergence of measurable functions
QUESTION [0 upvotes]: Let $f_{n}$ be a sequence of measurable functions in M(X,m), is that true that
{$ {x∈X∣lim f_{n}∈R}$} $ $  =  $⋃ _{M=1} ^∞⋂ _{N=1} ^∞  ⋃ _{n=N}^ ∞  ${x∈X∣ ∣f_{n} -f_{N} ∣< (1/M)}$  $   and that each member of the set {${x∈X∣ ∣f_{n} -f_{N} ∣< (1/M)}$} is an element of the sigma-algebra m? 
I´m trying to prove that if (f_{n}) is a sequence of funtions in M(X,m) then the set of ponits in X for wich limf_{n} exist in real extended is an element of the sigma-algebra m.

REPLY [1 votes]: As $\Bbb R$ is complete, then the sequence $(f_n(x))$ converge iff it is a Cauchy sequence.
and a sequence $f_n(x)$ is a Cauchy sequence iff for all $M\in \Bbb N^*$ there exists $N\in \Bbb N^*$, for all $n\geq N$, $\|f_n(x)-f_N(x)\|< \dfrac{1}{M}$ iff for all $M\in \Bbb N^*$ there exists $N\in \Bbb N^*$, for all $n\geq N$ $x\in \{\|f_n-f_N\|< \dfrac{1}{M}\}$ iff for all $M\in \Bbb N^*$ there exists $N\in \Bbb N^*$ $x\in\bigcap_{n\geq N}\{\|f_n-f_N\|< \dfrac{1}{M}\}$ iff for all $M\in \Bbb N^*$, $x\in \bigcup_{N\geq 1}\bigcap_{n\geq N}\{\|f_n-f_N\|< \dfrac{1}{M}\}$ iff 
$x\in \bigcap_{M\geq 1}\bigcup_{N\geq 1}\bigcap_{n\geq N}\{\|f_n-f_N\|< \dfrac{1}{M}\}$.
$f_N-f_n$ is a measurable function ,hence $\|f_N-f_n\|$ is measurable function, then $\{x \ ,\ \|f_N(x)-f_n(x)\|<\dfrac{1}{M}\}=\|f_N-f_n\|^{-1}(]-\infty,\frac{1}{M}[)$ is an element of the sigma-algebra (preimage of measurable subset by a measurable function).
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