Help in understanding this, convergence in measure. Let $(X,A,\mu)$ be a measurable space.
$g,f_1,f_2\ldots \in L^1(\mu)$ , $g\geq 0$.
$f:X\to C$ measurable.
Suppose $f_n$ converges to $f$ in measure ($\mu$) and $\forall n\in N |f_n|\leq g$.
Prove that :
$lim_{n\to \infty} \int_{X} f_n d\mu = \int_X f d\mu$.
I'm trying to show this with using the dominated convergence theorem of
Lebesgue since we have almost all the conditions .
But we have to show that $f_n$ converges to $f$ almost everywhere.
From the given information we can conclude that $f_n$ is a couchy sequence So by another theorem there is a sub sequence $f_{n_j}$ such that $f_{n_j}$ converges to $f$ almost everywhere.
I don't know how to finish it.Though I know that there is a theory in calculus which states:
$a_n$ converges iff  every subsequent of $a_n$ has a subsequent that converges.
But I don't know how to use it.
Can you help in this, and explain how it can be showed.

Prove it by contradiction. Suppose it is not true that $\int f_n d\mu \to \int f d\mu$. Then there exists $\epsilon >0$ and integers $1 \leq n_1<n_2<....$ such that $|\int f_{n_k} d\mu- \int f d\mu| \geq \epsilon$ for all $k$ $\,\,\, \cdots(1)$.
Now $(f_{n_k})$ converges to $f$ is measure because $f_n \to f$ in measure. Hence $(f_{n_k})$ has  a subsequence $(f_{n_{k_i}})$ converging almost everywhere to $f$. By DCT we get $\int f_{n_{k_i}} d\mu \to \int fd\mu$ But this contradicts (1).
