I have the following question about measure theory:
Let $(X,\mathcal A,\mu)$ be a measure space, and let $A$ be a measurable set. We say that $(f_n(x))_n$ convergences in measure $\mu$ to the function $f(x)$ on the set $A$, if the functions $f,f_n$ are measurable, finite a.e. in $A$, and for all $\delta>0$:
$ \lim_{k \to \infty} = \mu(\{x\in A : |f_k(x)-f(x)| \ge \delta \})= 0$
Show that if $f_n \to f$ pointwise in $A$, and $(f_n)_n$ converges in measure to some function, than $f_n \to f$ in measure on $A$.
The definition for convergence in measure here is a bit like the local convergence in measure, only here we don't assume that $A$ is of finite measure.
At first, I tried to use the claim that if $f_n \to g$ in measure then there exists a subsequence that convergences pointwise a.e. to $g$. However, for this proof, I need to use the Borel-Cantelli Lemma, which we haven't learned yet. Also, I don't think it's applicable to the definition of converges to measure that we use.
Perhaps there's a trick here or something obvious that I'm missing, but I could use a hint on this.