Convergence a.e. of this characteristic function let $f \geq 0$ be measurable on $E \subset \mathbb{R^n}$.
Now for any $\ n \geq 1$ there is a subset $E_n$ of $E$ with $\ m(E - E_n) < \frac{1}{n}$
I want to show that if I define: $f_n = f \chi_{_{E_n}}$
then $\lim\limits_{n \rightarrow \infty}f \chi_{_{E_n}} = f \chi_{_{E}} = f \ a.e.$
Is this fact obvious by construction and from the givens since measure of the $E_{_n}^{^c}$ gets smaller and converges to $0$

No this claim is incorrect. Let $f\equiv1$, $E=[0,1]$, and $F_n$ a sequence of intervals in $E$ with $|F_n|=1/(2n)$ whose union covers $E$ infinitely-many times. (This is possible because the harmonic series diverges.) Then let $E_n=E\setminus F_n$. Clearly $f\chi_{E_n}$ converges nowhere on $E$.
